CONTOH SOAL 6 BARISAN DAN DERET

 

$\begin{aligned}11.\quad &\textrm{Diberikan}\\ &\quad A=1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\cdots\\ &\textrm{dan}\\ &\quad B=1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\displaystyle \frac{1}{9^{\displaystyle 4}}+\cdots\\ &\textrm{Nyatakan}\quad \displaystyle \frac{A}{B}\quad \textrm{sebagai pecahan} \end{aligned}$

Jawaban: $\begin{aligned}&\textrm{Perhatikan bahwa}\\ &A=1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{6^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots\\ &\textrm{Selanjutnya deret di ataskita bagi menjadi dua bagian, yaitu}:\\&A=\underset{\textrm{suku-suku genap}}{\underbrace{\left( \displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\displaystyle \frac{1}{6^{\displaystyle 4}}+\displaystyle \frac{1}{8^{\displaystyle 4}}+\cdots \right)}}+\underset{\textrm{suku-suku ganjil}}{\underbrace{\left( 1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots \right)}}\\ &A=\displaystyle \frac{1}{2^{\displaystyle 4}}\left( 1+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{4^{\displaystyle 4}}+\cdots \right)+\left( 1+\displaystyle \frac{1}{3^{\displaystyle 4}}+\displaystyle \frac{1}{5^{\displaystyle 4}}+\displaystyle \frac{1}{7^{\displaystyle 4}}+\cdots \right)\\ &A=\displaystyle \frac{1}{2^{\displaystyle 4}}A+B\Leftrightarrow A=\displaystyle \frac{1}{16}A+B\Leftrightarrow \displaystyle \frac{15}{16}A=B\Leftrightarrow \displaystyle \frac{A}{B}=\displaystyle \frac{16}{15}\\ &\textrm{Jadi, perbandingan}\quad \displaystyle \frac{A}{B}=\displaystyle \frac{16}{15} \end{aligned}$

CONTOH SOAL 5 BARISAN DAN DERET

 

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

Jawaban: $\begin{aligned}&\textrm{Suku-suku dalam dalam deret dapat dinyatakan dengan}:10^{\displaystyle k}-1\\ &\circ \quad99=10^{\displaystyle 2}-1\\ &\circ \quad 999=10^{\displaystyle 3}-1\\ &\circ \quad 9999=10^{\displaystyle 4}-1\\ &\circ \qquad \vdots\\ &\circ \quad \textrm{Suku ke}-50\: \textrm{dengan 51 digit 9}=10^{\displaystyle 51}-1\\ &\textrm{Maka persamaan}\quad X\quad \textrm{dapat dituliskan sebagai}:\\ &X=\left( 10^{\displaystyle 2}-1 \right)+\left( 10^{\displaystyle 3}-1 \right)+\left( 10^{\displaystyle 4}-1 \right)+\cdots +\left( 10^{\displaystyle 51}-1 \right)\\ &\,\quad =10^{\displaystyle 51}+10^{\displaystyle 50}+10^{\displaystyle 49}+\cdots +10^{\displaystyle 4} +10^{\displaystyle 3} +10^{\displaystyle 2}-(1\times 50)\\ &\,\quad =\underset{50\: \textrm{buah angka}\: 1}{\underbrace{111\cdots 111}00}-(1\times 50)\\ &\,\quad =\underset{49\: \textrm{buah angka}\: 1}{\underbrace{111\cdots 111}050}\\ &\textrm{Jadi, banyak digit 1 pada bilangan}\:\: X\:\: \textrm{sebanyak 49} \end{aligned}$

$\begin{aligned}10.\quad&\text{Diberikan}\:\: S=\displaystyle \frac{1+2}{2}+\displaystyle \frac{1+2+3}{2^{\displaystyle 2}}+\displaystyle \frac{1+2+3+4}{2^{\displaystyle 3}}+\cdots \\ &\textrm{Tentukan nilai}\:\: S\end{aligned}$

Jawaban: $\begin{aligned}&\text{Diketahui}\\ &S=\displaystyle \frac{1+2}{2}+\displaystyle \frac{1+2+3}{2^{\displaystyle 2}}+\displaystyle \frac{1+2+3+4}{2^{\displaystyle 3}}+\displaystyle \frac{1+2+3+4+5}{2^{\displaystyle 4}}\cdots \\ &\Leftrightarrow S=\displaystyle \frac{3}{2}+\displaystyle \frac{6}{2^{\displaystyle 2}}+\displaystyle \frac{10}{2^{\displaystyle 3}}+\displaystyle \frac{15}{2^{\displaystyle 4}}+\displaystyle \frac{21}{2^{\displaystyle 5}}+\displaystyle \frac{28}{2^{\displaystyle 6}}+\cdots \\ &\Leftrightarrow \displaystyle \frac{1}{2}S=\displaystyle \frac{3}{2^{\displaystyle 2}}+\displaystyle \frac{6}{2^{\displaystyle 3}}+\displaystyle \frac{10}{2^{\displaystyle 4}}+\displaystyle \frac{15}{2^{\displaystyle 5}}+\displaystyle \frac{21}{2^{\displaystyle 6}}+\cdots \\ &\text{Untuk}\quad S-\displaystyle \frac{1}{2}S=\displaystyle \frac{3}{2}+\displaystyle \frac{6-3}{2^{\displaystyle 2}}+\displaystyle \frac{10-6}{2^{\displaystyle 3}}+\displaystyle \frac{15-10}{2^{\displaystyle 4}}+\displaystyle \frac{21-15}{2^{\displaystyle 5}}+\cdots \\ &\,\:\qquad\qquad \Leftrightarrow \displaystyle \frac{1}{2}S=\displaystyle \frac{3}{2}+\displaystyle \frac{3}{2^{\displaystyle 2}}+\displaystyle \frac{4}{2^{\displaystyle 3}}+\displaystyle \frac{5}{2^{\displaystyle 4}}+\displaystyle \frac{6}{2^{\displaystyle 5}}+\cdots \\ &\textrm{Saat}\quad \displaystyle \frac{1}{2}\left( \displaystyle \frac{1}{2}S \right)=\displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2^{\displaystyle 2}}+\displaystyle \frac{3}{2^{\displaystyle 3}}+\displaystyle \frac{4}{2^{\displaystyle 4}}+\displaystyle \frac{5}{2^{\displaystyle 5}}+\cdots \\ &\textrm{dan}\: \left( \displaystyle \frac{1}{2}S-\displaystyle \frac{1}{4}S \right)=\displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2}+0+\displaystyle \frac{1}{2^{\displaystyle 3}}+\displaystyle \frac{1}{2^{\displaystyle 4}}+\displaystyle \frac{1}{2^{\displaystyle 5}}+\cdots \\ &\Leftrightarrow \displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2}+\underset{\textrm{deret geometri tak hingga}}{\underbrace{\displaystyle \frac{\displaystyle \frac{1}{2^{\displaystyle 3}}}{1-\displaystyle \frac{1}{2}}}}\\ &\Leftrightarrow \displaystyle \frac{1}{4}S=\displaystyle \frac{3}{2}+\displaystyle \frac{1}{4}\qquad (\textrm{masing-masing ruas dikali 4})\\ &\Leftrightarrow S=6+1=7\\ &\textrm{Jadi, nilai}\quad S=7\end{aligned}$

CONTOH SOAL 4 BARISAN DAN DERET

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

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}\\$.

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

Jawaban: $\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 DAN DERET

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

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

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

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

PERSIAPAN MENJELANG TKA MATEMATIKA SMA-MA 2026

 

CBT MA FUTUHIYAH JEKETRO

Persiapan TKA MA Futuhiyah Jeketro

Nama Lengkap Siswa

Kelas / Nomor Ujian

TKA Matematika Wajib Class XII

90:00

Soal No. 1

Memuat soal...

Siswa: -

No: -

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Navigasi Soal

Hasil Ujian CBT TKA Matematika Wajib

Nama: - | ID: -

NILAI AKHIR

0

Skala 0 - 100

Jawaban Benar

0

Jawaban Salah

0

Tidak Dijawab

0

Pembahasan dan Analisis Soal

CONTOH SOAL 2 BARISAN DAN DERET

 

$\begin{aligned}3.\quad &(\textbf{LM UGM ke-19 Th.2007 Tk. SMA})\\ &\textrm{Diketahui barisan geometri dengan suku pertama}\quad a\\ &\textrm{dan rasio}\quad r.\:\: \textrm{Untuk sebarang}\quad n\quad \textrm{genap didefinisikan}\\ &\textrm{jumlahan}\quad S_{\displaystyle n}=U_{\displaystyle 1}+U_{\displaystyle 2}+U_{\displaystyle 3}+...+U_{\displaystyle n}\quad \textrm{dan}\\ &\widehat{S}_{\displaystyle n}=U_{\displaystyle 1}-U_{\displaystyle 2}+U_{\displaystyle 3}-U_{\displaystyle 4}+...+U_{\displaystyle n-1}-U_{\displaystyle n}.\quad \textrm{Nilai}\\ &r\quad \textrm{yang mungkin agar}\quad \displaystyle \frac{\widehat{S}_{\displaystyle n}}{S_{\displaystyle n}}>3?\\ &\text{a}.\quad -1< r<0\\ &\text{b}.\quad -1< r<\displaystyle -\frac{1}{2}\\ &\text{c}.\quad -1< r<\displaystyle -\frac{1}{3}\\ &\text{d}.\quad \displaystyle -\frac{1}{3}< r<0\\ &\text{e}.\quad \textrm{tergantung oleh nilai}\quad a\\ \end{aligned}$

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

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$\begin{aligned}4.\quad&(\textbf{OLIMPIADE SAINS PORSEMA Th.2012})\\ &\textrm{Jumlah 50 suku pertama dari deret berikut}\\ &\log 5+\log 55+ \log 605+\log 6655\:+\:...\quad \textrm{adalah}\: ....\\ &\text{a}.\quad \log \left( 55^{\displaystyle 1155} \right)\qquad\qquad\qquad\text{d}.\quad \log \left( 275^{\displaystyle 1150} \right)\\ &\text{b}.\quad \log \left( 5^{\displaystyle 25}.11^{\displaystyle 1225} \right)\qquad\quad\quad\text{e}.\quad 1150\log 5\\ &\text{c}.\quad \log \left( 25^{\displaystyle 25}.11^{\displaystyle 1225} \right)\\\end{aligned}$

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

CONTOH SOAL 1 BARISAN DAN DERET

 

$\begin{aligned}1.\quad&(\textbf{LM UGM ke-25 Th 2014 Tk.SMA})\\ &\textrm{Diberikan dua buah barisan aritmetika}\quad 1,4,7,...\quad \textrm{dan}\\ &2,7,12,...\:. \: \textrm{Jika}\:\:S\:\: \textrm{merupakan himpunan yang terdiri}\\ &\textrm{dari gabungan 2014 suku pertama kedua barisan tersebut},\\ &\textrm{Banyak anggota himpuan}\:\: S\:\: \textrm{adalah}\: ....\\ &\text{a}.\quad 3625\qquad\qquad\qquad\qquad \text{d}.\quad 4015\\ &\text{b}.\quad 3875\qquad \text{c}.\quad 4014\qquad \text{e}.\quad 4028\\ \end{aligned}$

Jawaban: $\begin{aligned}&\textbf{Jawab}:\quad \text{a}\\ &\textrm{Perhatikan bahwa}\\ &\textrm{Barisan pertama}:a_{1}=1,b_{1}=3,A_{n}=3n-2,A_{2014}=6040\\ &\textrm{Barisan kedua}:a_{2}=2,b_{2}=5,B_{m}=5m-3,B_{2014}=10067\\ &\textrm{Menentukan irisan kedua barisan}\: \left( A\cap B \right):\\ &\textrm{Barisan gabungan}:a_{12}=7,b_{12}=KPK(3,5)=15,\\ &\qquad\qquad C_{12=k=gabungan}=7+(k-1)15=15k-8\\ &\textrm{Maksimum suku}\le 6040\quad(\textrm{pilih yang terpendek})\\ &\qquad\qquad 15k-8\le 6040\Leftrightarrow 15k\le 6048\Leftrightarrow k\le 403,2\\ &\qquad\qquad \textrm{karena}\:\: k\:\: \textrm{bulat, maka nilai terbesar}\:k=403\\ &\textrm{Menghitung banyaknya anggota}\:\: S,\:\: \textrm{maka}:\\ &n(S)=n\left( A\cup B \right)=n(A)+n(B)-n\left( A\cap B \right)\\&\:\qquad=2014+2014-403=4028-403=3625\end{aligned}$

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$\begin{aligned}2.\quad&(\textbf{LM UGM Ke-25 Th 2014 Tk.SMA})\\ &\textrm{Hitunglah nilai dari}\\ &5-\displaystyle \frac{10}{3}+\frac{20}{9}-\frac{40}{27}+\frac{80}{81}-...\\ &\text{a}.\quad 5\qquad\qquad\qquad\:\: \text{d}.\quad 2\\ &\text{b}.\quad 4\qquad \text{c}.\quad 3\qquad \text{e}.\quad 1\\ \end{aligned}$

Jawaban: $\begin{aligned}&\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui deret geometri tak hingga yang konvergen}\\ & \textrm{dengan}\\ &\begin{cases} a & =5 \quad \\ r & =\displaystyle \frac{U_{2}}{U_{1}}=\displaystyle \frac{\displaystyle -\frac{10}{3}}{5}=\displaystyle -\frac{2}{3} \end{cases}\\ &\textrm{Sehingga jumlahnya adalah:}\\ &S_{\displaystyle \infty }=\displaystyle \frac{a}{1-r}=\displaystyle \frac{5}{1-\left( -\displaystyle \frac{2}{3} \right)}=\displaystyle \frac{5}{\displaystyle \frac{5}{3}}=3 \end{aligned}$

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BARISAN DAN DERET

 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

1. Astuti, A.N., Miyanto, Ngapiningsih. 2020. Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Yogyakarta: PT. PENERBIT INTAN PARIWARA
2. Kartini, Suprpto, Subandi, Setiyadi, U. 2005.Matematika Program Studi Ilmu ALam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
3. Noormandiri. 2017. Matematika Kelompok Peminatan Matematika dan Ilmu-Ilmu ALam untuk SMA/MA Kelas XII. Jakarta: ERLANGGA
4. 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.
5. 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)

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

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

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

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

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