Lanjutan Limit Fungsi Trigonometri

 $\text{C. Menentukan Nilai Limit Fungsi Trigonometri}$

Dalam bahasan ini yang akan dibahas adalah nilai limit mendekati $a$ atau nilai $x$ di sekitar $a$. Ada 3 cara yang populer digunakan untuk menentukan nilai limit fungsi trigonometri ini dengan salah satunya yang paling sering digunakan adalah substitusi langsung di antara cara-cara penyelesaian lainnya. Jika dengan cara substitusi langsung nantinya mendfapatkan nilai bentuk tak tentu yaitu ${\displaystyle \frac{0}{0}}$, maka cara Anda harus menggunakan cara yang lainnya sampai Anda temukan nilai limitnya. Selanjutnya 3 cara yang dimaksud di atas adalah sebagai berikut:

$\text{C. 1 dengan substitusi langsung}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}(\sin x+\tan x)\\ \text{b}& \underset{x\to \pi }{\text{lim}}\left({\displaystyle \sin x+\cos x}\right)\\ \text{c}& \underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{lim}}\left({\displaystyle \frac{\sin x+\cos x}{\tan x}}\right)\\ \text{d}& \underset{x\to {\displaystyle 0}}{\text{lim}}\left({\displaystyle \frac{1+\cos 2x}{1+2\cos x}}\right)\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan substitusi langsung didapatkan}\\ \text{a}.& \underset{x\to 0}{\text{lim}}(\sin x+\tan x)\\ & =\sin 0+\tan 0=0+0=0\\ \text{b}.& \underset{x\to \pi }{\text{lim}}\left({\displaystyle \sin x+\cos x}\right)\\ & =\sin \pi +\cos \pi =0+(-1)=-1\\ \text{c}.& \underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{lim}}\left({\displaystyle \frac{\sin x+\cos x}{\tan x}}\right)\\ & =\left({\displaystyle \frac{\sin {\displaystyle \frac{\pi }{4}+\cos {\displaystyle \frac{\pi }{4}}}}{\tan {\displaystyle \frac{\pi }{4}}}}\right)={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{2}+{\displaystyle \frac{1}{2}\sqrt{2}}}}{1}}\\ & ={\displaystyle \frac{\sqrt{2}}{1}=\sqrt{2}}\\ \text{d}.& \underset{x\to 0}{\text{lim}}\left({\displaystyle \frac{1+\cos 2x}{1+2\cos x}}\right)\\ & =\left({\displaystyle \frac{1+\cos 2(0)}{1+2\cos (0)}}\right)={\displaystyle \frac{1+1}{1+2.1}={\displaystyle \frac{2}{3}}}\end{array}\end{array}$

$\text{C. 2 dengan menyederhanakan}$

Langkah ini ditempuh setelah langkah substitusi langsung tidak memungkinkan atau ketemu bentuk tak tentu  ${\displaystyle \frac{0}{0}}$.

Gunakanlah identitas-identitas trigonometri yang Anda dapatkan di kelas XI  dan akan sering digunakan nantinya di antaranya, yaitu:

$\begin{array}{rl}\bullet \sin 2x& =2\sin x\cos x\\ \bullet \cos 2x& ={\cos }^{2}x-{\sin }^{2}x\\ & =1-2{\sin }^{2}x\\ & =2{\cos }^{2}x-1\\ \bullet \tan 2x& ={\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}\end{array}$

Demikian juga

$\begin{array}{rl}\bullet \sin A+\sin B& =2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \sin A-\sin B& =2\cos \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \cos A+\cos B& =2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \cos A-\cos B& =-2\sin \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\end{array}$.

Masih banyak bentuk identitas trigonometri selain di atas, karenanya sekiranya perlu maka hafalkanlah

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}}\\ \text{b}& \underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}}\\ \text{c}& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}}\\ \text{d}& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan menyederhankan bentuk trigonometri}\\ \text{ak}& \text{an didapatkan bentuk yang lebih sederhana}\\ \text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{\sin 0}{\sin 0}={\displaystyle \frac{0}{0},\text{maka perlu disederhanakan}}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2\sin x\cos x}{\sin x}=\underset{x\to 0}{\text{lim}}{\displaystyle 2\cos x}}}\\ & \text{baru digunakan substitusinya}\\ & =2\cos 0=2.1=2\\ \text{b}.& \underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}}\\ & \text{Sama seperti langkah di atas, yaitu}:\\ & =\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}=\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{2\sin 2x\cos 2x}{\sin x\cos x}}}\\ & =\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{4\sin x\cos x\cos 2x}{\sin x\cos x}=\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle 4\cos 2x}}\\ & =4\cos 2\left({\displaystyle \frac{\pi }{2}}\right)=4\cos \pi =4.(-1)=-4\\ \text{c}.& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{1-{\cos }^{2}0}{{\tan }^{2}0}={\displaystyle \frac{1-1}{0}=\frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}=\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{\left({\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\right)}}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle {\cos }^{2}x={\cos }^{2}0=1^2=1}\\ \text{d}.& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{{\sin }^{2}0}{1-\cos 0}={\displaystyle \frac{0}{1-1}=\frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}=\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{1-\cos x}}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{(1+\cos x)(1-\cos x)}{1-\cos x}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle (1+\cos x)}\\ & =1+\cos 0=1+1=2\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ \text{b}& \underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan menyederhankan bentuk trigonometri}\\ \text{ak}& \text{an didapatkan bentuk yang lebih sederhana}\\ \text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{\cos 0-\cos 0}{1-\cos 0}={\displaystyle \frac{1-1}{1-1}={\displaystyle \frac{0}{0},}}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin \left({\displaystyle \frac{x+3x}{2}}\right)\sin \left({\displaystyle \frac{x-3x}{2}}\right)}{1-(1-2{\sin }^{2}x)}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin 2x\sin (-x)}{2{\sin }^{2}x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2\sin 2x\sin x}{2\sin x.\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2(2\sin x\cos x)\sin x}{2\sin x.\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle 2\cos x}\\ & \text{baru digunakan substitusinya}\\ & ={\displaystyle 2\cos 0=2.1=2}\\ \text{b}.& \underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{1-\tan x}{\sin x-\cos x}=\frac{1-1}{\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}}={\displaystyle \frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}=\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\left({\displaystyle \frac{\sin x}{\cos x}}\right)}{\sin x-\cos x}}}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{-(\sin x-\cos x)}{\cos x(\sin x-\cos x)}}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{-1}{\cos x}={\displaystyle \frac{-1}{\cos \left({\displaystyle \frac{\pi }{4}}\right)}={\displaystyle \frac{-1}{\frac{1}{\sqrt{2}}}=-\sqrt{2}}}}\end{array}\end{array}$

$\text{C. 3 dengan rumus limit fungsi trigonometri}$

Berikut rumus limit fungsi trigonometri yang akan kita gunakan

$\begin{array}{rl}\bullet \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin ax}{ax}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{ax}{\sin ax}=\frac{a}{a}=1}\\ \bullet \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan ax}{ax}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{ax}{\tan ax}=\frac{a}{a}=1}\end{array}$.

BUKTINYA ada di sini

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{7x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 9x}}\\ \text{c}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{\sin 3x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{7x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{4}{7}\times {\displaystyle \frac{\sin 4x}{4x}={\displaystyle \frac{4}{7}}}}}\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 9x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2}{9}\times {\displaystyle \frac{9x}{\tan 9x}={\displaystyle \frac{2}{9}}}}}\\ \text{c}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{\sin 3x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{8x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{3x}{\sin 3x}\times \frac{8}{3}={\displaystyle \frac{8}{3}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{24x^2}{8{\sin }^{2}x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{5x^2}{15\tan (-9x)\sin 3x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{24x^2}{8{\sin }^{2}x}={\displaystyle \frac{24}{8}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}}}}}\\ & =3\times 1\times 1\\ & =3\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{5x^2}{15\tan (-9x)\sin 3x}}\\ & ={\displaystyle \frac{5}{15}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan (-9x)}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan 3x}}}}\\ & ={\displaystyle \frac{1}{3}\times (-\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{9x}{\tan 9x}\times \frac{1}{9}})\times \times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{3x}{\tan 3x}\times \frac{1}{3}}}\\ & =-{\displaystyle \frac{1}{3}\times \frac{1}{9}\times \frac{1}{3}=-\frac{1}{81}}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 6x}{x^2}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos 4x-\cos 2x}{x^2}}\\ \text{c}& \underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x^2-5x+6)\sin (x-3)}{{(x^2-7x+12)}^2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 6x}{x^2}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-(1-2{\sin }^{2}3x)}{x^2}}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2{\sin }^{2}3x}{x^2}=2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}}}}\\ & =2\times 3\times 3=18\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos 4x-\cos 2x}{x^2}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin \left({\displaystyle \frac{4x+2x}{2}}\right)\sin \left({\displaystyle \frac{4x-2x}{2}}\right)}{x^2}}\\ & =-2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x.\sin x}{x^2}}\\ & =-2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{x}=-2\times 3\times 1=-6}}\\ \text{c}.& \underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x^2-5x+6)\sin (x-3)}{{(x^2-7x+12)}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x-2)(x-3)\sin (x-3)}{{((x-3)(x-4))}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x-2)(x-3)\sin (x-3)}{(x-3)(x-3)(x-4{)}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle {\displaystyle \frac{x-2}{(x-4{)}^2}\times \underset{x\to 3}{\text{lim}}{\displaystyle \frac{\sin (x-3)}{(x-3)}}}}\\ & ={\displaystyle \frac{(3-2)}{(3-4{)}^2}\times 1={\displaystyle \frac{1}{(-1{)}^2}=\frac{1}{1}=1}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tentukanlah nilai limit dari}\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin (4x-2)}{\tan 2x-1}}\\ \\ & \text{Jawab}:\\ & \text{Kita misalkan}a=2x-1,\\ & \text{ketika}x\to {\displaystyle \frac{1}{2},\text{maka akan didapatkan}a\to 0}\\ & (\text{textbf{dibaca}: saat nilai}x\text{mendekati}{\displaystyle \frac{1}{2},\text{maka nilai}}\\ & a\text{akan mendekati nilai}0).\\ & \text{Selanjutnya kita buatkan penyesuaian, yaitu}:\\ & \begin{array}{rl}\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin (4x-2)}{\tan 2x-1}}& =\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin 2(2x-1)}{\tan 2x-1}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2a}{\tan a}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2a}{2a}\times 2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{a}{\tan a}}}\\ & =2\times 1=2\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selidikilah limit fungsi berikut, apakah}\\ & \text{memiliki nilai limit atau tidak}\\ & \text{a}.\underset{x\to 0}{\text{lim}}2x-1\\ & \text{b}.\underset{x\to 0}{\text{lim}}{x}^2-x-2\\ & \text{c}.\underset{x\to 1}{\text{lim}}f(x),\text{dengan}f(x)=\{\begin{array}{ll}2x& ;x<1\\ 4x-1& ;x\ge 1\end{array}\\ & \text{ d}.\underset{x\to 2}{\text{lim}}\sqrt{f(x)},\text{dengan}f(x)=\{\begin{array}{ll}4x-1& ;x<2\\ 2x+5& ;x\ge 2\end{array}\\ & \text{e}.\underset{x\to 0}{\text{lim}}(x+\cos x)\\ & \text{f}.\underset{x\to 0}{\text{lim}}x\tan x\\ & \text{g}.\underset{x\to {\displaystyle \frac{\pi }{2}}}{\text{lim}}(\sin x+2\cos x)\\ & \text{h}.\underset{x\to {\displaystyle \frac{\pi }{2}}}{\text{lim}}(2\tan x-\sin 2x)\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 5x}{x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{4x}{\sin x}}\\ \text{c}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 6x}{8x}}\\ \text{d}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 7x}}\\ \text{e}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{\sin 2x}}\\ \text{f}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{\tan 8x}}\\ \text{9}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{{\sin }^{2}5x}{2x^2}}\\ \text{h}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 5x\tan 6x}{x\tan 7x}}\\ \text{i}& \underset{x\to y}{\text{lim}}{\displaystyle \frac{\sin x-\sin y}{x-y}}\\ \text{j}& \underset{x\to 2}{\text{lim}}{\displaystyle \frac{\sin (x^2-4)}{x-2}}\\ \text{k}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin mx-\sin nx}{\cos mx-\cos nx}}\\ \text{l}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 3x\cos x}{x^2}}\end{array}\\ \end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. 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.
3. Yuana, A.R., Indriyastuti. 2017. Perspektif Matematika untuk Kelas XII SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI PUSTAKA MANDIRI.

Limit Fungsi Trigonometri

 $\text{A. Pendahuluan}$

Mengingat kembali definisi limit yang telah dipelajari sebelumnya di kelas XI, yaitu limit fungsi aljabar $f(x)$ yang didefinisikan dengan:

$\begin{array}{rl}\underset{x\to a}{\text{lim}}f(x)=L& \mathbf{\text{adalah}}\text{Jika}x\text{mendekati}a\\ & \text{dengan tidak sama dengan}a,\\ & \text{maka nilai}f(x)\text{mendekati}L\end{array}$.

Perhatikan definisi di atas istilah  $x\text{mendekati}a$ dituliskan dengan simbol  $(x\to a)$. Suatu nilai limit dianggap ada jika nilai $f(x)$ mendekati  $a$ dari arah kiri sama dengan nilai $f(x)$ mendekati  $a$ dari arah kanan dengan nilai yang sama misalnya $L$. Jika disimbolkan pernyataan ini menjadi berikut

$\begin{array}{rl}\underset{x\to a^{-}}{\text{lim}}f(x)=\underset{x\to a^{+}}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}f(x)=L& \end{array}$.

$\begin{array}{rl}\text{Perlu di}& \text{perhatikan bahwa didekati dari}\\ \bullet \mathbf{\text{kiri}}& \text{disimbolkan dengan}\underset{x\to a^{-}}{\text{lim}}f(x),\text{dan}\\ \bullet \mathbf{\text{kan}}& \mathbf{\text{an}}\text{disimbolkan dengan}\underset{x\to a^{+}}{\text{lim}}f(x)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari}f(x)={\displaystyle \frac{x^2-4}{x-2}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika fungsi}{\displaystyle \frac{x^2-4}{x-2}}\\ & \text{di sekitar}x=2\text{sebagaimana dalam tabel}\\ & \text{berikut}\end{array}\end{array}$.

$\begin{array}{rl}.& \text{Jadi, nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}=2\text{atau dapat dikatakan}}\\ & \text{nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}\mathbf{\text{ada}}}\\ & \text{meskipun nilai substitusi langsung}x=2\text{yaitu}\\ & f(0)={\displaystyle \frac{0^2-0}{0-0}=\frac{0}{0}\text{berupa bentuk tak}}\\ & \text{tentu. Berikut ilustrasinya}\end{array}$

$\begin{array}{l}\\ 2.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 5}{\text{lim}}f(x),\text{untuk}f(x)=\{\begin{array}{ll}x& \text{saat}x<5\\ 5-x& \text{saat}x\ge 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 5^{-},\text{maka}\underset{x\to 5^{-}}{\text{lim}}f(x)=x\\ & \text{atau}\underset{x\to 5^{-}}{\text{lim}}f(x)=5\\ & \text{boleh juga dituliskan dengan}\\ & \underset{x\to 5^{-}}{\text{lim}}f(x)=x=5.\text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 5^{+},\\ & \text{maka}\underset{x\to 5^{+}}{\text{lim}}f(x)=\underset{x\to 5^{+}}{\text{lim}}(5-x)=5-5=0.\\ & \text{Karena nilai}\underset{x\to 5^{-}}{\text{lim}}f(x)\ne \underset{x\to 5^{+}}{\text{lim}}f(x),\text{maka}\\ & \text{nilai atau harga}\underset{x\to 5}{\text{lim}}f(x)\mathbf{\text{tidak ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 0}{\text{lim}}f(x),\text{untuk}f(x)=\cos x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 0^{-},\text{maka}\underset{x\to 0^{-}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& -0,5& -0,4& -0,3& -0,2& -0,1& 0\\ \cos x& ...& ...& 0,999986& 0,999994& 0,9999985& 1\\ \hline\end{array}\\ & \text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 0^{+},\\ & \text{maka}\underset{x\to 0^{+}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& 0& 0,1& 0,2& 0,3& 0,4& 0,5\\ \cos x& 1& 0,9999985& 0,999994& 0,999986& ...& ...\\ \hline\end{array}\\ & \text{Karena nilai}\underset{x\to 0^{-}}{\text{lim}}f(x)=\underset{x\to 0^{+}}{\text{lim}}f(x)=1,\text{maka}\\ & \text{nilai}\underset{x\to 0}{\text{lim}}\cos x\mathbf{\text{ ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\text{B. Sifat-Sifat Limit Fungsi}$

$\begin{array}{rl}& \text{Misalkan}f\text{dan}g\text{adalah fungsi-fungsi yang}\\ & \text{mempunyai nilai limit di titik sekitar}x=a\\ & \text{atau}(x\to a)\text{dan}c\text{adalah suatu konstanta}\\ & \text{serta}n\text{adalah suatu bilangan bulat positif},\\ & \text{maka berlaku sifat-sifat berikut}:\\ & \begin{array}{l}\\ 1.& \underset{x\to a}{\text{lim}}{\displaystyle c=c}\\ 2.& \underset{x\to a}{\text{lim}}{\displaystyle x^n=a^n}\\ 3.& \underset{x\to a}{\text{lim}}c.f(x)=c.\underset{x\to a}{\text{lim}}f(x)\\ 4.& \underset{x\to a}{\text{lim}}(f(x)\pm g(x))=\underset{x\to a}{\text{lim}}f(x)\pm \underset{x\to a}{\text{lim}}g(x)\\ 5.& \underset{x\to a}{\text{lim}}(f(x)\times g(x))=\underset{x\to a}{\text{lim}}f(x)\times \underset{x\to a}{\text{lim}}g(x)\\ 6.& \underset{x\to a}{\text{lim}}{\displaystyle \frac{f(x)}{g(x)}={\displaystyle \frac{\underset{x\to a}{\text{lim}}f(x)}{\underset{x\to a}{\text{lim}}g(x)}}}\\ 7.& \underset{x\to a}{\text{lim}}{(f(x))}^n={[\underset{x\to a}{\text{lim}}f(x))]}^n\\ 8.& \underset{x\to a}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)},\text{dengan}\underset{x\to a}{\text{lim}}f(x)\ge 0\\ & \text{dan}n\text{genap}\end{array}\end{array}$