Lanjutan Limit Fungsi Trigonometri

 $\color{blue}\textrm{C. Menentukan Nilai Limit Fungsi Trigonometri}$

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:

$\color{blue}\textrm{C. 1 dengan substitusi langsung}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

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

$\color{blue}\textrm{C. 2 dengan menyederhanakan}$

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:

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

Demikian juga

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

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

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

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

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

$\color{blue}\textrm{C. 3 dengan rumus limit fungsi trigonometri}$

Berikut rumus limit fungsi trigonometri yang akan kita gunakan

$\begin{aligned}\bullet \: \: \: \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin ax}{ax}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ax}{\sin ax}=\frac{a}{a}=1\\ \bullet \: \: \: \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan ax}{ax}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ax}{\tan ax}=\frac{a}{a}=1 \end{aligned}$.

BUKTINYA ada di sini

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah nilai limit berikut}\\ &\begin{array}{lll} \textrm{a}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 4x}{7x}\\ \textrm{b}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 2x}{\tan 9x}\\ \textrm{c}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 8x}{\sin 3x} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 4x}{7x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{4}{7}\times \displaystyle \frac{\sin 4x}{4x}=\displaystyle \frac{4}{7}\\ \textrm{b}.&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{2x}{\tan 9x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{2}{9}\times \displaystyle \frac{9x}{\tan 9x}=\displaystyle \frac{2}{9}\\ \textrm{c}.&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 8x}{\sin 3x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 8x}{8x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{3x}{\sin 3x}\times \frac{8}{3}=\displaystyle \frac{8}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah nilai limit berikut}\\ &\begin{array}{lll} \textrm{a}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{24x^{2}}{8\sin ^{2}x}\\ \textrm{b}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 5x^{2}}{15\tan (-9x)\sin 3x} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{24x^{2}}{8\sin ^{2}x}=\displaystyle \frac{24}{8}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}\\ &=3\times 1\times 1\\ &=3\\ \textrm{b}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 5x^{2}}{15\tan (-9x)\sin 3x}\\ &=\displaystyle \frac{5}{15}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ x}{\tan (-9x)}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ x}{\tan 3x}\\ &=\displaystyle \frac{1}{3}\times \left ( -\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 9x}{\tan 9x}\times \frac{1}{9} \right )\times \times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 3x}{\tan 3x}\times \frac{1}{3}\\ &=-\displaystyle \frac{1}{3}\times \frac{1}{9}\times \frac{1}{3}=-\frac{1}{81} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukanlah nilai limit berikut}\\ &\begin{array}{lll} \textrm{a}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1-\cos 6x}{x^{2}}\\ \textrm{b}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ \cos 4x-\cos 2x}{x^{2}}\\ \textrm{c}&\underset{x\rightarrow 3 }{\textrm{lim}}\: \displaystyle \frac{ \left (x^{2}-5x+6 \right )\sin \left ( x-3 \right )}{\left ( x^{2}-7x+12 \right )^{2}} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1-\cos 6x}{x^{2}}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1-\left ( 1-2\sin ^{2}3x \right )}{x^{2}}\\ &=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{2\sin ^{2}3x}{x^{2}}=2\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x}{x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x}{x}\\ &=2\times 3\times 3=18\\ \textrm{b}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ \cos 4x-\cos 2x}{x^{2}}\\ &=\underset{x\rightarrow 0 }{\textrm{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\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x.\sin x}{x^{2}}\\ &=-2\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x}{x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{x}=-2\times 3\times 1=-6\\ \textrm{c}.\: \: &\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ \left (x^{2}-5x+6 \right )\sin \left ( x-3 \right )}{\left ( x^{2}-7x+12 \right )^{2}}\\ &=\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ (x-2)(x-3)\sin \left ( x-3 \right )}{\left ( (x-3)(x-4) \right )^{2}}\\ &=\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ (x-2)(x-3)\sin \left ( x-3 \right )}{(x-3)(x-3)(x-4)^{2}}\\ &=\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \displaystyle \frac{x-2}{(x-4)^{2}}\times \underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ \sin \left ( x-3 \right )}{(x-3)} \\ &=\displaystyle \frac{(\color{red}3\color{black}-2)}{(\color{red}3\color{black}-4)^{2}}\times 1=\displaystyle \frac{1}{(-1)^{2}}=\frac{1}{1}=1 \end{aligned} \end{array}$.

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

$\LARGE\colorbox{magenta}{LATIHAN SOAL}$.

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

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

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.