Contoh Soal 1 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 1.& \text{Hasil dari}{\displaystyle \frac{3^4{.5}^3{.7}^2}{27.125.49}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 9\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{3^4{.5}^3{.7}^2}{27.125.49}}& ={\displaystyle \frac{3^4{.5}^3{.7}^2}{3^3{.5}^3{.7}^2}}\\ & =3^{4-3}{.5}^{3-3}{.7}^{2-2}\\ & =3^1{.5}^0{.7}^0\\ & =3.1.1\\ & =3\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Bentuk sederhana dari}{\displaystyle \frac{5a^4{b}^2}{a^2{b}^{-3}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle a{b}^2}\\ \\ \text{b}.& {\displaystyle a^2{b}^2}\\ \\ \text{c}.& {\displaystyle 5a^2{b}^5}\\ \\ \text{d}.& 5a^4{b}^5\\ \\ \text{e}.& 5a^5{a}^2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{5a^4{b}^2}{a^2{b}^{-3}}}& =5a^{4-2}{b}^{2-(-3)}\\ & =5a^2{b}^5\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Bentuk sederhana dari}{\left({\displaystyle \frac{a{b}^2}{a^2{b}^3}}\right)}^4\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{b}}\\ \\ \text{b}.& {\displaystyle \frac{1}{ab}}\\ \\ \text{c}.& {\displaystyle \frac{1}{a^2{b}^2}}\\ \\ \text{d}.& {\displaystyle \frac{1}{a{b}^2}}\\ \\ \text{e}.& {\displaystyle \frac{1}{a^4{b}^4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{a{b}^2}{a^2{b}^3}}\right)}^4& ={\left(a^{1-2}{b}^{2-3}\right)}^4\\ & ={\left(a^{-1}{b}^{-1}\right)}^4\\ & ={\left({\displaystyle \frac{1}{ab}}\right)}^4\\ & ={\displaystyle \frac{1}{a^4{b}^4}}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Bentuk sederhana dari}{\displaystyle \frac{3^{\frac{5}{6}}\times {12}^{\frac{7}{12}}}{2^{-\frac{1}{4}}\times 6^{\frac{2}{3}}}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.6^{\frac{1}{4}}& & \text{d}.{\left({\displaystyle \frac{2}{3}}\right)}^{\frac{3}{4}}\\ \text{b}.6^{\frac{3}{4}}& \text{c}.6^{\frac{2}{3}}& \text{e}.{\left({\displaystyle \frac{3}{2}}\right)}^{\frac{3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{3^{\frac{5}{6}}\times {12}^{\frac{7}{12}}}{6^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}& ={\displaystyle \frac{3^{\frac{5}{6}}\times (3\times 4{)}^{\frac{7}{12}}}{(2\times 3{)}^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 4^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times {\left(2^2\right)}^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 2^{\frac{14}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & =2^{(\frac{14}{12}-\frac{2}{3}+\frac{1}{4})}\times 3^{(\frac{5}{6}+\frac{7}{12}-\frac{2}{3})}\\ & =2^{(\frac{14-8+3}{12})}\times 3^{(\frac{10+7-8}{12})}\\ & =2^{\frac{9}{12}}\times 3^{\frac{9}{12}}\\ & =(2\times 3{)}^{\frac{9}{12}}\\ & =6^{\frac{9}{12}}\\ & =6^{\frac{3}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Bentuk sederhana dari}{\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{-3}}{a^{-1}{b}^{-\frac{3}{2}}}}\right)}^{\frac{3}{2}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a}{b}}\\ \\ \text{b}.& {\displaystyle \frac{b}{a}}\\ \\ \text{c}.& {\displaystyle ab}\\ \\ \text{d}.& \sqrt[a]{b}\\ \\ \text{e}.& \sqrt[4]{{\left({\displaystyle \frac{a}{b}}\right)}^9}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{-3}}{a^{-1}{b}^{-\frac{3}{2}}}}\right)}^{\frac{3}{2}}& ={\displaystyle \frac{a^{\frac{3}{4}}{b}^{\frac{-9}{2}}}{a^{-\frac{3}{2}}{b}^{-\frac{9}{4}}}}\\ & =a^{\frac{3}{4}+\frac{3}{2}}{b}^{\frac{-9}{2}+\frac{9}{4}}\\ & =a^{\frac{3+6}{4}}{b}^{\frac{-18+9}{4}}\\ & =a^{\frac{9}{4}}{b}^{-\frac{9}{4}}\\ & ={\left({\displaystyle \frac{a}{b}}\right)}^{\frac{9}{4}}\\ & =\sqrt[4]{{\left({\displaystyle \frac{a}{b}}\right)}^9}\end{array}\end{array}$

Fungsi Eksponen

 $\text{A. Bilangan Pangkat Positif}$

Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $m$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat dedefinisikan dengan:

$a^m=\underset{m}{\underbrace{a\times a\times \times a\times ...\times a}}$

$\begin{array}{rl}\text{Bilangan}& :\\ a& \text{disebut basis atau bilangan pokok}\\ n& \text{disebut sebagai bilangan pangkat/eksponen}\end{array}$

$\text{ Contoh Soal}$

$(1).3^4=3\times 3\times 3\times 3=81$

$(2).5^4=5\times 5\times 5\times 5=625$

$(3).2^6=2\times 2\times 2\times 2\times 2\times 2=64$

$(4).6^7=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

$(5).(-3{)}^3=(-3)\times (-3)\times (-3)=-27$

$(6).(-2{)}^6=(-2)\times (-2)\times (-2)\times (-2)\times (-2)\times (-2)=64$

$(7).{\left({\displaystyle \frac{1}{5}}\right)}^3=\left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)={\displaystyle \frac{1}{125}}$

$(8).{(-{\displaystyle \frac{1}{2}})}^4=(-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{1}{32}}$

$\text{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{array}{r}\\ 1.& a^m.a^n=a^{m+n}\\ 2.& a^m:a^n=a^{m-n}\\ 3.& {\left(a^m\right)}^n=a^{m.n},\text{syarat}a\ne 0\\ 4.& {\left(ab\right)}^n=a^n.b^n\\ 5.& {\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n},\text{syarat}b\ne 0\end{array}$

Beberpa hal yang perlu diketahui juga, yaitu

$\begin{array}{r}\\ 1.& (a+b{)}^2=a^2+2ab+b^2\\ 2.& (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3\\ 3.& {(a+\frac{1}{a})}^2=a^2+2+{\displaystyle \frac{1}{a^2},\text{syarat}a\ne 0}\end{array}$

$\text{ Contoh Soal}$

$\begin{array}{r}\\ (1).& 2^6\times 2^4\times 2^7=2^{6+4+7}=2^{17}\\ (2).& 2^5\times 3^5\times 7^5={(2.3.7)}^5={\left(42\right)}^5\\ (3).& {\displaystyle \frac{a^3.a^7.a^6}{a^9}={\displaystyle \frac{a^{3+7+6}}{a^9}=\frac{a^{16}}{a^9}=a^{16-9}=a^7,\text{syarat}a\ne 0}}\end{array}$

$\begin{array}{rl}(4).{\displaystyle \frac{3^7{.7}^3.2}{{\left(42\right)}^3}}& =\frac{2^1{.3}^7{.7}^3}{{\left(2.3.7\right)}^3}=\frac{2^1{.3}^7{.7}^3}{2^3{.3}^3{.7}^3}\\ & =2^{1-3}{.3}^{7-3}{.7}^{3-3}=2^{-2}{.3}^4{.7}^0\\ & =\frac{1}{2^2}{.3}^4.1=\frac{3^4}{2^2}\end{array}$

$\begin{array}{rl}(5).{\displaystyle \frac{2^{2020}+2^{2021}+2^{2022}}{7}}& ={\displaystyle \frac{{1.2}^{2020}+2^1{.2}^{2020}+2^2{.2}^{2020}}{7}}\\ & =\frac{(1+2+4){.2}^{2020}}{7}\\ & =\frac{{7.2}^{2020}}{7}\\ & =2^{2020}\end{array}$

$\begin{array}{rl}(6){\displaystyle \frac{{\left(2^{n+2}\right)}^2-2^2{.2}^{2n}}{2^n{.2}^{n+2}}}& ={\displaystyle \frac{2^{2(n+2)}-2^2{.2}^{2n}}{2^n{.2}^n{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}{.2}^{2.2}-2^2{.2}^{2n}}{2^{n+n}{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}(2^4-2^2)}{2^{2n}{.2}^2}}\\ & ={\displaystyle \frac{(2^4-2^2)}{2^2}=\frac{16-4}{4}}\\ & ={\displaystyle \frac{12}{4}=3}\end{array}$

$\text{C. Bentuk Akar}$

Bilangan bentuk akar di sini adalah kebalikan dari bilangan bentuk pangkat. Bilangan bentuk akar selanjutnya disebut bilangan irasional. Sebagai contoh $\sqrt{2}$, $\sqrt{3}$, $\sqrt{8}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$, $\sqrt[3]{7}$ dan tapi ingat $\sqrt{4}$ dan  $\sqrt[3]{8}$ serta  $\sqrt[3]{27}$ adalah bukan bentuk akar, karena nantinya akan menghasilkan masing-masing 2 dan 3 serta 3.

$\begin{array}{rl}& \\ 1.& a^{\frac{1}{n}}=\sqrt[n]{a}\\ 2.& a^{\frac{m}{n}}=\sqrt[n]{a^m}\\ 3.& a^{\frac{1}{2}}=\sqrt[2]{a^1}=\sqrt{a}\end{array}$

Sifat-sifat yang berlaku pada operasi bilangan bentuk akar

$\begin{array}{rl}& \\ 1.& a\sqrt[n]{c}+b\sqrt[n]{c}=(a+b)\sqrt[n]{c}\\ 2.& a\sqrt[n]{c}-b\sqrt[n]{c}=(a-b)\sqrt[n]{c}\\ 3.& \sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.& \sqrt[n]{a^n}=a\\ 5.& a\sqrt[n]{c}xb\sqrt[n]{d}=ab\sqrt[n]{cd}\\ 6.& \frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.& \sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.& \sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}\end{array}$

Merasionalkan penyebut

Jika suatu pecahan penyebutnya mengandung bilangan irasional atau bentuk akar, maka penyebut ini dapat dibuat menjadi bilangan rasional. Perhatikanlah langkah berikut

$\begin{array}{rl}1.& {\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{\left(\sqrt{b^2}\right)}=\frac{a}{b}\sqrt{b}}\\ 2.& {\displaystyle \frac{a}{\sqrt[3]{b}}=\frac{a}{\sqrt[3]{b}}\times \frac{\sqrt[3]{b^2}}{\sqrt[3]{b^2}}=\frac{a\sqrt[3]{b^2}}{\left(\sqrt[3]{b^3}\right)}=\frac{a}{b}\sqrt[3]{b^2}}\\ 3.& {\displaystyle \frac{a}{\sqrt[5]{b^3}}={\displaystyle \frac{a}{\sqrt[5]{b^3}}\times \frac{\sqrt[5]{b^2}}{\sqrt[5]{b^2}}=\frac{a\sqrt[5]{b^2}}{\sqrt[5]{b^5}}=\frac{a}{b}\sqrt[5]{b^2}}}\end{array}$

Merasionalkan di atas adalah contoh bebrapa contoh model merasionalkan jika berjenis tunggal tetapi jika nanti jenisnya lebih dari itu, maka perhatikanlah simulasi contoh berikut

$\begin{array}{rl}& \\ 1.& {\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c(a-\sqrt{b})}{a^2-b}}\\ 2.& {\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c(a+\sqrt{b})}{a^2-b}}\\ 3.& {\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c(\sqrt{a}-\sqrt{b})}{a-b}}\end{array}$

Perhatikanlah simulasi contoh di atas, bentuk $a+\sqrt{b}$ memiliki bentuk sekawan (irasional juga) $a-\sqrt{b}$, demikian juga bentuk $\sqrt{a}+\sqrt{b}$ memiliki sekawan $\sqrt{a}-\sqrt{b}$. Disamping itu ada bentuk khusus yatu bentuk  $\sqrt[3]{a}+\sqrt[3]{b}$ memiliki bentuk sekawan $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$.

Contoh Soal 5 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 21.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}(x-\sqrt{x^2-10x})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -10}\\ \text{b}.{\displaystyle -5}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 5}\\ \text{e}.{\displaystyle 10}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(x-\sqrt{x^2-10x})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2}-\sqrt{x^2-10x})\\ & \text{Selanjutnya gunakan formula}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})\\ & ={\displaystyle \frac{b-p}{2\sqrt{a}},\text{maka}}\\ & ={\displaystyle \frac{0-(-10)}{2\sqrt{1}}}\\ & =\frac{10}{2}\\ & =5\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}5\tan {\displaystyle \frac{1}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \mathrm{\infty }}\\ \text{b}.& {\displaystyle 5}\\ \text{c}.& {\displaystyle \sqrt{3}}\\ \text{d}.& 1\\ \text{e}.& {\displaystyle 0}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}5\tan {\displaystyle \frac{1}{x}}& =\underset{u\to 0}{\text{Lim}}5\tan {\displaystyle u}\\ & =5\tan 0\\ & =5.\mathrm{\infty }\\ & =\mathrm{\infty }\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}15x\tan {\displaystyle \frac{4}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0}\\ \text{b}.& {\displaystyle \frac{1}{4}}\\ \text{c}.& {\displaystyle 4}\\ \text{d}.& {\displaystyle \frac{11}{4}}\\ \text{e}.& {\displaystyle 60}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}15x\tan {\displaystyle \frac{4}{x}}& =\cdots \\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}\text{maka}x={\displaystyle \frac{1}{u}}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{15}{u}.\tan 4u}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{15\tan 4u}{u}}\\ & =15\times 4\\ & =60\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{x}^2{\sin }^2\left({\displaystyle \frac{ab}{x}}\right)\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle ab}\\ \text{b}.& {\displaystyle a^{2}b}\\ \text{c}.& {\displaystyle a{b}^2}\\ \text{d}.& {\displaystyle (ab{)}^2}\\ \text{e}.& {\displaystyle \frac{1}{(ab{)}^2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{x}^2\sin {\displaystyle \frac{ab}{x}}& =\underset{u\to 0}{\text{Lim}}{\left({\displaystyle \frac{1}{u}}\right)}^2{\sin }^2{\displaystyle abu}\\ & =\underset{u\to 0}{\text{Lim}}\left({\displaystyle \frac{{\sin }^{2}abu}{u^2}}\right)\\ & =(ab{)}^2\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{{\displaystyle \frac{x}{100}}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\mathrm{\infty }}\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{{\displaystyle \frac{x}{100}}}}& =100\times \underset{0}{\underbrace{\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sin 2x}{x}}}}\\ & =100\times 0\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 26.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{18}{x\sin {\displaystyle \frac{3}{x}}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -54}\\ \text{b}.& {\displaystyle -6}\\ \text{c}.& {\displaystyle {\displaystyle \frac{1}{6}}}\\ \text{d}.& {\displaystyle 6}\\ \text{e}.& {\displaystyle 54}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{18}{x\sin {\displaystyle \frac{3}{x}}}}& =\cdots \\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}\text{maka}x={\displaystyle \frac{1}{u}}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{18}{{\displaystyle \frac{1}{u}\sin 3u}}}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{18u}{\sin 3u}}\\ & ={\displaystyle \frac{18}{3}}\\ & =6\end{array}\end{array}$.

$\begin{array}{l}\\ 27.& \text{Nilai dari}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{4x\sin {\displaystyle \frac{2}{x}}}{2}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -4}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& {\displaystyle {\displaystyle \frac{1}{2}}}\\ \text{d}.& {\displaystyle 2}\\ \text{e}.& {\displaystyle 4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{4x\sin {\displaystyle \frac{2}{x}}}{2}}& =\cdots \\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}\text{maka}x={\displaystyle \frac{1}{u}}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{4}{u}.\sin 2u}}{2}}\\ & =\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{4\sin 2u}{2u}}\\ & ={\displaystyle \frac{4\times 2}{2}}\\ & =4\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Nilai dari}\\ & \underset{x\to -\mathrm{\infty }}{\text{Lim}}{\displaystyle x\cos \frac{1}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\mathrm{\infty }}\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\underset{x\to -\mathrm{\infty }}{\text{Lim}}{\displaystyle x\cos \frac{1}{x}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (-x)\cos \frac{1}{(-x)}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (-x)\cos \frac{1}{(x)}}\\ & =-\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (x)\cos \frac{1}{(x)}}\\ & \{\begin{array}{ll}u& ={\displaystyle \frac{1}{x}}\\ x& \to \mathrm{\infty },\text{maka}{\displaystyle u\to 0}\end{array}\\ & =-\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{1}{u}\cos u}\\ & =-\underset{u\to 0}{\text{Lim}}{\displaystyle \frac{\cos u}{u}}\\ & =-{\displaystyle \frac{1}{0}}\\ & =-\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 29.& \text{Asimtot tegak dari fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x=2\text{dan}x=4\\ \text{b}.& x=2\text{dan}x=3\\ \text{c}.& x=3\text{dan}x=4\\ \text{d}.& x=3\text{saja}\\ \text{e}.& x=2\text{saja}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Asimtot tegak fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}}\\ & \text{terjadi saat penyebut}=0.\\ & \text{Sehingga}{x}^2-5x+6=0\\ & \iff (x-2)(x-3)=0,\text{maka}\\ & x=2\text{atau}x=3\\ & \therefore \text{asimtot tegak fungsi}\\ & f(x)={\displaystyle \frac{x^2-6x-8}{x^2-5x+6}}\\ & \text{adalah}x=2\text{dan}x=3\end{array}\end{array}$

$\begin{array}{l}\\ 30.& \text{Asimtot datar dari fungsi}\\ & g(x)={\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& y=-3\\ \text{b}.& y=-1\\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Asim}& \text{tot datar dari fungsi}\\ g(x)& ={\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\text{untuk}}\\ g(x)& ={\displaystyle \frac{(6x^2-8x+2)}{(-2x^2+5x-2)}\text{terjadi saat}}\\ y& ={\displaystyle \frac{6}{-2}=-3}\\ & \mathbf{\text{atau dapat juga dicari}}\mathbf{\text{dengan}}\\ y& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(6x^2-8x+2)}{(-2x^2+5x-2)}\times {\displaystyle \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{6-{\displaystyle \frac{8}{x}+\frac{2}{x^2}}}{-2+{\displaystyle \frac{5}{x}-\frac{2}{x^2}}}}\\ & ={\displaystyle \frac{6-0+0}{-2+0-0}}\\ & ={\displaystyle \frac{6}{-2}}\\ & =-3\end{array}\end{array}$

Contoh Soal 4 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 16.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{4}{9}}\\ \\ \text{c}.& {\displaystyle \frac{1}{2}}\\ \\ \text{d}.& 1\\ \\ \text{e}.& {\displaystyle \frac{3}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4x^2-2x}-\sqrt{x^2+1}}{\sqrt{9x^2-1}}\times {\displaystyle \frac{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^2}}}{\sqrt{9-\frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}}\\ & ={\displaystyle \frac{2-1}{3}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x^4+2x^3-5x+2021}{2x^3-4x^2+2020}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{9}}\\ \\ \text{b}.{\displaystyle \frac{3}{2}}\\ \\ \text{c}.{\displaystyle 0}& \\ \\ \text{d}.{\displaystyle 1}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x^4+2x^3-5x+2021}{2x^3-4x^2+2020}}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3x^4}{x^4}+\frac{2x^3}{x^4}-\frac{5x}{x^4}+\frac{2021}{x^4}}}{{\displaystyle \frac{2x^3}{x^4}-\frac{4x^2}{x^4}+\frac{2020}{x^4}}}}}\\ & ={\displaystyle \frac{3+{\displaystyle \frac{2}{x}-\frac{5}{x^2}+\frac{2021}{x^4}}}{{\displaystyle \frac{4}{x}-\frac{4}{x^2}+\frac{2020}{x^4}}}}\\ & ={\displaystyle \frac{3+0-0+0}{0-0+0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+3x+4}{3x^2+2x+3}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{3}}\\ \\ \text{b}.{\displaystyle \frac{1}{3}}\\ \\ \text{c}.{\displaystyle 0}\\ \\ \text{d}.{\displaystyle 3}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+3x+4}{3x^2+2x+3}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2}{x^2}+\frac{3x}{x^2}+\frac{4}{x^2}}}{{\displaystyle \frac{3x^2}{x^2}+\frac{2x}{x^2}+\frac{3}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{1+{\displaystyle \frac{3}{x}+\frac{4}{x^2}}}{3+{\displaystyle \frac{2}{x}+\frac{3}{x^2}}}}\\ & ={\displaystyle \frac{1+0+0}{3+0+0}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x}{9x^2+x+1}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 3}\\ \\ \text{b}.{\displaystyle 1}\\ \\ \text{c}.{\displaystyle \frac{1}{3}}\\ \\ \text{d}.{\displaystyle 0}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x}{9x^2+x+1}}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3x}{x^2}}}{{\displaystyle \frac{9x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3}{x}}}{{\displaystyle \frac{9x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}}\\ & ={\displaystyle \frac{0}{9+0+0}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2-2x-8}-\sqrt{x^2+2x+1})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2}& & \\ \text{b}.{\displaystyle -1}\\ \text{c}.{\displaystyle -\frac{1}{2}}\\ \text{d}.{\displaystyle 0}\\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2-2x-8}-\sqrt{x^2+2x+1})\\ & =\mathrm{\infty }-\mathrm{\infty }=\mathbf{\text{tidak diperbolehkan}}\\ & \text{Selanjutnya gunakan formula}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})={\displaystyle \frac{b-p}{2\sqrt{a}},\text{maka}}\\ & ={\displaystyle \frac{-2-2}{2\sqrt{1}}}\\ & =\frac{-4}{2}\\ & =-2\end{array}\end{array}$

Contoh Soal 3 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 11.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{3x+1}-\sqrt{3x-2})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 4\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{3x+1}-\sqrt{3x-2})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{3x+1}-\sqrt{3x-2})\times {\displaystyle \frac{(\sqrt{3x+1}+\sqrt{3x-2})}{(\sqrt{3x+1}+\sqrt{3x-2})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}\frac{(3x+1)-(3x-2)}{(\sqrt{3x+1}+\sqrt{3x-2})}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3}{(\sqrt{3x+1}+\sqrt{3x-2})}\times \frac{{\displaystyle \frac{1}{\sqrt{x}}}}{{\displaystyle \frac{1}{\sqrt{x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{3}{\sqrt{x}}}}{(\sqrt{{\displaystyle \frac{3x}{x}+\frac{1}{x}}}+\sqrt{{\displaystyle \frac{3x}{x}-\frac{2}{x}}})}}\\ & ={\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{4x^2+6x+8}-\sqrt{4x^2-8x+7})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \\ \text{c}.& {\displaystyle \frac{3}{2}}\\ \\ \text{d}.& {\displaystyle \frac{7}{2}}\\ \\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{4x^2+6x+8}-\sqrt{4x^2-8x+7})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{4x^2+6x+8}-\sqrt{4x^2-8x+7})\\ & \times {\displaystyle \frac{(\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7})}{(\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(4x^2+6x+8)-(4x^2-8x+7)}{\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{14x+1}{\sqrt{4x^2+6x+8}+\sqrt{4x^2-8x+7}}\times \frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{14+{\displaystyle \frac{1}{x}}}{\sqrt{{\displaystyle \frac{4x^2}{x^2}+\frac{6x}{x^2}+\frac{8}{x^2}}}+\sqrt{{\displaystyle \frac{4x^2}{x^2}-\frac{8x}{x^2}+\frac{7}{x^2}}}}}\\ & ={\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}}\\ & ={\displaystyle \frac{14}{2+2}=\frac{7}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 8\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})\\ & \times {\displaystyle \frac{(\sqrt{2x^2+3x-1}+\sqrt{x^2-5x+3})}{(\sqrt{2x^2+3x-1}+\sqrt{x^2-5x+3})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(2x^2+3x-1)-(x^2-5x+3)}{(\sqrt{2x^2+3x-1}+\sqrt{x^2-5x+3})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{x^2+8x-4}{(\sqrt{2x^2+3x-1}-\sqrt{x^2-5x+3})}\times \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2}{x^2}+\frac{8x}{x^2}-\frac{4}{x^2}}}{\sqrt{{\displaystyle \frac{2x^2}{x^4}+\frac{3x}{x^4}-\frac{1}{x^4}}}+\sqrt{{\displaystyle \frac{x^2}{x^4}-\frac{5x}{x^4}+\frac{3}{x^4}}}}}\\ & ={\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}}\\ & ={\displaystyle \frac{1}{0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 4\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})\\ & \times {\displaystyle \frac{(\sqrt{x^2+3x+1}+\sqrt{3x^2+2x+5})}{(\sqrt{x^2+3x+1}+\sqrt{3x^2+2x+5})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^2+3x+1)-(3x^2+2x+5)}{(\sqrt{x^2+3x+1}+\sqrt{3x^2+2x+5})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{-2x^2+x-4}{(\sqrt{x^2+3x+1}-\sqrt{3x^2+2x+5})}\times \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{-2x^2}{x^2}+\frac{x}{x^2}-\frac{4}{x^2}}}{\sqrt{{\displaystyle \frac{x^2}{x^4}+\frac{3x}{x^4}+\frac{1}{x^4}}}+\sqrt{{\displaystyle \frac{3x^2}{x^4}+\frac{2x}{x^4}+\frac{5}{x^4}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{-2+{\displaystyle \frac{1}{x}-\frac{4}{x^2}}}{\sqrt{{\displaystyle \frac{1}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}}}+\sqrt{{\displaystyle \frac{3}{x^2}+\frac{2}{x^3}+\frac{5}{x^4}}}}}\\ & ={\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}}\\ & ={\displaystyle \frac{-2}{0}}\\ & =-\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}((3x-2)-\sqrt{9x^2-2x+5})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \\ \text{b}.& -{\displaystyle \frac{5}{3}}\\ \\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \\ \text{d}.& 1\\ \\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}((3x-2)-\sqrt{9x^2-2x+5})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{(3x-2{)}^2}-\sqrt{9x^2-2x+5})\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{(9x^2-12x+4}-\sqrt{9x^2-2x+5})\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}(\sqrt{(a{x}^2+bx+c}-\sqrt{p{x}^2+qx+r})\\ & \text{Jika dikerjakan dengan rumus singkat}\\ & \text{maka}\{\begin{array}{c}a=p=3\\ b=-12\\ q=-2\end{array}\\ & ={\displaystyle \frac{b-q}{2\sqrt{a}}}\\ & ={\displaystyle \frac{-12-(-2)}{2\sqrt{9}}}\\ & ={\displaystyle \frac{-10}{6}}\\ & =-\frac{5}{3}\end{array}\end{array}$

Contoh Soal 2 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

 $\begin{array}{l}\\ 6.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}-\sqrt{4x+2021})\times \frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{{\displaystyle \frac{1}{x}(\sqrt{8x-2020}+\sqrt{4x+2021})}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{8x}{x^2}-\frac{2020}{x^2}}}+\sqrt{{\displaystyle \frac{4x}{x^2}+\frac{2021}{x^2}}})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{8}{x}-{\displaystyle \frac{2020}{x}}}}+\sqrt{{\displaystyle \frac{4}{x}+{\displaystyle \frac{2021}{x}}}})}}\\ & ={\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}}\\ & ={\displaystyle \frac{4}{0}}\\ & =\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}+\sqrt{4x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{8x-2020}+\sqrt{4x+2021})\\ & =\sqrt{\mathrm{\infty }}+\sqrt{\mathrm{\infty }}\\ & =\mathrm{\infty }\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai yang memenuhi}\\ & \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{4x-2020}-\sqrt{8x+2021})\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& 0\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& \mathrm{\infty }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{4x-2020}-\sqrt{8x+2021})\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}(\sqrt{4x-2020}-\sqrt{8x+2021})\times \frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{1}{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4-{\displaystyle \frac{4041}{x}}}{{\displaystyle \frac{1}{x}(\sqrt{4x-2020}+\sqrt{8x+2021})}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{4x}{x^2}-\frac{2020}{x^2}}}+\sqrt{{\displaystyle \frac{8x}{x^2}+\frac{2021}{x^2}}})}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{-4-{\displaystyle \frac{4041}{x}}}{(\sqrt{{\displaystyle \frac{4}{x}-{\displaystyle \frac{2020}{x}}}}+\sqrt{{\displaystyle \frac{8}{x}+{\displaystyle \frac{2021}{x}}}})}}\\ & ={\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}}\\ & ={\displaystyle \frac{-4}{0}}\\ & =-\mathrm{\infty }\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}=....\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle 1}\\ \text{b}.{\displaystyle 1}\\ \text{c}.{\displaystyle 2}\\ \text{d}.{\displaystyle 4}\\ \text{e}.{\displaystyle 8}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}\times {\displaystyle \frac{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{4x^2+3x-(4x^2-5x)}{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{3x+5x}{\sqrt{4x^2+3x}+\sqrt{4x^2-5x}}\times {\displaystyle \frac{\left({\displaystyle \frac{1}{x}}\right)}{\left(\sqrt{{\displaystyle \frac{1}{x^2}}}\right)}}}\\ & ={\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}}\\ & ={\displaystyle \frac{8}{4}}\\ & =2\end{array}\end{array}$

$\begin{array}{rl}\text{ada cara lain yang lebih sede}& \text{rhana, yaitu:}\\ .\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}& =\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}\\ & \{\begin{array}{ll}a& =4\\ b& =3\\ p& =-4\end{array}\\ \text{Jika}& \\ \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{a{x}^2+bx+c}& -\sqrt{a{x}^2+px+q}={\displaystyle \frac{b-p}{2\sqrt{a}}}\\ \text{Sehingga}& \\ \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}-\sqrt{4x^2-5x}& ={\displaystyle \frac{3-(-5)}{2\sqrt{4}}}\\ & ={\displaystyle \frac{8}{2.2}}\\ & =2\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}+\sqrt{4x^2-5x}=....\\ & \begin{array}{l}\\ \text{a}.\mathrm{\infty }\\ \text{b}.{\displaystyle 1}\\ \text{c}.{\displaystyle 2}\\ \text{d}.{\displaystyle 4}\\ \text{e}.{\displaystyle 8}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \underset{x\to \mathrm{\infty }}{\text{Lim}}\sqrt{4x^2+3x}+\sqrt{4x^2-5x}\\ & =\sqrt{\mathrm{\infty }}+\sqrt{\mathrm{\infty }}=\mathrm{\infty }\end{array}$

☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝

$\begin{array}{rl}\text{Sebagai}& \mathbf{\text{CATATAN}}\text{di sini}\\ \text{Sifat-sif}& \text{at bilangan tak hingga}\\ (1)& \mathrm{\infty }+\mathrm{\infty }=\mathrm{\infty }\\ (2)& -\mathrm{\infty }+(-\mathrm{\infty })=-\mathrm{\infty }\\ (3)& \mathrm{\infty }\times \mathrm{\infty }=\mathrm{\infty }\\ (4)& -\mathrm{\infty }+(-\mathrm{\infty })=\mathrm{\infty }\\ (5)& k.\mathrm{\infty }=\mathrm{\infty },k\text{positif}\\ (6)& k.(-\mathrm{\infty })=-\mathrm{\infty },k\text{positif}\\ (7)& k.\mathrm{\infty }=-\mathrm{\infty },k\text{negatif}\\ (8)& k.(-\mathrm{\infty })=\mathrm{\infty },k\text{negatif}\\ \text{yang ha}& \text{rus dihindari}\\ (1)& \mathrm{\infty }-\mathrm{\infty },\text{bentuk tak tentu}\\ (2)& {\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }},-{\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }},\text{dan}\frac{0}{0}}}\end{array}$

Contoh Soal 1 Limit di Ketakhinggan (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 1.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}=....}\\ \\ & \text{a}.{\displaystyle 8}\\ & \text{b}.{\displaystyle 4}\\ & \text{c}.2\\ & \text{d}.1\\ & \text{e}.{\displaystyle \frac{1}{2}}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8x^2+x-2020}{2x^2-2021x}\times {\displaystyle \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{{\displaystyle \frac{8x^2}{x^2}+\frac{x}{x^2}-\frac{2020}{x^2}}}{{\displaystyle \frac{2x^2}{x^2}-\frac{2021x}{x^2}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{8+{\displaystyle \frac{1}{x}-\frac{2020}{x^2}}}{2-{\displaystyle \frac{2021}{x}}}}\\ & ={\displaystyle \frac{8+{\displaystyle \frac{1}{\mathrm{\infty }}-\frac{2020}{{\mathrm{\infty }}^2}}}{2-{\displaystyle \frac{2021}{\mathrm{\infty }}}}}\\ & ={\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nilai}\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2019}{\sqrt{9x^2-2020x}}=....}\\ \\ & \text{a}.{\displaystyle 3}\\ & \text{b}.{\displaystyle 1}\\ & \text{c}.{\displaystyle \frac{1}{3}}\\ & \text{d}.-{\displaystyle \frac{1}{3}}\\ & \text{e}.{\displaystyle -3}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{x+2021}{\sqrt{9x^2-2020x}}\times {\displaystyle \frac{\left({\displaystyle \frac{1}{x}}\right)}{(-\sqrt{{\displaystyle \frac{1}{x^2}}})}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{{\displaystyle \frac{x}{x}+\frac{2021}{x}}}{-\sqrt{{\displaystyle \frac{9x^2}{x^2}-\frac{2020x}{x^2}}}}}\\ & =\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1+{\displaystyle \frac{2021}{x}}}{-\sqrt{9-{\displaystyle \frac{2020}{x}}}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{2021}{\mathrm{\infty }}}}{-\sqrt{9-{\displaystyle \frac{2020}{\mathrm{\infty }}}}}={\displaystyle \frac{1}{-\sqrt{9}}}}\\ & =-{\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=....}\\ \\ & \text{a}.{\displaystyle 1}\\ & \text{b}.{\displaystyle 4}\\ & \text{c}.9\\ & \text{d}.16\\ & \text{e}.25\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{e}}\\ & \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times {\displaystyle \frac{{\displaystyle \frac{1}{5^x}}}{{\displaystyle \frac{1}{5^x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{2{\left({\displaystyle \frac{2}{5}}\right)}^x+3{\left({\displaystyle \frac{3}{5}}\right)}^x+4{\left({\displaystyle \frac{4}{5}}\right)}^x+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{{\displaystyle \frac{1}{2}{\left({\displaystyle \frac{2}{5}}\right)}^x+\frac{1}{3}{\left({\displaystyle \frac{3}{5}}\right)}^x+\frac{1}{4}{\left({\displaystyle \frac{4}{5}}\right)}^x+\frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{0+0+0+5{\left({\displaystyle \frac{5}{5}}\right)}^x}{0+0+0+{\displaystyle \frac{1}{5}{\left({\displaystyle \frac{5}{5}}\right)}^x}}}\\ & ={\displaystyle \frac{5.1}{{\displaystyle \frac{1}{5}.1}}}\\ & =25\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \mathbf{\text{(USM UGM Mat IPA)}}\\ & \text{Nilai}\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-x-1)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{5}{3}}\\ \\ \text{b}.{\displaystyle \frac{2}{3}}\\ \\ \text{c}.-{\displaystyle \frac{1}{3}}\\ \\ \text{d}.-{\displaystyle \frac{2}{3}}\\ \\ \text{e}.-{\displaystyle \frac{5}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\end{array}$

$\begin{array}{rl}.\underset{x\to \mathrm{\infty }}{\text{Lim}}& {\displaystyle (\sqrt[3]{x^3-2x^2}-x-1)}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\sqrt[3]{x^3-2x^2}-\sqrt[3]{{(x+1)}^3})}\\ & \text{ingat bentuk}a-b=(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})\\ & \text{dan untuk}\{\begin{array}{ll}a& =(x^3-2x^2)\\ & \\ b& ={(x+1)}^3\end{array}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^3-2x^2)-{(x+1)}^3}{\sqrt[3]{{(x^3-2x^2)}^2}+\sqrt[3]{(x^3-2x^2){(x+1)}^3}+\sqrt[3]{{\left({(x+1)}^3\right)}^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{(x^3-2x^2)-(x^3+3x^2+3x+1)}{{(x^3-2x^2)}^{\frac{2}{3}}+{(x^6+...)}^{\frac{1}{3}}+{(x+1)}^{\frac{6}{3}}}}\\ & =\underset{x\to \mathrm{\infty }}{\text{Lim}}{\displaystyle \frac{-5x^2+...}{{(x^3-2x^2)}^{\frac{2}{3}}+{(x^6+...)}^{\frac{1}{3}}+{(x+1)}^{\frac{6}{3}}}}\\ & ={\displaystyle \frac{-5}{1+1+1}}\\ & =-{\displaystyle \frac{5}{3}}\end{array}$

$\begin{array}{l}\\ 5.& \text{Nilai}\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 1}\\ \\ \text{b}.{\displaystyle \frac{3}{2}}\\ \\ \text{c}.{\displaystyle 2}\\ \\ \text{d}.{\displaystyle \frac{5}{2}}\\ \\ \text{e}.{\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)})}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle ((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{k}-\frac{1}{k+1}))}\\ & =\underset{k\to \mathrm{\infty }}{\text{Lim}}{\displaystyle (1-\frac{1}{k+1})}\\ & ={\displaystyle (1-\frac{1}{\mathrm{\infty }+1})}\\ & ={\displaystyle 1-\frac{1}{\mathrm{\infty }}}\\ & =1-0\\ & =1\end{array}\end{array}$

Limit di Ketakhinggaan (Kelas XII Matematika Peminatan)

 Untuk $n$ bilangan bulat positif, dan $k$ suatu konstanta, serta fungsi  $f$ dan $g$ adalah dua buah fungsi yang memiliki nilai limit saat  $x$  mendekati ketakhinggaan $(x\to \mathrm{\infty })$ , maka berlakulah sifat-sifat berikut:

$\begin{array}{l}\\ 1.& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1}{x}=0\mathbf{\text{dan}}\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1}{x}=0}}\\ 2.& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1}{x^n}=0\mathbf{\text{dan}}\underset{x\to -\mathrm{\infty }}{\text{lim}}{\displaystyle \frac{1}{x^n}=0}}\\ 3.& \underset{x\to \mathrm{\infty }}{\text{lim}}k=k\\ 4.& \underset{x\to \mathrm{\infty }}{\text{lim}}k.f(x)=k.\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)\\ 5.& \underset{x\to \mathrm{\infty }}{\text{lim}}(f(x)+g(x))=\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)+\underset{x\to \mathrm{\infty }}{\text{lim}}g(x)\\ 6.& \underset{x\to \mathrm{\infty }}{\text{lim}}(f(x)-g(x))=\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)-\underset{x\to \mathrm{\infty }}{\text{lim}}g(x)\\ 7.& \underset{x\to \mathrm{\infty }}{\text{lim}}(f(x)\times g(x))=\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)\times \underset{x\to \mathrm{\infty }}{\text{lim}}g(x)\\ 8.& \underset{x\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{f(x)}{g(x)}={\displaystyle \frac{\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)}{\underset{x\to \mathrm{\infty }}{\text{lim}}g(x)}}}\\ 9.& \underset{x\to \mathrm{\infty }}{\text{lim}}{(f(x))}^n={[\underset{x\to \mathrm{\infty }}{\text{lim}}f(x))]}^n\\ 10.& \underset{x\to \mathrm{\infty }}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)},\text{dengan}\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)\ge 0\end{array}$

Contoh Soal 4 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 16.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-4}<\sqrt{2x+8}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -4<x\le {\displaystyle \frac{2}{3}}\\ \text{b}.& -4<x<3\\ \text{c}.& {\displaystyle \frac{2}{3}\le x<3}\\ \text{d}.& 2<x\le 4\\ \text{e}.& -4\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \sqrt{6x-4}}& <\sqrt{2x+8}\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-4& <2x+8\\ 6x-2x& <8+4\\ 4x& <12\\ x& <3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-4& \ge 0\\ 6x& \ge 4\\ x& \ge {\displaystyle \frac{2}{3}}\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 2x+8& \ge 0\\ 2x& \ge -8\\ x& \ge -4\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{x+3}>\sqrt{12-2x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 3<x\le 6\\ \text{b}.& -3<x\le 6\\ \text{c}.& -6<x\le 3\\ \text{d}.& -6<x\le -3\\ \text{e}.& x<3\text{atau}x>6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle {\displaystyle \sqrt{x+3}}}& >\sqrt{12-2x}\\ 1.\text{Kuadrat}& \text{kan}\\ x+3& >12-2x\\ x+2x& >12-3\\ 3x& >9\\ x& >3\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ x+3& \ge 0\\ x& \ge -3\\ 3.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 12-2x& \ge 0\\ 2x-12& \le 0\\ 2x& \le 12\\ x& \le 6\end{array}\end{array}$

$\begin{array}{l}\\ 18.& (\mathbf{\text{SBMPTN 2013 Mat Das}})\\ & \text{Jika}1<m<2,\text{maka semua nilai}\\ & x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2+4x}{-x^2+3x-3m}>0\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x>-3\\ \text{b}.& x<-4\\ \text{c}.& -4<x<0\\ \text{d}.& x<-4\text{atau}x>0\\ \text{e}.& x<-3\text{atau}x>-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}1.& \text{Diketahui bahwa}:{\displaystyle \frac{x^2+4x}{-x^2+3x-3m}>0}\\ & \text{dengan kondisi}1<m<2\\ & \text{Perhatikanlah penyebutnya yang}\\ & \text{mengandung bilangan}m\text{yang terletak}\\ & \text{pada interval}:1<m<2.\\ 2.& \text{Kita cek kondisi penyebutnya dengan}\\ & \text{menentukan}Diskriminan(D)-\text{nya}\\ & \text{yaitu}:\\ & a{x}^2+bx+c\{\begin{array}{l}a>0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit positif}\\ a<0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit negatif}\end{array}\\ & \text{Karena penyebutnya}:-x^2+3x-3m,\\ & \text{dengan}a=-1,b=3,\mathrm{\&}c=-3m,\text{maka}\\ & D=3^2-4(-1)(-3m)=9-12m\\ 3.& \text{Karena nilai}m\text{berada pada}1<m<2\\ & \text{maka}\\ & 1<m<2\\ & \iff 12<12m<24\\ & \iff -12>-12m>-24\\ & \iff 9-12>9-12m>-13\\ & \iff -3>9-12m>-13\\ & \iff -13<9-12m<-3\\ & \text{Ini berarti nilai}D\text{negatif, sehingga}\\ & \text{berakibat penyebut berupa}-x^2+3x-3m\\ & \text{adalah wilayah}definitnegatif\\ 4.& \text{Selanjutnya pemfaktoran pertidaksamaan}\\ & \bullet \text{semula}\\ & {\displaystyle \frac{x(x+4)}{\underset{definitnegatif}{\underbrace{-x^2+3x-3m}}}>0\iff \frac{x(x+4)}{-}>0}\\ & \bullet \text{akan berubah menjadi}\\ & x(x+4)<0\\ & \text{pembuat nol-nya adalah}:x(x+4)=0\\ & \text{maka}x=-4\text{atau}x=0,\text{sehingga}\\ & \text{interval nilai}x-\text{nya}:-4<x<0\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}1<a<2,\text{maka semua nilai}\\ & x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{-x^2+2ax-6}{x^2+3x}\le 0\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x<-3\text{atau}x>0\\ \text{b}.& x<-3\text{atau}x\ge -2\\ \text{c}.& x\le -2\text{atau}x\ge 2\\ \text{d}.& -3<x<0\\ \text{e}.& -2\le x<0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}1.& \text{Diketahui bahwa}:{\displaystyle \frac{-x^2+2ax-6}{x^2+3x}\le 0}\\ & \text{untuk membedakan}a\text{pada persamaan}\\ & \text{kuadrat dengan}a\text{di atas, selanjutnya}\\ & \text{kita menuliskan}a\text{di atas dengan}:m\\ & \text{karena}1<a<2\text{diubah}:1<m<2\\ & \text{Perhatikanlah pembilang yang}\\ & \text{mengandung bilangan}m\text{yang terletak}\\ & \text{pada interval}:1<m<2.\\ 2.& \text{Kita cek kondisi pembilangnya dengan}\\ & \text{menentukan}Diskriminan(D)-\text{nya}\\ & \text{yaitu}:\\ & a{x}^2+bx+c\{\begin{array}{l}a>0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit positif}\\ a<0\mathrm{\&}D=b^2-4ac<0\\ \Rightarrow \text{definit negatif}\end{array}\\ & \text{Karena pebilangnya}:-x^2+2mx-6,\\ & \text{dengan}a=-1,b=2m,\mathrm{\&}c=-6,\text{maka}\\ & D=(2m{)}^2-4(-1)(-6)=4m^2-24\\ 3.& \text{Karena nilai}m\text{berada pada}1<m<2\\ & \text{maka}\\ & 1<m<2\\ & \iff 1^2<m^2<2^2\iff 1<m^2<4\\ & \iff 4<4m^2<16\\ & \iff 4-24<4m^2-24<16-24\\ & \iff -20<4m^2-24<-8\\ & \text{Ini berarti nilai}D\text{negatif, sehingga}\\ & \text{berakibat pembilangnya berupa}-x^2+2mx-6\\ & \text{adalah wilayah}definitnegatif\\ 4.& \text{Selanjutnya pemfaktoran pertidaksamaan}\\ & \bullet \text{semula}\\ & {\displaystyle \frac{\stackrel{definitnegatif}{\overbrace{-x^2+2mx-6}}}{x(x+3)}\le 0\iff \frac{-}{x(x+3)}\le 0}\\ & \bullet \text{akan berubah menjadi}\\ & x(x+3)>0\\ & \text{pembuat nol-nya adalah}:x(x+3)=0\\ & \text{maka}x=-3\text{atau}x=0,\text{sehingga}\\ & \text{interval nilai}x-\text{nya}:x<-3\text{atau}x>0\end{array}\end{array}$

$\begin{array}{l}\\ 20.& (\text{KSM 2019})\\ & \text{Banyaknya selesaian real dari persamaan}\\ & 2x^2-7x+6=15\lfloor {\displaystyle \frac{1}{x}}\rfloor \lfloor x\rfloor \\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 3\\ \text{d}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{tidak ada}}\\ & \end{array}$

$.\begin{array}{rl}\text{Solu}& \text{si jawaban merujuk dari blog Pak Anang}\\ \text{And}& \text{a bisa mengunjunginya di blognya beliau}\end{array}$

$.\begin{array}{rl}\text{di}& \end{array}$  sini (Soal dan Pembahasan KSM 2019 oleh Pak Anang)

Contoh Soal 3 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 11.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x-5}\le x\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 1<x<0\\ \\ \text{b}.& x<1\text{atau}x\ge 5\\ \\ \text{c}.& {\displaystyle \frac{5}{6}\le x\le 1\text{atau}x\ge 5}\\ \\ \text{d}.& {\displaystyle \frac{5}{6}\le x<1\text{atau}5<x<6}\\ \\ \text{e}.& x\ge 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{6x-5}& \le x\\ 1.\text{Kuadrat}& \text{kan}\\ 6x-5& \le x^2\\ -x^2+6x-5& \le 0\\ x^2-6x+5& \ge 0\\ (x-1)(x-5)& \ge 0\\ x\le 1\text{atau}& x\ge 5\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x-5& \ge 0\\ 6x& \ge 5\\ x& \ge {\displaystyle \frac{5}{6}}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{6x+6}>6\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>7\\ \text{b}.& x\ge 7\\ \text{c}.& x<7\\ \text{d}.& x>1\\ \text{e}.& x\ge 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\sqrt{6x+6}& >6\\ 1.\text{Kuadrat}& \text{kan}\\ 6x+6& >36\\ x+1& >6\\ x& >7\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 6x+6& \ge 0\\ 6x& \ge 6\\ x& \ge {\displaystyle \frac{6}{6}}\\ x& \ge 1\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Penyelesaian pertidaksamaan}\\ & x+2>{\displaystyle \sqrt{10-x^2}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\le x\le \sqrt{10}\\ \text{b}.& 1<x\le \sqrt{10}\\ \text{c}.& -3<x\le \sqrt{10}\\ \text{d}.& -\sqrt{10}\le x\le \sqrt{10}\\ \text{e}.& x<-3\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x+2& >{\displaystyle \sqrt{10-x^2}}\\ 1.\text{Kuadrat}& \text{kan}\\ x^2+4x+4& >10-x^2\\ 2x^2+4x& +4-10>0\\ 2x^2+4x-& 6>0\\ x^2+2x-& 3>0\\ (x+3)& (x-1)>0\\ x<-3& \text{atau}x>1\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 10-x^2& \ge 0\\ x^2-10& \le 0\\ (x-\sqrt{10})& (x+\sqrt{10})\le 0\\ -\sqrt{10}\le x& \le \sqrt{10}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \sqrt{3x+7}\ge x-1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -1<x<6\\ \text{b}.& -1\le x<6\\ \text{c}.& x\ge -{\displaystyle \frac{7}{3}}\\ \text{d}.& -{\displaystyle \frac{7}{3}\le x\le 6}\\ \text{e}.& -{\displaystyle \frac{7}{3}\le x\le 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{3x+7}& \ge x-1\\ 1.\text{Kuadrat}& \text{kan}\\ 3x+7& \ge x^2-2x+1\\ -x^2+3x& +2x+7-1\ge 0\\ -x^2+5x+& 6\ge 0\\ x^2-5x-& 6\le 0\\ (x+1)& (x-6)\le 0\\ -1\le x& \le 6\\ 2.\text{Di bawa}& \text{h tanda akar}\ge 0\\ 3x+7& \ge 0\\ 3x& \ge -7\\ x& \ge -{\displaystyle \frac{7}{3}}\end{array}\end{array}$

 $\begin{array}{l}\\ 15.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{2x-8}<\sqrt{x+5}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& x\ge -5\\ \text{b}.& x<-13\text{atau}x\ge -4\\ \text{c}.& x<13\\ \text{d}.& 4\le x<13\\ \text{e}.& -5\le x\le 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{2x-8}& <\sqrt{x+5}\\ (1)\text{kuadratkan}& \\ 2x-8& <x+5\\ x& <13\\ (2)2x-8\ge 0& \\ x& \ge 4\\ (3)x+5\ge 0& \\ x& \ge 5\\ \text{perhatikan}& \text{lah garis bilangannya berikut}\\ 1& \begin{array}{c}\\ & & & & & & & & & & \\ \text{cline}1-7& & & & & & & & & & \text{X}\\ & & & & & & 13& & & & \end{array}\\ \\ 2& \begin{array}{c}\\ & & & & & & & & & & \\ \text{cline}4-11& & & & & & & & & & \text{X}\\ & & & 4& & & & & & & \end{array}\\ \\ 3& \begin{array}{c}\\ & & & & & & & & & & \\ \text{cline}6-11& & & & & & & & & & \text{X}\\ & & & & & 5& & & & & \end{array}\end{array}\end{array}$

Contoh Soal 2 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

 $\begin{array}{l}\\ 6.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{2x+6}{x-4}\le 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -10<x<4\\ \text{b}.& -10\le x<4\\ \text{c}.& -4<x\le 10\\ \text{d}.& x\le -10\text{atau}x\ge 4\\ \text{e}.& x<-10\text{atau}x\ge 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2x+6}{x-4}}& \le 1\\ {\displaystyle \frac{2x+6}{x-4}-1}& \le 0\\ {\displaystyle \frac{2x+6-(x-4)}{x-4}}& \le 0\\ {\displaystyle \frac{x+10}{x-4}}& \le 0\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai}x\text{yang memenuhi}{\displaystyle \frac{x^2-x}{x+3}\ge 1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-3\text{atau}-1\le x\le 3\\ \text{b}.& -3<x\le -1\text{atau}x\ge 3\\ \text{c}.& -3\le x\le 3\\ \text{d}.& -3\le x\le -1\text{atau}x\ge 3\\ \text{e}.& -3\le x\le -1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-x}{x+3}}& \ge 1\\ {\displaystyle \frac{x^2-x}{x+3}}& -1\ge 0\\ {\displaystyle \frac{x^2-x-(x+3)}{x+3}}& \ge 0\\ {\displaystyle \frac{x^2-2x-3}{x+3}}& \ge 0\\ {\displaystyle \frac{(x-3)(x+1)}{x+3}}& \ge 0\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai}x\text{yang memenuhi}\\ & x+2+{\displaystyle \frac{1}{x+4}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x<-4\text{atau}x\ge -3\\ \text{b}.& x<-4\text{atau}x>-3\\ \text{c}.& -4\le x\le -3\\ \text{d}.& x>-4\\ \text{e}.& -4\le x\le -3\text{atau}x>-3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}x+2+{\displaystyle \frac{1}{x+4}}& >0\\ {\displaystyle \frac{(x+2)(x+4)+1}{(x+4)}}& >0\\ {\displaystyle \frac{x^2+6x+8+1}{x+4}}& >0\\ {\displaystyle \frac{x^2+6x+9}{x+4}}& >0\\ {\displaystyle \frac{(x+3{)}^2}{(x+4)}}& >0\\ x& >-4\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Nilai}x\text{yang memenuhi}\\ & x+3<{\displaystyle \frac{x^2+6x+11}{x}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|x<-3{\displaystyle \frac{2}{3}\text{atau}x>0,x\in \mathbb{R}}\}\\ \text{b}.& \{x|0\le x\le 11,x\in \mathbb{R}\}\\ \text{c}.& \{x|x<-11\text{atau}x>0,x\in \mathbb{R}\}\\ \text{d}.& \{x|x<0\text{atau}x>11,x\in \mathbb{R}\}\\ \text{e}.& \{x|x\le 11\text{atau}x>0,x\in \mathbb{R}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}x+3<{\displaystyle \frac{x^2+6x+11}{x}}& \\ x+3-{\displaystyle \frac{x^2+6x+11}{x}}& <0\\ {\displaystyle \frac{x(x+3)-(x^2+6x+11)}{x}}& <0\\ {\displaystyle \frac{x^2+3x-x^2-6x-11}{x}}& <0\\ {\displaystyle \frac{-3x-11}{x}}& <0\\ {\displaystyle \frac{3x+11}{x}}& >0\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai}x\text{berikut yang tidak memenuhi}\\ & {\displaystyle \frac{x-3}{x^2+2x+1}\le 0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& -1\\ \text{c}.& 1\\ \text{d}.& 2\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{x-3}{x^2+2x+1}\le 0}& \\ {\displaystyle \frac{(x-3)}{(x+1{)}^2}\le 0}& \\ \text{Pembuat nol}& \\ \{\begin{array}{ll}x& =3,\text{boleh digunakan}\\ x& =-1,\text{textrm{tetapi}}x\ne -1,\end{array}& \\ \text{sehingga}-1\text{tidak digunakan}& \\ \begin{array}{c}\\ & & & & & & & & & & \\ & -& -& -& -& -& & +& +& & \\ & & -1& & & & 3& & & & \\ & & & & & & & & & & \end{array}& \end{array}\end{array}$

Contoh Soal 1 Pertidaksamaan Rasional dan Irasional Satu Variabel (Kelas X Matematika Wajib)

$\begin{array}{l}\\ 1.& \text{Diketahui pertidaksamaan}{\displaystyle \frac{x+10}{x-9}\le 0}\\ & \text{dan diberikan beberapa nilai berikut}\\ & (\text{i})x=-6(\text{iii})x=-14\\ & (\text{ii})x=-10(\text{iv})x=-18\\ & \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & \text{di atas adalah ditunjukkan oleh}....\\ & \begin{array}{l}\\ \text{a}.& (\text{i})\text{dan}(\text{ii})\\ \text{b}.& (\text{i})\text{dan}(\text{iii})\\ \text{c}.& (\text{ii})\text{dan}(\text{iii})\\ \text{d}.& (\text{ii})\text{dan}(\text{iv})\\ \text{e}.& (\text{iii})\text{dan}(\text{iv})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x+10}{x-9}}& \le 0\\ \text{HP}=& \{x|-10\le x<9,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{6}{x-3}<\frac{8}{x-2}\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 2\le x<6\\ \text{b}.& 2\le x<3\\ \text{c}.& 2<x<3\text{atau}x>6\\ \text{d}.& x<3\text{atau}3<x<6\\ \text{e}.& x<2\text{atau}x>3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{6}{x-3}<\frac{8}{x-2}}& \\ {\displaystyle \frac{6}{x-3}-\frac{8}{x-2}}& <0\\ {\displaystyle \frac{6(x-2)-8(x-3)}{(x-3)(x-2)}}& <0\\ {\displaystyle \frac{6x-8x-12+24}{(x-2)(x-3)}}& <0\\ {\displaystyle \frac{-2x+12}{(x-2)(x-3)}}& <0\\ {\displaystyle \frac{2x-12}{(x-2)(x-3)}}& >0\\ {\displaystyle \frac{2(x-6)}{(x-2)(x-3)}}& >0\\ \text{HP}=& \{x|2<x<3\text{atau}x>6,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Penyelesaian pertidaksamaan}\\ & {\displaystyle \frac{x^2-81}{x^2}\ge 0\text{ adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x\le -9\text{atau}x\ge 9\\ \text{b}.& -9\le x<0\text{atau}x\ge 9\\ \text{c}.& -9\le x<0\text{atau}0<x\le 9\\ \text{d}.& -9<x\le 9\\ \text{e}.& x\in \mathbb{R}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-81}{x^2}}& \ge 0\\ {\displaystyle \frac{(x+9)(x-9)}{x^2}}& \ge 0\\ \text{HP}=& \{x|x\le -9\text{atau}x\ge 9,x\in \mathbb{R}\}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2-4}{x+2}>0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& x>2\\ \text{b}.& -2\le x<2\\ \text{c}.& x<-2\text{atau}x>2\\ \text{d}.& x<-2\text{atau}-2<x<2\\ \text{e}.& x\ge -2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2-4}{x+2}}& >0\\ {\displaystyle \frac{(x+2)(x-2)}{(x+2)}}& >0\\ (x-2)& >0\\ x& >2\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Nilai}x\text{yang memenuhi pertidaksamaan}\\ & {\displaystyle \frac{x^2+x-30}{2x^2+13x-45}<0\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& \{x|-9<x<5,x\in \mathbb{R}\}\\ \text{b}.& \{x|-6<x<5,x\in \mathbb{R}\}\\ \text{c}.& \{x|-9<x<-6\text{atau}x<5,x\in \mathbb{R}\}\\ \text{d}.& \{x|-9<x<-6\text{atau}{\displaystyle \frac{5}{2}<x<5,x\in \mathbb{R}}\}\\ \text{e}.& \{x|x<-9\text{atau}-6<x<{\displaystyle \frac{5}{2}\text{atau}x<5,x\in \mathbb{R}}\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{x^2+x-30}{2x^2+13x-45}}& <0\\ {\displaystyle \frac{(x+6)(x-5)}{(x+9)(2x-5)}}& <0\\ \text{Cukup jelas}& \end{array}\end{array}$

Pertidaksamaan Rasional dan Irasional Satu Variabel

 A. Bentuk Umum Pertidaksamaan Rasional

$\begin{array}{rl}& \mathbf{\text{Pertidak samaan Rasional}}\\ & \\ & \{\begin{array}{ll}\mathbf{\text{A}}& \begin{array}{rl}& \\ \{\begin{array}{ll}{\displaystyle \frac{f(x)}{g(x)}}& <0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \le 0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& >0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \ge 0\end{array}& \begin{array}{cc}\mathbf{\text{misalnya}}& \{\begin{array}{ll}{\displaystyle \frac{x-1}{x+3}}& <0\\ \\ {\displaystyle \frac{x-1}{x+3}}& \le 0\\ \\ {\displaystyle \frac{x-1}{x+3}}& >0\\ \\ {\displaystyle \frac{x-1}{x+3}}& \ge 0\end{array}\end{array}\\ & \end{array}\\ \mathbf{\text{B}}& \begin{array}{rl}& \\ \{\begin{array}{ll}{\displaystyle \frac{f(x)}{g(x)}}& <0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \le 0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& >0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \ge 0\end{array}& \begin{array}{cc}\mathbf{\text{misalnya}}& \{\begin{array}{ll}{\displaystyle \frac{x^2-4}{x+6}}& <0\\ \\ {\displaystyle \frac{x^2-4}{x+6}}& \le 0\\ \\ {\displaystyle \frac{x^2-4}{x+6}}& >0\\ \\ {\displaystyle \frac{x^2-4}{x+6}}& \ge 0\end{array}\end{array}\\ & \end{array}\\ \mathbf{\text{C}}& \begin{array}{rl}& \\ \{\begin{array}{ll}{\displaystyle \frac{f(x)}{g(x)}}& <0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \le 0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& >0\\ \\ {\displaystyle \frac{f(x)}{g(x)}}& \ge 0\end{array}& \begin{array}{cc}\mathbf{\text{misalnya}}& \{\begin{array}{ll}{\displaystyle \frac{x^2+2x-3}{x^2-4}}& <0\\ \\ {\displaystyle \frac{x^2+2x-3}{x^2-4}}& \le 0\\ \\ {\displaystyle \frac{x^2+2x-3}{x^2-4}}& >0\\ \\ {\displaystyle \frac{x^2+2x-3}{x^2-4}}& \ge 0\end{array}\end{array}\\ & \end{array}\end{array}\\ & \end{array}$

B. Menyelesaikan Pertidaksamaan Rasional

Misal pada bentuk A di atas, maka penyelesaiannya adalah

Jika nantinya berupa pertidaksamaan yang mengandung kuadrat, maka gunakan materi sebelumnya yang berkaitan dengan pertidaksamaan yang mengandung bentuk kuadrat.

Lihat di sini

C. Bentuk Umum Pertidaksamaan Irasional

Bentuk pertama

$$\begin{gathered} \mathbf{\text{Bentuk Umum}}1: \\ \sqrt{f(x)}\cdots A\{\begin{array}{l}\sqrt{f(x)}<A\\ \\ \sqrt{f(x)}\le A\\ \\ \sqrt{f(x)}>A\\ \\ \sqrt{f(x)}\ge A\end{array} \end{gathered}$$

Bentuk kedua

$$\begin{gathered} \mathbf{\text{Bentuk Umum}}2: \\ \sqrt{f(x)}\cdots \sqrt{g(x)}\{\begin{array}{l}\sqrt{f(x)}<\sqrt{g(x)}\\ \\ \sqrt{f(x)}\le \sqrt{g(x)}\\ \\ \sqrt{f(x)}>\sqrt{g(x)}\\ \\ \sqrt{f(x)}\ge \sqrt{g(x)}\end{array} \end{gathered}$$

D. Penyelesaian pada pertidaksamaan irasional

1. Kuadartakan masing-masing ruas
2. Dibawah tanda akar (numerus) haruslah $\ge 0$
3. Himpunan penyelesaian berupa irisan dari penyelesaian yang didapatkan