A. BENTUK DASAR
\(\displaystyle \lim_{x \rightarrow \: 0}\dfrac{\sin ax}{bx} = \dfrac ab\)
\(\displaystyle \lim_{x\rightarrow \: 0}\dfrac{\tan ax}{bx} = \dfrac ab\)
\(\displaystyle \lim_{x\rightarrow \: m}\dfrac{\sin a(x - m)}{b(x - m)} = \dfrac ab\)
\(\displaystyle \lim_{x\rightarrow \: m}\dfrac{\tan a(x - m)}{b(x - m)} = \dfrac ab\)
Contoh 01
\begin{equation*} \begin{split} & \lim_{x\rightarrow \: 0} \: \frac{\sin4x}{3x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{\cancelto {4} {\sin4x}}{\cancelto {3} {3x}} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac {4}{3}} \end{split} \end{equation*}
Contoh 02
\begin{equation*} \begin{split} & \lim_{x\rightarrow \: 0} \: \frac{2x}{\tan 5x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{\cancelto {2} {2x}}{\cancelto {5} {\tan 5x}} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac {2}{5}} \end{split} \end{equation*}
Contoh 03
\begin{equation*} \begin{split} & \lim_{x\rightarrow \: 0} \: \frac{\sin 6x}{\tan 2x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{\cancelto {6} {\sin 6x}}{\cancelto {2} {\tan 2x}} \\\\ & \bbox[5px, border: 2px solid magenta] {3} \end{split} \end{equation*}
Contoh 04
\begin{equation*} \begin{split} & \lim_{x \rightarrow \: 2} \: \frac{3x - 6}{\tan(x-2)} \\\\ & \lim_{x \rightarrow \: 2} \: \frac{3(x - 2)}{\tan(x-2)} \\\\ & \lim_{x \rightarrow \: 2} \: \frac{3 \cancel{(x - 2)}}{\cancel{\tan(x-2)}} \\\\ & \bbox[5px, border: 2px solid magenta] {3} \end{split} \end{equation*}
B. KALI SEKAWAN
\(\displaystyle \lim_{x \rightarrow a} \dfrac {1}{\sqrt{x + m} - \sqrt{x + n}} \times \dfrac {\sqrt{x + m} + \sqrt{x + n}}{\sqrt{x + m} + \sqrt{x + n}} = \dfrac {\sqrt{x + m} + \sqrt{x + n}}{(x + m) - (x + n)}\)
Penjelasan
\((a + b)(a - b) = a^2 - b^2\)
\((\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b\)
\((\sqrt{x + m} + \sqrt{x + n})(\sqrt{x + m} - \sqrt{x + n}) = (\sqrt{x + m})^2 - (\sqrt{x + n})^2 = (x + m) - (x + n)\)
Contoh
\begin{equation*} \begin{split} & \lim_{x \rightarrow \: 0} \: \frac{\sin4x}{\sqrt{x+2}-\sqrt{2}} \quad {\color {blue} \times \frac {\sqrt{x+2} + \sqrt{2}}{\sqrt{x+2} + \sqrt{2}}} \\\\ & \lim_{x \rightarrow \: 0} \: \frac {\sin 4x \:.\: (\sqrt{x+2} + \sqrt{2})}{x + 2 - 2} \\\\ & \lim_{x \rightarrow \: 0} \: \frac{\sin 4x \:.\: (\sqrt{x+2} + \sqrt{2})}{x} \\\\ & \lim_{x \rightarrow \: 0} \: \frac{\cancel {\sin 4x} \:.\: (\sqrt{x+2} + \sqrt{2})}{\cancel {x}} \\\\ & \lim_{x \rightarrow \: 0} \: 4 \:.\: (\sqrt{x+2} + \sqrt{2}) \\\\ & 4 \:.\: (\sqrt{0+2} + \sqrt{2}) \\\\ & 4 \:.\: 2 \sqrt {2} \\\\ & \bbox[5px, border: 2px solid magenta] {8 \sqrt{2}} \end{split} \end{equation*}
Keterangan:
Dengan rumus \((a + b)(a - b) = a^2 - b^2\), maka:
\begin{equation*} \begin{split} & (\sqrt{x+2}+\sqrt{2})(\sqrt{x+2}-\sqrt{2}) = x + 2 - 2 = x \end{split} \end{equation*}
C. MENGUBAH FUNGSI
Beberapa soal limit trigonometri dapat diselesaikan dengan mengubah fungsi trigonometri, seperti beberapa contoh berikut:
Contoh 01
\begin{equation*} \begin{split} & \lim_{x \rightarrow \: \pi} \: \frac{\sin x}{\pi-x} \\\\ & \lim_{x \rightarrow \: \pi} \: \frac{\sin (\pi - x)}{\pi-x} \\\\ & \lim_{x \rightarrow \: \pi} \: \frac{\cancel{\sin (\pi - x)}}{\cancel{\pi-x}} \\\\ & \bbox[5px, border: 2px solid magenta] {1} \end{split} \end{equation*}
Contoh 02
\begin{equation*} \begin{split} & \lim_{x \rightarrow \: \frac 12 \pi} \: \frac {\cos x}{x - \frac 12 \pi} \\\\ & \lim_{x \rightarrow \: \frac 12 \pi} \: \frac {\sin (\frac 12 \pi - x)}{-(\frac 12 \pi - x)} \\\\ & \lim_{x \rightarrow \: \frac 12 \pi} \: \frac {\cancel{\sin (\frac 12 \pi - x)}}{- \cancel{(\frac 12 \pi - x)}} \\\\ & \bbox[5px, border: 2px solid magenta] {- 1} \end{split} \end{equation*}
Contoh 03
\begin{equation*} \begin{split} & \lim_{x \rightarrow \: 1} \: (x-1) \:.\: \cot \pi x \\\\ & \lim_{x \rightarrow \: 1} \: \frac {x - 1}{\tan \pi x} \\\\ & \lim_{x \rightarrow \: 1} \: \frac {x - 1}{-\tan (\pi - \pi x)} \\\\ & \lim_{x \rightarrow \: 1} \: \frac {x - 1}{-\tan \pi(1 - x)} \\\\ & \lim_{x \rightarrow \: 1} \: \frac {\cancel{x - 1}}{- \cancel{\tan \pi (1 - x)}} \\\\ & \bbox[5px, border: 2px solid magenta] {- \frac {1}{\pi}} \end{split} \end{equation*}
D. SUDUT RANGKAP
Rumus trigonometri sudut rangkap yang sering digunakan:
\(\sin 2x = 2 \sin x \cos x\)
\(\cos 2x = 2 \cos^2 x - 1\)
\(\cos 2x = 1 - 2 \sin^2 x\)
\(\tan 2x = \dfrac {2 \tan x}{1 - \tan^2 x}\)
\(\sin 3x = 3 \sin x - 4 \sin^3 x\)
\(\cos 3x = 4 \cos^3 x - 3 \cos x\)
Contoh 01
\begin{equation*} \begin{split} & \lim_{x\rightarrow \: 0} \: \frac{2x \:\tan4x}{\cos2x-1} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2x \:\tan4x}{1 - 2 \sin^2 x -1} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2x \:\tan4x}{- 2 \sin^2 x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2x}{-2 \sin x} \:.\: \frac {\tan4x}{\sin x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2 \cancel{x}}{-2 \cancel{\sin x}} \:.\: \frac {\cancel {\tan 4x}}{\cancel {\sin x}} \\\\ & -1 \:.\: 4 \\\\ & \bbox[5px, border: 2px solid magenta] {-4} \end{split} \end{equation*}
Contoh 02
\begin{equation*} \begin{split} & \lim_{x\rightarrow \: 0} \: \frac{2\sin x-\sin2x}{x^2 \: \tan x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2\sin x- 2 \sin x \cos x}{x^2 \: \tan x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2\sin x \left[1 - \cos x \right]}{x^2 \: \tan x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2\sin x \left[1 - (1 - 2 \sin^2 \tfrac 12 x) \right]}{x^2 \: \tan x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2\sin x \left[1 - 1 + 2 \sin^2 \tfrac 12 x \right]}{x^2 \: \tan x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2\sin x \:.\: 2 \sin^2 \tfrac 12 x}{x^2 \: \tan x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2\sin x}{x} \:.\: \frac {2 \sin \tfrac 12 x}{x} \:.\: \frac { \sin \tfrac 12 x}{\tan x} \\\\ & 2 \:.\: 1 \:.\: \tfrac 12 \\\\ & \bbox[5px, border: 2px solid magenta] {1} \end{split} \end{equation*}
E. PENJUMLAHAN FUNGSI
\(\sin A + \sin B = 2 \sin \frac 12 (A + B) \cos \frac 12 (A - B)\)
\(\sin A - \sin B = 2 \cos \frac 12 (A + B) \sin \frac 12 (A - B)\)
\(\cos A + \cos B = 2 \cos \frac 12 (A + B) \cos \frac 12 (A - B)\)
\(\cos A - \cos B = -2 \sin \frac 12 (A + B) \sin \frac 12 (A - B)\)
\(\tan A + \tan B = \tan (A + B) \:.\: [1 - \tan A \tan B ]\)
\(\tan A - \tan B = \tan (A - B) \:.\: [1 + \tan A \tan B ]\)
Contoh 01
\begin{equation*} \begin{split} & \lim_{x\rightarrow \: 0} \: \frac{\sin x - \sin 3x}{x \cos x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2 \:.\: \cos \frac 12 (x + 3x) \:.\: \sin \frac 12 (x - 3x)}{x \cos x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2 \:.\: \cos 2x \:.\: \sin (-x)}{x \cos x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{-2 \:.\: \cos 2x \:.\: \sin x}{x \cos x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{-2 \:.\: \cos 2x \:.\: \cancel {\sin x}}{\cancel{x} \cos x} \\\\ & \frac {-2 \cos 0}{\cos 0} \\\\ & \bbox[5px, border: 2px solid magenta] {-2} \end{split} \end{equation*}
Contoh 02
\begin{equation*} \begin{split} & \lim_{x\rightarrow \: 0} \: \frac{\cos x-\cos 3x}{1-\cos 2x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{-2 \:.\: \sin \frac 12 (x + 3x) \:.\: \sin \frac 12 (x - 3x)}{1- (1 - 2 \:.\: \sin^2 x)} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{-2 \:.\: \sin 2x \:.\: \sin (-x)}{1- 1 + 2 \:.\: \sin^2 x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2 \:.\: \sin 2x \:.\: \sin x}{2 \:.\: \sin x \:.\: \sin x} \\\\ & \lim_{x\rightarrow \: 0} \: \frac{2 \:.\: \cancel{\sin 2x} \:.\: \cancel{\sin x}}{2 \:.\: \cancel{\sin x} \:.\: \cancel{\sin x}} \\\\ & \bbox[5px, border: 2px solid magenta] {2} \end{split} \end{equation*}
F. LIMIT TAK HINGGA
Limit trigonometri bentuk tak hingga dapat diselesaikan dengan mengubah variabelnya menjadi mendekati 0.
Contoh
\(\displaystyle \lim_{x \rightarrow \: \sim} \: x \:.\: \sin{\frac{1}{x}}\)
Misalkan \(y = \dfrac 1x\) sehingga \(x = \dfrac 1y\)
Untuk \(x \rightarrow \: \sim\) maka \(y \rightarrow 0\)
Penyelesaian
\begin{equation*} \begin{split} & \lim_{x \rightarrow \: \sim} \: x \:.\: \sin{\frac{1}{x}} \\\\ & \lim_{y \rightarrow \: 0} \: \frac 1y \:.\: \sin y \\\\ & \lim_{y \rightarrow \: 0} \: \frac {\sin y}{y} \\\\ & \bbox[5px, border: 2px solid magenta] {1} \end{split} \end{equation*}