PAS MATEMATIKA BLOG

CONTOH SOAL 5 BARISAN

$\begin{aligned}9.\quad&\textrm{Bilangan}\quad X=\underset{50}{\underbrace{99+999+9999+\cdots +999\cdots 999}}\:. \: \textrm{Berapa banyak}\\ &\textrm{digit 1 muncul pada bilangan}\quad X\\ &\end{aligned}$

CONTOH SOAL 4 BARISAN

$\begin{aligned}7.\quad&\textrm{Perhatikanlah hubungan berikut}\\ &\qquad\qquad\begin{matrix} 6^{\displaystyle 2}-5^{\displaystyle 2}=11 \\ 56^{\displaystyle 2}-45^{\displaystyle 2}=1111 \\ 556^{\displaystyle 2}-445^{\displaystyle 2}=111111 \\ 5556^{\displaystyle 2}-4445^{\displaystyle 2}=11111111 \\ \vdots \quad\qquad\qquad \vdots \end{matrix}\\ &\textrm{Tulislah pola yang ada pada hubungan di atas dan buktikan}\\& \end{aligned}$.

Bukti: $\begin{aligned} &\textrm{Perhatikan bahwa}:\\ &\textbf{Pola suku pertama}:\:\textrm{ angka 6 yang didahului oleh}\quad n\quad \textrm{buah}\\ &\textrm{angka 5. Polanya dapat dituliskan sebagai}:\\ &A_{\displaystyle n}=\underset{\textrm{n buah 5}}{\underbrace{55\cdots 56}}=\displaystyle \frac{5.\left( 10^{\displaystyle n+1}-1 \right)}{9}+1=\displaystyle \frac{5.10^{\displaystyle n+1}+4}{9}\\ &\textbf{Pola suku kedua}:\:\textrm{ angka 5 yang didahului oleh}\quad n\quad \textrm{buah}\\ &\textrm{angka 4. Polanya dapat dituliskan sebagai}:\\ &B_{\displaystyle n}=\underset{\textrm{n buah 4}}{\underbrace{44\cdots 45}}=\displaystyle \frac{4.\left( 10^{\displaystyle n+1}-1 \right)}{9}+1=\displaystyle \frac{4.10^{\displaystyle n+1}+5}{9}\\ &\textbf{Hasil(ruas kanan)}\\ &\underset{\textrm{2n+2 buah 1}}{\underbrace{11\cdots 11}}=\displaystyle \frac{10^{\displaystyle 2n+2}-1}{9}\\ &\textbf{Pola umum}\\ &\left( \underset{\textrm{n }}{\underbrace{55\cdots 56}} \right)^{\displaystyle 2}-\left( \underset{\textrm{n}}{\underbrace{44\cdots 45}} \right)^{\displaystyle 2}=\underset{\textrm{2n+2}}{\underbrace{11\cdots 11}}\\ &\textrm{Ingat bentuk selisih kuadrat, yaitu}:a^{2}-b^{2}=(a+b)(a-b)\\ &A_{\displaystyle n}^{\displaystyle 2}-B_{\displaystyle n}^{\displaystyle 2}=\left( A_{\displaystyle n}+B_{\displaystyle n} \right)\left( A_{\displaystyle n}-B_{\displaystyle n} \right)\\ &=\left( \displaystyle \frac{5.10^{\displaystyle n+1}+4}{9}+\displaystyle \frac{4.10^{\displaystyle n+1}+5}{9} \right)\left( \displaystyle \frac{5.10^{\displaystyle n+1}+4}{9}-\displaystyle \frac{4.10^{\displaystyle n+1}+5}{9} \right)\\&=\left( \frac{9.10^{\displaystyle n+1}+9}{9} \right)\left( \displaystyle \frac{10^{\displaystyle n+1}-1}{9} \right)\\ &=\left( 10^{\displaystyle n+1}+1 \right)\left( \displaystyle \frac{10^{\displaystyle n+1}-1}{9} \right)\\ &=\displaystyle \frac{\left( 10^{\displaystyle n+1} \right)^{\displaystyle 2}-1}{9}\\ &=\displaystyle \frac{\left( 10^{\displaystyle 2n+2} \right)-1}{9}\quad \textrm{adalah deretan angka 1 sebanyak}\quad 2n+2\\ &=\underset{2n+2}{\underbrace{11\cdots 11}}\qquad \textbf{Terbukti}\end{aligned}\\$.

$\begin{aligned}8.\quad&\textrm{Barisan}\:\: \left\{ a_{\displaystyle n} \right\}_{n\ge 1}\:\: \textrm{didefinisikan dengan}\:\: a_{\displaystyle 1}=\displaystyle \frac{1}{2},\: a_{\displaystyle k+1}=a_{\displaystyle k}^{\displaystyle 2}+a_{\displaystyle k}\\ &\textrm{untuk semua}\:\: k\ge 1.\:\: \textrm{Tentukan bilangan bulat terbesar yang kurang}\\ &\textrm{dari atau sama dengan}\: \displaystyle \frac{1}{a_{\displaystyle 1}+1}+\frac{1}{a_{\displaystyle 2}+1}+\cdots +\frac{1}{a_{\displaystyle 2026}+1}\\\end{aligned}$

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$\begin{aligned}&\textrm{Diberikan barisan}:a_{\displaystyle 1},a_{\displaystyle 2},a_{\displaystyle 3},\cdots \:\: \textrm{dengan}\: a_{\displaystyle 1}=\displaystyle \frac{1}{2},a_{\displaystyle k+1}=a_{\displaystyle k}^{\displaystyle 2}+a_{\displaystyle k}\\ &a_{\displaystyle k+1}=a_{\displaystyle k}\left( a_{\displaystyle k}+1 \right)\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle k+1}}=\displaystyle \frac{1}{a_{\displaystyle k}\left( a_{\displaystyle k}+1 \right)}\\ &\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle k+1}}=\displaystyle \frac{1}{a_{\displaystyle k}}-\displaystyle \frac{1}{a_{\displaystyle k}+1}\Leftrightarrow \displaystyle \frac{1}{a_{\displaystyle k}+1}=\displaystyle \frac{1}{a_{\displaystyle k}}-\displaystyle \frac{1}{a_{\displaystyle k+1}}\\ &\textrm{Selanjutnya kembali ke deret pada soal}\\ &S_{\displaystyle n}=\displaystyle \sum_{k=1}^{n}\displaystyle \frac{1}{a_{\displaystyle k}+1}=\left( \displaystyle \frac{1}{a_{\displaystyle 1}}-\frac{1}{a_{\displaystyle 2}} \right)+\left( \displaystyle \frac{1}{a_{\displaystyle 2}}-\frac{1}{a_{\displaystyle 3}} \right)+\cdots +\left( \displaystyle \frac{1}{a_{\displaystyle n}}-\frac{1}{a_{\displaystyle n+1}} \right)\\ &\:\:\:\,\quad\quad\quad\quad\quad\quad\quad=\displaystyle \frac{1}{a_{\displaystyle 1}}-\displaystyle \frac{1}{a_{\displaystyle n+1}}=\displaystyle \frac{1}{\left( \displaystyle \frac{1}{2} \right)}-\displaystyle \frac{1}{a_{\displaystyle n+1}}=2-\displaystyle \frac{1}{a_{\displaystyle n+1}}\\ &S_{\displaystyle 2026}=\displaystyle \sum_{k=1}^{n}\displaystyle \frac{1}{a_{\displaystyle 2026}+1}=2-\displaystyle \frac{1}{a_{\displaystyle 2027}}\\ &\qquad\qquad\qquad\qquad\qquad\qquad(\textrm{dengan}\quad a_{\displaystyle 2027}\gt 1\Rightarrow 1\lt \displaystyle \frac{1}{a_{\displaystyle 2027}}\lt 2)\\ &\textrm{Jadi, bilangan bulat terbesar yang kurang dari atau sama dengan}\\& S_{\displaystyle 2026}=\left\lfloor 2-\displaystyle \frac{1}{a_{\displaystyle 2027}} \right\rfloor =1 \end{aligned}$

CONTOH SOAL 3 BARISAN

$\begin{aligned}5.\quad&(\textbf{KSM Matematika MA Tk. Kab/kota Th.2013})\\ &\textrm{Hasil penjumlahan dari tak hingga suku berbentuk}\\ &\displaystyle \frac{1}{5}+\frac{2}{5^{\displaystyle 2}}+\frac{1}{5^{\displaystyle 3}}+\frac{2}{5^{\displaystyle 4}}+\frac{1}{5^{\displaystyle 5}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\quad \textrm{adalah}\: ....\\\\ &\text{a}.\quad \displaystyle \frac{25}{24}\qquad\qquad\qquad\text{d}.\quad \displaystyle \frac{1}{4}\\ &\text{b}.\quad \displaystyle \frac{24}{25}\qquad\qquad\quad\quad\text{e}.\quad \displaystyle \frac{1}{12}\\ &\text{c}.\quad \displaystyle \frac{7}{24}\\\end{aligned}$

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$\begin{aligned}&\textrm{Perhatikan bahwa deret jumlah semua sukunya}:\\ &\displaystyle \frac{1}{5}+\frac{2}{5^{\displaystyle 2}}+\frac{1}{5^{\displaystyle 3}}+\frac{2}{5^{\displaystyle 4}}+\frac{1}{5^{\displaystyle 5}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\\ &\textrm{Selanjutnya deret kita partisi menjadi dua bagian, yaitu}:\\ &\textbf{Deret pertama untuk suku ganjil}\\ &S_{\displaystyle 1}=\displaystyle \frac{1}{5}+\frac{1}{5^{\displaystyle 3}}+\frac{1}{5^{\displaystyle 5}}\:+\:...\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=a=\displaystyle \frac{1}{5} \\\textrm{rasio}&=r=\displaystyle \frac{1}{25}\\ S_{\displaystyle \infty }&=\displaystyle \frac{a}{1-r}=\frac{\displaystyle \frac{1}{5}}{1-\displaystyle \frac{1}{25}}=\displaystyle \frac{5}{24} \end{cases}\\ &\textbf{Deret kedua untuk suku genap}\\ &S_{\displaystyle 2}=\frac{2}{5^{\displaystyle 2}}+\frac{2}{5^{\displaystyle 4}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=a=\displaystyle \frac{2}{5^{\displaystyle 2}}=\frac{2}{25} \\\textrm{rasio}&=r=\displaystyle \frac{1}{25}\\ S_{\displaystyle \infty }&=\displaystyle \frac{a}{1-r}=\frac{\displaystyle \frac{2}{25}}{1-\displaystyle \frac{1}{25}}=\displaystyle \frac{2}{24} \end{cases}\\ &\textbf{Jumlah totalnya adalah}:\\ &S_{\textrm{total}}=S_{\displaystyle 1}+S_{\displaystyle 2}=\displaystyle \frac{5}{24}+\frac{2}{24}=\displaystyle \frac{7}{24}\end{aligned}$

$\begin{aligned}6.\quad&(\textbf{KSM Matematika MA Tk. Kab/kota Th.2013})\\ &\textrm{nilai}\quad n\quad \textrm{terkecil yang memenuhi}\:\: \frac{1}{2^{\displaystyle n}}\lt 0,001\:\: \textrm{adalah}\: ....\\ &\text{a}.\quad \displaystyle 9\qquad\qquad\qquad\text{d}.\quad \displaystyle 522\\ &\text{b}.\quad \displaystyle 10\:\:\quad\quad\quad\quad\quad\text{e}.\quad \displaystyle 501\\ &\text{c}.\quad \displaystyle 11\\ \end{aligned}$

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$\begin{aligned}&\textbf{Jawab: b}\\ &\textrm{Perhatikan bahwa}:\\ &\frac{1}{2^{\displaystyle n}}\lt 0,001\Leftrightarrow \displaystyle \frac{1}{2^{\displaystyle n}}\lt \displaystyle \frac{1}{1000}\Leftrightarrow 2^{\displaystyle n}\gt 1000\\ &\textrm{Selanjutnya cukup kita uji untuk}\\&\begin{cases} n=9&\Rightarrow 2^{\displaystyle 9}=512\ngtr 1000 \\n=10&\Rightarrow 2^{\displaystyle 10}=1024\gt 1000\\ n=11&\Rightarrow 2^{\displaystyle 11}=2048\gt 1000\: (\textrm{bukan yang terkecil}) \end{cases}\\ &\textbf{Jadi,}\:\: n\:\: \textrm{terkecilnya adalah}=10\end{aligned}$

PERSIAPAN MENJELANG TKA MATEMATIKA SMA-MA 2026

CBT TKA MA Futuhiyah Jeketro

TKA Matematika Wajib Class XII

90:00

Soal No. 1

Memuat soal...

Hasil Ujian CBT TKA Matematika Wajib

Nama: - | ID: -

NILAI AKHIR

Skala 0 - 100

Jawaban Benar

Jawaban Salah

Tidak Dijawab

Pembahasan dan Analisis Soal

CONTOH SOAL 2 BARISAN

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$\begin{aligned}&\textbf{Jawab: b}\\&\textrm{Suku-suku barisan geometri adalah}:\: U_{\displaystyle n}=a.r^{\displaystyle n-1}\\ &\textrm{Untuk}\quad n\quad \textrm{genap, maka}:\\ &\textbf{Menentukan}\quad S_{\displaystyle n}\\ &S_{\displaystyle n}=U_{\displaystyle 1}+U_{\displaystyle 2}+U_{\displaystyle 3}+...+U_{\displaystyle n}=a+ar+ar^{\displaystyle 2}+...+ar^{n-1}\\ &\:\:\:\quad=a\left( \displaystyle \frac{1-r^{\displaystyle n}}{1-r} \right)\\ &\textbf{Menentukan}\quad \widehat{S}_{\displaystyle n}\\ &\widehat{S}_{\displaystyle n}=U_{\displaystyle 1}-U_{\displaystyle 2}+U_{\displaystyle 3}-U_{\displaystyle 4}+...+U_{\displaystyle n-1}-U_{\displaystyle n}\\ &\:\:\quad =a-ar+ar^{\displaystyle 2}+ar^{\displaystyle 4}+...+a.r^{n-1}\\ &\qquad \textrm{Karena}:\: (-r)^{\displaystyle n}=r^{\displaystyle n}\\ &\:\:\:\quad=a\left( \displaystyle \frac{1-r^{\displaystyle n}}{1+r} \right)\\ &\textbf{Menentukan rasio}\\ &\displaystyle \frac{\widehat{S}_{\displaystyle n}}{S_{\displaystyle n}}>3\Leftrightarrow \displaystyle \frac{a\left( \displaystyle \frac{1-r^{\displaystyle n}}{1+r} \right)}{a\displaystyle \left( \frac{1-r^{\displaystyle n}}{1-r} \right)}>3\Leftrightarrow \displaystyle \frac{1-r}{1+r}\gt 3\\ &\Leftrightarrow \displaystyle \frac{1-r}{1+r}-3\gt 0\\ &\Leftrightarrow \displaystyle \frac{1-r-3(r+1)}{1+r}\gt 0\\ &\Leftrightarrow \displaystyle \frac{-4r-2}{1+r}\gt 0\quad (\textrm{masing-masing ruas dikali dengan }-1)\\ &\Leftrightarrow \displaystyle \frac{4r+2}{1+r}\lt 0\\ &\textrm{Secara ketaksamaan wilayah}\quad r\quad \textrm{akan berada di}\\ &-1\lt r\lt \displaystyle -\frac{1}{2}\end{aligned}$

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$\begin{aligned}&\textbf{Jawab: c}\\&\textrm{Perhatikan bahwa suku-suku bagian numerusnya}\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=\log 5=\log 5.11^{\displaystyle 0} \\\textrm{suku ke}-2&= U_{\displaystyle 2}=\log 55=\log 5.11^{\displaystyle 1}\\ \textrm{suku ke}-3&= U_{\displaystyle 3}=\log 605=\log 5.11^{\displaystyle 2}\\ \textrm{suku ke}-4&= U_{\displaystyle 2}=\log 6655=\log 5.11^{\displaystyle 3}\\ ...\\ ...\\ \textrm{suku ke}-n&= U_{\displaystyle n}=\log \left( 5.11^{\displaystyle n-1} \right)=\log 5+(n-1)\log11 \end{cases}\\ &\textbf{Jumlah 50 suku pertamanya adalah}:\\ &S_{\displaystyle 50}=U_{\displaystyle 1}+U_{\displaystyle 2}+U_{\displaystyle 3}+U_{\displaystyle 4}+...+U_{\displaystyle 50}\\ &\qquad =\log\left( 5.11^{\displaystyle 0} \right)+\log\left( 5.11^{\displaystyle 1} \right)+\log\left( 5.11^{\displaystyle 2} \right)+...+\log\left( 5.11^{\displaystyle 49} \right)\\ &\qquad =\log\left( 5.11^{\displaystyle 0}\times 5.11^{\displaystyle 1}\times 5.11^{\displaystyle 2}\times 5.11^{\displaystyle 3}\times ...\times 5.11^{\displaystyle 49} \right)\\ &\qquad =\log\left( 5^{\displaystyle 50}.11^{\displaystyle 0+1+2+3+4+...+48+49} \right)\\ &\qquad =\log\left( 5^{\displaystyle 2\times 25}.11^{\displaystyle \frac{49\times 50}{2}} \right)\\ &\qquad =\log\left( 25^{\displaystyle 25}.11^{\displaystyle 1225} \right)\end{aligned}$

CONTOH SOAL 1 BARISAN

$\begin{aligned}1.\quad&(\textbf{LM UGM ke-25 Th 2014 Tk.SMA})\\ &\textrm{Diberikan dua buah barisan aritmetika}\quad 1,4,7,...\quad \textrm{dan}\\ &2,7,12,...\:. \: \textrm{Jika}\:\:S\:\: \textrm{merupakan himpunan yang terdiri}\\ &\textrm{dari gabungan 2014 suku pertama kedua barisan tersebut},\\ &\textrm{Banyak anggota himpuan}\:\: S\:\: \textrm{adalah}\: ....\\ &\text{a}.\quad 3625\qquad\qquad\qquad\qquad \text{d}.\quad 4015\\ &\text{b}.\quad 3875\qquad \text{c}.\quad 4014\qquad \text{e}.\quad 4028\\\\ &\textbf{Jawab}:\quad \text{a}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\textrm{Barisan pertama}:a_{1}=1,b_{1}=3,A_{n}=3n-2,A_{2014}=6040\\ &\textrm{Barisan kedua}:a_{2}=2,b_{2}=5,B_{m}=5m-3,B_{2014}=10067\\ &\textrm{Menentukan irisan kedua barisan}\: \left( A\cap B \right):\\ &\textrm{Barisan gabungan}:a_{12}=7,b_{12}=KPK(3,5)=15,\\ &\qquad\qquad C_{12=k=gabungan}=7+(k-1)15=15k-8\\ &\textrm{Maksimum suku}\le 6040\quad(\textrm{pilih yang terpendek})\\ &\qquad\qquad 15k-8\le 6040\Leftrightarrow 15k\le 6048\Leftrightarrow k\le 403,2\\ &\qquad\qquad \textrm{karena}\:\: k\:\: \textrm{bulat, maka nilai terbesar}\:k=403\\ &\textrm{Menghitung banyaknya anggota}\:\: S,\:\: \textrm{maka}:\\ &n(S)=n\left( A\cup B \right)=n(A)+n(B)-n\left( A\cap B \right)\\&\:\qquad=2014+2014-403=4028-403=3625\end{aligned} \end{aligned}$.

$\begin{aligned}2.\quad&(\textbf{LM UGM Ke-25 Th 2014 Tk.SMA})\\ &\textrm{Hitunglah nilai dari}\\ &5-\displaystyle \frac{10}{3}+\frac{20}{9}-\frac{40}{27}+\frac{80}{81}-...\\ &\text{a}.\quad 5\qquad\qquad\qquad\:\: \text{d}.\quad 2\\ &\text{b}.\quad 4\qquad \text{c}.\quad 3\qquad \text{e}.\quad 1\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui deret geometri tak hingga yang konvergen}\\ & \textrm{dengan}\\ &\begin{cases} a & =5 \quad \\ r & =\displaystyle \frac{U_{2}}{U_{1}}=\displaystyle \frac{\displaystyle -\frac{10}{3}}{5}=\displaystyle -\frac{2}{3} \end{cases}\\ &\textrm{Sehingga jumlahnya adalah:}\\ &S_{\displaystyle \infty }=\displaystyle \frac{a}{1-r}=\displaystyle \frac{5}{1-\left( -\displaystyle \frac{2}{3} \right)}=\displaystyle \frac{5}{\displaystyle \frac{5}{3}}=3 \end{aligned}$.

$\begin{aligned}3.\quad &(\textbf{LM UGM ke-19 Th.2007 Tk. SMA})\\ &\textrm{Diketahui barisan geometri dengan suku pertama}\quad a\\ &\textrm{dan rasio}\quad r.\:\: \textrm{Untuk sebarang}\quad n\quad \textrm{genap didefinisikan}\\ &\textrm{jumlahan}\quad S_{\displaystyle n}=U_{\displaystyle 1}+U_{\displaystyle 2}+U_{\displaystyle 3}+...+U_{\displaystyle n}\quad \textrm{dan}\\ &\widehat{S}_{\displaystyle n}=U_{\displaystyle 1}-U_{\displaystyle 2}+U_{\displaystyle 3}-U_{\displaystyle 4}+...+U_{\displaystyle n-1}-U_{\displaystyle n}.\quad \textrm{Nilai}\\ &r\quad \textrm{yang mungkin agar}\quad \displaystyle \frac{\widehat{S}_{\displaystyle n}}{S_{\displaystyle n}}>3?\\ &\text{a}.\quad -1< r<0\\ &\text{b}.\quad -1< r<\displaystyle -\frac{1}{2}\\ &\text{c}.\quad -1< r<\displaystyle -\frac{1}{3}\\ &\text{d}.\quad \displaystyle -\frac{1}{3}< r<0\\ &\text{e}.\quad \textrm{tergantung oleh nilai}\quad a\\\\ &\textbf{Jawab: b}\\ &\begin{aligned}&\textrm{Suku-suku barisan geometri adalah}:\: U_{\displaystyle n}=a.r^{\displaystyle n-1}\\ &\textrm{Untuk}\quad n\quad \textrm{genap, maka}:\\ &\textbf{Menentukan}\quad S_{\displaystyle n}\\ &S_{\displaystyle n}=U_{\displaystyle 1}+U_{\displaystyle 2}+U_{\displaystyle 3}+...+U_{\displaystyle n}=a+ar+ar^{\displaystyle 2}+...+ar^{n-1}\\ &\:\:\:\quad=a\left( \displaystyle \frac{1-r^{\displaystyle n}}{1-r} \right)\\ &\textbf{Menentukan}\quad \widehat{S}_{\displaystyle n}\\ &\widehat{S}_{\displaystyle n}=U_{\displaystyle 1}-U_{\displaystyle 2}+U_{\displaystyle 3}-U_{\displaystyle 4}+...+U_{\displaystyle n-1}-U_{\displaystyle n}\\ &\:\:\quad =a-ar+ar^{\displaystyle 2}+ar^{\displaystyle 4}+...+a.r^{n-1}\\ &\qquad \textrm{Karena}:\: (-r)^{\displaystyle n}=r^{\displaystyle n}\\ &\:\:\:\quad=a\left( \displaystyle \frac{1-r^{\displaystyle n}}{1+r} \right)\\ &\textbf{Menentukan rasio}\\ &\displaystyle \frac{\widehat{S}_{\displaystyle n}}{S_{\displaystyle n}}>3\Leftrightarrow \displaystyle \frac{a\left( \displaystyle \frac{1-r^{\displaystyle n}}{1+r} \right)}{a\displaystyle \left( \frac{1-r^{\displaystyle n}}{1-r} \right)}>3\Leftrightarrow \displaystyle \frac{1-r}{1+r}\gt 3\\ &\Leftrightarrow \displaystyle \frac{1-r}{1+r}-3\gt 0\\ &\Leftrightarrow \displaystyle \frac{1-r-3(r+1)}{1+r}\gt 0\\ &\Leftrightarrow \displaystyle \frac{-4r-2}{1+r}\gt 0\quad (\textrm{masing-masing ruas dikali dengan }-1)\\ &\Leftrightarrow \displaystyle \frac{4r+2}{1+r}\lt 0\\ &\textrm{Secara ketaksamaan wilayah}\quad r\quad \textrm{akan berada di}\\ &-1\lt r\lt \displaystyle -\frac{1}{2}\end{aligned} \end{aligned}$.

$\begin{aligned}4.\quad&(\textbf{OLIMPIADE SAINS PORSEMA Th.2012})\\ &\textrm{Jumlah 50 suku pertama dari deret berikut}\\ &\log 5+\log 55+ \log 605+\log 6655\:+\:...\quad \textrm{adalah}\: ....\\ &\text{a}.\quad \log \left( 55^{\displaystyle 1155} \right)\qquad\qquad\qquad\text{d}.\quad \log \left( 275^{\displaystyle 1150} \right)\\ &\text{b}.\quad \log \left( 5^{\displaystyle 25}.11^{\displaystyle 1225} \right)\qquad\quad\quad\text{e}.\quad 1150\log 5\\ &\text{c}.\quad \log \left( 25^{\displaystyle 25}.11^{\displaystyle 1225} \right)\\\\ &\textbf{Jawab: c}\\ &\begin{aligned}&\textrm{Perhatikan bahwa suku-suku bagian numerusnya}\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=\log 5=\log 5.11^{\displaystyle 0} \\\textrm{suku ke}-2&= U_{\displaystyle 2}=\log 55=\log 5.11^{\displaystyle 1}\\ \textrm{suku ke}-3&= U_{\displaystyle 3}=\log 605=\log 5.11^{\displaystyle 2}\\ \textrm{suku ke}-4&= U_{\displaystyle 2}=\log 6655=\log 5.11^{\displaystyle 3}\\ ...\\ ...\\ \textrm{suku ke}-n&= U_{\displaystyle n}=\log \left( 5.11^{\displaystyle n-1} \right)=\log 5+(n-1)\log11 \end{cases}\\ &\textbf{Jumlah 50 suku pertamanya adalah}:\\ &S_{\displaystyle 50}=U_{\displaystyle 1}+U_{\displaystyle 2}+U_{\displaystyle 3}+U_{\displaystyle 4}+...+U_{\displaystyle 50}\\ &\qquad =\log\left( 5.11^{\displaystyle 0} \right)+\log\left( 5.11^{\displaystyle 1} \right)+\log\left( 5.11^{\displaystyle 2} \right)+...+\log\left( 5.11^{\displaystyle 49} \right)\\ &\qquad =\log\left( 5.11^{\displaystyle 0}\times 5.11^{\displaystyle 1}\times 5.11^{\displaystyle 2}\times 5.11^{\displaystyle 3}\times ...\times 5.11^{\displaystyle 49} \right)\\ &\qquad =\log\left( 5^{\displaystyle 50}.11^{\displaystyle 0+1+2+3+4+...+48+49} \right)\\ &\qquad =\log\left( 5^{\displaystyle 2\times 25}.11^{\displaystyle \frac{49\times 50}{2}} \right)\\ &\qquad =\log\left( 25^{\displaystyle 25}.11^{\displaystyle 1225} \right)\end{aligned}\end{aligned}$.

BARISAN

Materi Barisan dan Deret

materi pola bilangan - induksi matematika
pola bilangan dan barisan serta deret aritmetika
pola bilangan dan barisan serta deret geometri
berikut contoh soal induksi matematika dan pola bilangan
contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5
materi barisan dan deret aritmetika (hitung)
lanjutan materi barisan dan deret geometri (ukur)
berikut contoh soalnya
contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5, contoh 6, contoh 7.
barisan dan deret aritmetika dan geometri sekaligus
contoh soal selingan (diselipkan soal untuk kompetisi) yang terkait barisan dan deret: contoh 1, contoh 2, contoh 3, contoh 4.

CONTOH 10-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

$\begin{aligned}46.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-1-x}=\:....\\ &\text{a}.\quad \displaystyle \frac{2}{3}\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 3\\ &\text{b}.\quad 1\qquad\qquad \text{c}.\quad 2\qquad\qquad \text{e}.\quad 4\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Ingat bentuk}\\ & \underset{x\rightarrow \infty }{\textrm{Lim}}\: \sqrt{\displaystyle ax^{\displaystyle 2}+bx+c}-\sqrt{\displaystyle ax^{\displaystyle 2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ &\textrm{Sehingga soal untuk di atas}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-1-x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-(1+x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1}}{\sqrt{\displaystyle x^{\displaystyle 2}+x}-\sqrt{\displaystyle x^{\displaystyle 2}+2x+1}}\\ &=\displaystyle \frac{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( \sqrt{\displaystyle x^{\displaystyle 2}-3x+1}-\sqrt{\displaystyle x^{\displaystyle 2}-x-1} \right)}{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( \sqrt{\displaystyle x^{\displaystyle 2}+x}-\sqrt{\displaystyle x^{\displaystyle 2}+2x+1} \right)}\\ &=\displaystyle \frac{\left( \displaystyle \frac{-3-(-1)}{2\sqrt{1}} \right)}{\left( \displaystyle \frac{1-2}{2\sqrt{1}} \right)}=\displaystyle \frac{-2}{-1}=2\\\end{aligned}$

$\begin{aligned}47.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x+1}+3^{\displaystyle x+1}}{2^{\displaystyle x-1}+3^{\displaystyle x-1}}=\:....\\ &\text{a}.\quad \displaystyle 1\:\:\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 9\\ &\text{b}.\quad 3\qquad\qquad \text{c}.\quad 4\qquad\qquad \text{e}.\quad 16\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x+1}+3^{\displaystyle x+1}}{2^{\displaystyle x-1}+3^{\displaystyle x-1}}=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2^{\displaystyle x}.2+3^{\displaystyle x}.3}{\displaystyle \frac{2^{\displaystyle x}}{2}+\displaystyle \frac{3^{\displaystyle x}}{3}}\times \frac{\displaystyle \frac{1}{3^{\displaystyle x}}}{\displaystyle \frac{1}{3^{\displaystyle x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+3}{\displaystyle \frac{1}{2}.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+\displaystyle \frac{1}{3}}\\ & \textrm{Perhatikan bahwa saat}\: x\longrightarrow \infty \:,\: \textrm{maka}\:\: \left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}\longrightarrow 0\\ &\textrm{Sehingga}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{2.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+3}{\displaystyle \frac{1}{2}.\left( \displaystyle \frac{2}{3} \right)^{\displaystyle x}+\displaystyle \frac{1}{3}}= \displaystyle \frac{2.\left( 0 \right)+3}{\displaystyle \frac{1}{2}.\left( 0 \right)+\displaystyle \frac{1}{3}}=\displaystyle \frac{3}{\displaystyle \frac{1}{3}}=9 \end{aligned}$.

$\begin{aligned}48.\quad &\textrm{Nilai}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left\lfloor x \right\rfloor}{x}=\:....\\\\ &\textrm{dengan}\:\left\lfloor x \right\rfloor= \textrm{bilangan bulat terbesar yang }\\ &\textrm{kurang dari atau sama dengan}\:\:\: x\\\\ &\text{a}.\quad \displaystyle \frac{1}{2}\qquad\qquad\qquad\qquad\qquad \text{d}.\quad 1\\ &\text{b}.\quad 2\qquad\qquad \text{c}.\quad 0\qquad\qquad \text{e}.\quad \textrm{tidak ada}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Berdasarkan sifat fungsi tangga, untuk setiap bilangan}\\ &\textrm{riil}\:\:\: x\:\:\: \textrm{berlaku}:\\ & x-1<\left\lfloor x \right\rfloor\le x\\ &\text{Untuk}\quad x>0,\:\: \textrm{bagilah seluruh ruas dengan}\:\:\: x\\ &\displaystyle \frac{x-1}{x}<\displaystyle \frac{\left\lfloor x \right\rfloor}{x}\le \displaystyle \frac{x}{x}\Leftrightarrow \left( 1-\displaystyle \frac{1}{x} \right)<\displaystyle \frac{\left\lfloor x \right\rfloor}{x}\le 1\\ &\textrm{Selanjutnya kita hitung nilai limit batas kiri dan kanan}\\ &\text{ketika}:x\longrightarrow \infty \:(\textrm{ingat ini bukan limit kiri dan kanan})\\ &\circ \quad \textrm{Batas kiri}\\ &\qquad \underset{x\rightarrow \infty }{\textrm{Lim}}\: \left( 1-\displaystyle \frac{1}{x} \right)=1-\displaystyle \frac{1}{\infty }=1-0=1\\ &\circ \quad \textrm{Batas kanan}\\ &\qquad \underset{x\rightarrow \infty }{\textrm{Lim}}\: 1=1\\ &\textrm{Berdasarkan teorema apit (Squeeze Theorem), karena}\\ &\textrm{batas kiri sama dengan batas kanan, maka nilai}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left\lfloor x \right\rfloor}{x}=1 \end{aligned} \end{aligned}$.

CONTOH 9-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

$\begin{array}{ll}\\ 41.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{18}{x\sin \displaystyle \frac{3}{x}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -54 \\ \textrm{b}.&\displaystyle -6\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{6}\\ \textrm{d}.&\displaystyle 6\\ \textrm{e}.&\displaystyle 54 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{18}{x\sin \displaystyle \frac{3}{x}}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18}{\displaystyle \frac{1}{u}\sin 3u}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{18u}{\sin 3u}\\ &=\displaystyle \frac{18}{3}\\ &=6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 42.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x\sin \displaystyle \frac{2}{x}}{2}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4 \\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&\displaystyle \displaystyle \frac{1}{2}\\ \textrm{d}.&\displaystyle 2\\ \textrm{e}.&\displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x\sin \displaystyle \frac{2}{x}}{2}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{4}{u}.\sin 2u}{2}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{4\sin 2u}{2u}\\ &=\displaystyle \frac{4\times 2}{2}\\ &=4 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 43.&\textrm{Nilai dari}\\ &\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\underset{x\rightarrow -\infty }{\textrm{Lim}}\: \: \displaystyle x\cos \frac{1}{x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(-x)}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (-x)\cos \frac{1}{(x)}\\ &=-\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle (x)\cos \frac{1}{(x)}\\ &\begin{cases} u & =\displaystyle \frac{1}{x} \\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{u}\cos u\\ &=-\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\cos u}{u}\\ &=-\displaystyle \frac{1}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 44.&\textrm{Asimtot tegak dari fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&x=2\: \: \textrm{dan}\: \: x=4 \\ \textrm{b}.&x=2\: \: \textrm{dan}\: \: x=3\\ \textrm{c}.&x=3\: \: \textrm{dan}\: \: x=4\\ \textrm{d}.&x=3\: \: \textrm{saja}\\ \textrm{e}.&x=2\: \: \textrm{saja} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ & \textrm{terjadi saat penyebut} =0.\\ &\textrm{Sehingga}\: \: x^{2}-5x+6=0\\ &\Leftrightarrow (x-2)(x-3)=0,\: \: \textrm{maka}\\ & x=2\: \: \textrm{atau}\: \: x=3\\ &\therefore \: \: \textrm{asimtot tegak fungsi}\\ &f(x)=\displaystyle \frac{x^{2}-6x-8}{x^{2}-5x+6}\\ &\textrm{adalah}\: \: x=2\: \: \textrm{dan}\: \: x=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Asimtot datar dari fungsi}\\ &g(x)=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&y=-3 \\ \textrm{b}.&y=-1\\ \textrm{c}.&\displaystyle \frac{1}{3}\\ \textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}\textrm{Asim}&\textrm{tot datar dari fungsi}\\ g(x)&=\displaystyle \frac{(2x-2)(3x-1)}{(1-2x)(x-2)}\: \: \textrm{untuk}\\ g(x)&=\displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\: \: \textrm{terjadi saat}\\ y&=\displaystyle \frac{6}{-2}=-3\\ &\textbf{atau dapat juga dicari}\: \textbf{dengan}\\ y&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{(6x^{2}-8x+2)}{(-2x^{2}+5x-2)}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{6-\displaystyle \frac{8}{x}+\frac{2}{x^{2}}}{-2+\displaystyle \frac{5}{x}-\frac{2}{x^{2}}}\\ &=\displaystyle \frac{6-0+0}{-2+0-0}\\ &=\displaystyle \frac{6}{-2}\\ &=-3 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

Astuti, A.N., Miyanto, Ngapiningsih. 2020. Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Yogyakarta: PT. PENERBIT INTAN PARIWARA
Kartini, Suprpto, Subandi, Setiyadi, U. 2005.Matematika Program Studi Ilmu ALam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
Noormandiri. 2017. Matematika Kelompok Peminatan Matematika dan Ilmu-Ilmu ALam untuk SMA/MA Kelas XII. Jakarta: ERLANGGA
Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
Tim. 2020. Modul Matematika (Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII). Tangerang: RAHMA GEMILANG.

CONTOH 8-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

$\begin{array}{l}\\ 36.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -10 \\ \textrm{b}.\quad \displaystyle -5\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \displaystyle 5\\ \textrm{e}.\quad \displaystyle 10 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( x-\sqrt{x^{2}-10x} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}}-\sqrt{x^{2}-10x} \right )\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )\\ &=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{0-(-10)}{2\sqrt{1}}\\ &=\frac{10}{2}\\ &=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \infty \\ \textrm{b}.&\displaystyle 5\\ \textrm{c}.&\displaystyle \sqrt{3}\\ \textrm{d}.&1\\ \textrm{e}.&\displaystyle 0 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 5\tan \displaystyle \frac{1}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: 5\tan \displaystyle u\\ &=5\tan 0\\ &=5.\infty \\ &=\infty \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle \frac{4}{x}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0\\ \textrm{b}.&\displaystyle \frac{1}{4}\\ \textrm{c}.&\displaystyle 4\\ \textrm{d}.&\displaystyle \frac{11}{4}\\ \textrm{e}.&\displaystyle 60 \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: 15x\tan \displaystyle \frac{4}{x}&=\cdots \\ &\begin{cases} u & =\displaystyle \frac{1}{x} \quad \textrm{maka}\quad x=\displaystyle \frac{1}{u}\\ x & \rightarrow \infty ,\: \: \textrm{maka}\: \: \displaystyle u\rightarrow 0 \end{cases}\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15}{u}.\tan 4u\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{15\tan 4u}{u}\\ &=15\times 4\\ &=60 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin^{2} \left (\displaystyle \frac{ab}{x} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ab \\ \textrm{b}.&\displaystyle a^{2}b\\ \textrm{c}.&\displaystyle ab^{2}\\ \textrm{d}.&\displaystyle (ab)^{2}\\ \textrm{e}.&\displaystyle \frac{1}{(ab)^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: x^{2}\sin \displaystyle \frac{ab}{x}&=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{1}{u} \right )^{2}\sin^{2} \displaystyle abu\\ &=\underset{u\rightarrow 0 }{\textrm{Lim}}\: \: \left ( \displaystyle \frac{\sin ^{2}abu}{u^{2}} \right )\\ &=(ab)^{2} \end{aligned} \end{array}$

$\begin{array}{l}\\ 40.&\textrm{Nilai dari}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \textrm{d}.&\displaystyle 1\\ \textrm{e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{\displaystyle \frac{x}{100}}&=100\times \underset{0}{\underbrace{\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sin 2x}{x}}}\\ &=100\times 0\\ &=0 \end{aligned} \end{array}$

CONTOH 7-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

$\begin{array}{ll}\\ 31.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3} \\\\ \textrm{b}.&\displaystyle \frac{4}{9}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4x^{2}-2x}-\sqrt{x^{2}+1}}{\sqrt{9x^{2}-1}}\times \displaystyle \frac{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-\frac{2}{x}}-\sqrt{1+\frac{1}{x^{2}}}}{\sqrt{9-\frac{1}{x^{2}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{4-0}-\sqrt{1+0}}{\sqrt{9-0}}\\ &=\displaystyle \frac{2-1}{3}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020} =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{9}\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 0\quad &\\\\ \textrm{d}.\quad \displaystyle 1\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x^{4}+2x^{3}-5x+2021}{2x^{3}-4x^{2}+2020}\\ &=\displaystyle \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x^{4}}{x^{4}}+\frac{2x^{3}}{x^{4}}-\frac{5x}{x^{4}}+\frac{2021}{x^{4}}}{\displaystyle \frac{2x^{3}}{x^{4}}-\frac{4x^{2}}{x^{4}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+\displaystyle \frac{2}{x}-\frac{5}{x^{2}}+\frac{2021}{x^{4}}}{\displaystyle \frac{4}{x}-\frac{4}{x^{2}}+\frac{2020}{x^{4}}}\\ &=\displaystyle \frac{3+0-0+0}{0-0+0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{4}{3} \\\\ \textrm{b}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{c}.\quad \displaystyle 0\\\\ \textrm{d}.\quad \displaystyle 3\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{b}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}+3x+4}{3x^{2}+2x+3}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}+\frac{4}{x^{2}}}{\displaystyle \frac{3x^{2}}{x^{2}}+\frac{2x}{x^{2}}+\frac{3}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{1+\displaystyle \frac{3}{x}+\frac{4}{x^{2}}}{3+\displaystyle \frac{2}{x}+\frac{3}{x^{2}}}\\ &= \displaystyle \frac{1+0+0}{3+0+0}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{l}\\ 34.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 3 \\\\ \textrm{b}.\quad \displaystyle 1\\\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad \displaystyle 0\\\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{d}\\ &\begin{aligned}\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x}{9x^{2}+x+1}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3x}{x^{2}}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{3}{x}}{\displaystyle \frac{9x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}}\\ &= \displaystyle \frac{0}{9+0+0}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 35.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -2 &&\\ \textrm{b}.\quad \displaystyle -1\\ \textrm{c}.\quad \displaystyle -\frac{1}{2}\\ \textrm{d}.\quad \displaystyle 0\\ \textrm{e}.\quad \displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{x^{2}-2x-8}-\sqrt{x^{2}+2x+1} \right )\\ &=\infty -\infty =\textbf{tidak diperbolehkan}\\ &\textrm{Selanjutnya gunakan formula}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \left ( \sqrt{ax^{2}+bx+c}-\sqrt{ax^{2}+px+q} \right )=\displaystyle \frac{b-p}{2\sqrt{a}},\quad \textrm{maka}\\ &=\displaystyle \frac{-2-2}{2\sqrt{1}}\\ &=\frac{-4}{2}\\ &=-2 \end{aligned} \end{array}$

CONTOH 6-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

$\begin{array}{ll}\\ 26.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.& 2\\ \textrm{d}.& 4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{3x+1}-\sqrt{3x-2} \right )\times \displaystyle \frac{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \frac{(3x+1)-(3x-2)}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{3}{\left (\sqrt{3x+1}+\sqrt{3x-2} \right )}\times \frac{\displaystyle \frac{1}{\sqrt{x}}}{\displaystyle \frac{1}{\sqrt{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{3}{\sqrt{x}}}{\left (\sqrt{\displaystyle \frac{3x}{x}+\frac{1}{x}}+\sqrt{\displaystyle \frac{3x}{x}-\frac{2}{x}} \right )}\\ &=\displaystyle \frac{0}{\sqrt{3+0}+\sqrt{3-0}}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\\\ \textrm{c}.& \displaystyle \frac{3}{2}\\\\ \textrm{d}.& \displaystyle \frac{7}{2}\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{4x^{2}+6x+8}-\sqrt{4x^{2}-8x+7} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}{\left (\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 4x^{2}+6x+8 \right )-\left ( 4x^{2}-8x+7 \right )}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14x+1}{\sqrt{4x^{2}+6x+8}+\sqrt{4x^{2}-8x+7}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{14+\displaystyle \frac{1}{x}}{\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}+\frac{6x}{x^{2}}+\frac{8}{x^{2}}}+\sqrt{\displaystyle \frac{4x^{2}}{x^{2}}-\frac{8x}{x^{2}}+\frac{7}{x^{2}}}}\\ &=\displaystyle \frac{14+0}{\sqrt{4+0+0}-\sqrt{4-0+0}}\\ &=\displaystyle \frac{14}{2+2}=\frac{7}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&1\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \textrm{d}.&8\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( 2x^{2}+3x-1 \right )-\left ( x^{2}-5x+3 \right )}{\left (\sqrt{2x^{2}+3x-1}+\sqrt{x^{2}-5x+3} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{x^{2}+8x-4}{\left (\sqrt{2x^{2}+3x-1}-\sqrt{x^{2}-5x+3} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{x^{2}}{x^{2}}+\frac{8x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{2x^{2}}{x^{4}}+\frac{3x}{x^{4}}-\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{x^{2}}{x^{4}}-\frac{5x}{x^{4}}+\frac{3}{x^{4}}}}\\ &=\displaystyle \frac{1+0-0}{\sqrt{0+0+0}+\sqrt{0-0+0}}\\ &=\displaystyle \frac{1}{0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 29.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )\\ &\qquad\qquad\times \displaystyle \frac{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\left ( x^{2}+3x+1 \right )-\left ( 3x^{2}+2x+5 \right )}{\left (\sqrt{x^{2}+3x+1}+\sqrt{3x^{2}+2x+5} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2x^{2}+x-4}{\left (\sqrt{x^{2}+3x+1}-\sqrt{3x^{2}+2x+5} \right )}\times \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{-2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{x^{2}}{x^{4}}+\frac{3x}{x^{4}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3x^{2}}{x^{4}}+\frac{2x}{x^{4}}+\frac{5}{x^{4}}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{-2+\displaystyle \frac{1}{x}-\frac{4}{x^{2}}}{\sqrt{\displaystyle \frac{1}{x^{2}}+\frac{3}{x^{3}}+\frac{1}{x^{4}}}+\sqrt{\displaystyle \frac{3}{x^{2}}+\frac{2}{x^{3}}+\frac{5}{x^{4}}}}\\ &=\displaystyle \frac{-2+0-0}{\sqrt{0+0+0}+\sqrt{0+0+0}}\\ &=\displaystyle \frac{-2}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\\\ \textrm{b}.&-\displaystyle \frac{5}{3}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&1\\\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left ((3x-2)-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(3x-2)^{2}}-\sqrt{9x^{2}-2x+5} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(9x^{2}-12x+4}-\sqrt{9x^{2}-2x+5} \right )\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \left (\sqrt{(ax^{2}+bx+c}-\sqrt{px^{2}+qx+r} \right )\\ &\textrm{Jika dikerjakan dengan rumus singkat}\\ &\color{black}\textrm{maka}\quad \left\{\begin{matrix} a=p=3\\ b=-12\: \\ q=-2\: \: \: \end{matrix}\right.\\ &=\displaystyle \frac{b-q}{2\sqrt{a}}\\ &=\displaystyle \frac{-12-(-2)}{2\sqrt{9}}\\ &=\displaystyle \frac{-10}{6}\\ &=-\frac{5}{3} \end{aligned} \end{array}$

CONTOH 5-LIMIT FUNGSI (LIMIT DI KETAKHINGGAN)

$\begin{array}{ll}\\ 21.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}-\sqrt{4x+2021} \right )\times \frac{\sqrt{8x-2020}+\sqrt{4x+2021}}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(8x-2020)-(4x+2021)}{\sqrt{8x-2020}+\sqrt{4x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4x-4041}{\sqrt{8x-2020}+\sqrt{4x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{8x-2020}+\sqrt{4x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{4x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{8}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{4}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{4}{0}\\ &=\infty \end{aligned} \end{array}$

$\begin{array}{l}\\ 22.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{8x-2020}+\sqrt{4x+2021} \right )\\ &=\sqrt{\infty }+\sqrt{\infty }\\ &=\infty \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai yang memenuhi}\\ &\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&\infty \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\:\left ( \sqrt{4x-2020}-\sqrt{8x+2021} \right )\times \frac{\sqrt{4x-2020}+\sqrt{8x+2021}}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{(4x-2020)-(8x+2021)}{\sqrt{4x-2020}+\sqrt{8x+2021}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4x-4041}{\sqrt{4x-2020}+\sqrt{8x+2021}}\times \frac{\displaystyle \frac{1}{x}}{\displaystyle \frac{1}{x}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\displaystyle \frac{1}{x}\left (\sqrt{4x-2020}+\sqrt{8x+2021} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4x}{x^{2}}-\frac{2020}{x^{2}}}+\sqrt{\displaystyle \frac{8x}{x^{2}}+\frac{2021}{x^{2}}} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{-4-\displaystyle \frac{4041}{x}}{\left (\sqrt{\displaystyle \frac{4}{x}-\displaystyle \frac{2020}{x}}+\sqrt{\displaystyle \frac{8}{x}+\displaystyle \frac{2021}{x}} \right )}\\ &=\displaystyle \frac{-4-0}{\sqrt{0-0}+\sqrt{0+0}}\\ &=\displaystyle \frac{-4}{0}\\ &=-\infty \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 1\\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\times \displaystyle \frac{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{4x^{2}+3x-(4x^{2}-5x)}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{3x+5x}{\sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left ( \sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\displaystyle \frac{3+5}{\sqrt{4}+\sqrt{4}}\\ &=\displaystyle \frac{8}{4}\\ &=2 \end{aligned} \end{array}$

$\begin{aligned}\textrm{ada cara lain yang lebih sede}&\textrm{rhana, yaitu:}\\ .\qquad\: \, \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}&=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}\\ &\begin{cases} a & = 4\\ b & =3 \\ p & = -4 \end{cases}\\ \textrm{Jika}\quad &\\ \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{ax^{2}+bx+c}&-\sqrt{ax^{2}+px+q}=\displaystyle \frac{b-p}{2\sqrt{a}}\\ \textrm{Sehingga}\quad&\\ \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}-\sqrt{4x^{2}-5x}&=\displaystyle \frac{3-(-5)}{2\sqrt{4}}\\ &=\displaystyle \frac{8}{2.2}\\ &=2 \end{aligned}$

$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \infty \\ \textrm{b}.\quad \displaystyle 1\\ \textrm{c}.\quad \displaystyle 2\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 8 \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \sqrt{4x^{2}+3x}+\sqrt{4x^{2}-5x}\\ &=\sqrt{\infty }+\sqrt{\infty }=\infty \end{array}$

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

CBT MA FUTUHIYAH JEKETRO

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Jawaban Benar

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