Limit Fungsi Trigonometri

Dalam menyelesaikan limit fungsi trigonometri memerlukan beberapa cara sebagai berikut:

A. Metode Substitusi

Menentukan nilai limit fungsi trigonometri dengan cara substitusi secara langsung berlaku jika hasil yang diperoleh terdefinisi atau bukan bentuk tak tentu $\left( \frac{0}{0} \right)$.

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Contoh 1.
Tentukan nilai $\lim\limits_{x \to \frac{\pi }{3}}\frac{2\tan x}{\sin^2x-\cos^2x}$.
Penyelesaian:
$\begin{align}\lim\limits_{x \to \frac{\pi }{3}}\frac{2\tan x}{\sin^2x-\cos^2x} &= \frac{2\tan \frac{\pi }{3}}{\sin^2\frac{\pi }{3}-\cos^2\frac{\pi }{3}} \\ &= \frac{2\cdot \sqrt{3}}{\left( \frac{1}{2}\sqrt{3} \right)^2-\left( \frac{1}{2} \right)^2} \\ &= \frac{2\sqrt{3}}{\frac{3}{4}-\frac{1}{4}} \\ &= \frac{2\sqrt{3}}{\frac{2}{4}} \\ &= 4\sqrt{3} \end{align}$

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Contoh 2.
Tentukan nilai $\lim\limits_{x \to \frac{\pi }{4}}\frac{\tan^2x}{2\sin x-2\cos x}$.
Penyelesaian:
$\begin{align}\lim\limits_{x \to \frac{\pi }{4}}\frac{\tan^2x}{2\sin x-2\cos x} &= \frac{\tan^2\frac{\pi }{4}}{2\sin \frac{\pi }{4}-2\cos \frac{\pi }{4}} \\ &= \frac{1^2}{2\cdot \frac{1}{2}\sqrt{2}-2\cdot \frac{1}{2}\sqrt{2}} \\ &= \frac{1}{0} \\ &= \infty \end{align}$

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B. Metode Penyederhanaan

Jika setelah mencoba menyelesaian dengan cara substitusi langsung memperoleh nilai dalam bentuk tak tentu $\frac{0}{0}$, maka salah satunya dengan menggunakan metode penyederhanaan.

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Contoh 1.
Tentukan nilai $\lim\limits_{x \to \frac{\pi }{4}}\frac{4\cos 2x}{\cos x-\sin x}$.
Penyelesaian:
Jika menggunakan metode substitusi diperoleh:
$\begin{align}\lim\limits_{x \to \frac{\pi }{4}}\frac{4\cos 2x}{\cos x-\sin x} &= \frac{4\cos \left( 2\cdot \frac{\pi }{4} \right)}{\cos \frac{\pi }{4}-\sin \frac{\pi }{4}} \\ &= \frac{4\cos \left( \frac{\pi }{2} \right)}{\cos \frac{\pi }{4}-\sin \frac{\pi }{4}} \\ &= \frac{4\cdot 0}{\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}} \\ &= \frac{0}{0}\,(\text{tak tentu}) \end{align}$
Oleh karena itu, untuk menyelesaikan limit fungsi trigonometri tersebut diperlukan metode penyederhanaan sebagai berikut:
$\lim\limits_{x \to \frac{\pi }{4}}\frac{4\cos 2x}{\cos x-\sin x}$
= $\lim\limits_{x \to \frac{\pi }{4}}\frac{4(\cos^2x-\sin^2x)}{\cos x-\sin x}$
= $\lim\limits_{x \to \frac{\pi }{4}}\frac{4(\cos x+\sin x)(\cos x-\sin x)}{\cos x-\sin x}$
= $\lim\limits_{x \to \frac{\pi }{4}}4(\cos x+\sin x)$
= $4\left( \cos \frac{\pi }{4}+\sin \frac{\pi }{4} \right)$
= $4\left( \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2} \right)$
= $4\sqrt{2}$

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Contoh 2.
Tentukan nilai $\lim\limits_{x \to \frac{\pi }{2}}\frac{\sin 2x}{\cos x}$.
Penyelesaian:
Jika menggunakan metode substitusi diperoleh:
$\begin{align}\lim\limits_{x \to \frac{\pi }{2}}\frac{\sin 2x}{\cos x} &= \frac{\sin \left( 2\cdot \frac{\pi }{2} \right)}{\cos \frac{\pi }{2}} \\ &= \frac{\sin \pi }{\cos \frac{\pi }{2}} \\ &= \frac{0}{0}\,(\text{tak tentu}) \end{align}$
Oleh karena itu, untuk menyelesaikan limit fungsi trigonometri tersebut diperlukan metode penyederhanaan sebagai berikut:
$\begin{align}\lim\limits_{x \to \frac{\pi }{2}}\frac{\sin 2x}{\cos x} &= \lim\limits_{x \to \frac{\pi }{2}}\frac{2\sin x\cdot \cos x}{\cos x} \\ &= \lim\limits_{x \to \frac{\pi }{2}}2\sin x \\ &= 2\sin \frac{\pi }{2} \\ &= 2\cdot 1 \\ &= 2 \end{align}$

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C. Menggunakan Rumus Limit Trigonometri

Rumus Dasar

1. $\lim\limits_{x \to 0}\frac{\sin ax}{bx}=\lim\limits_{x \to 0}\frac{ax}{\sin bx}=\frac{a}{b}$
2. $\lim\limits_{x \to 0}\frac{\tan ax}{bx}=\lim\limits_{x \to 0}\frac{ax}{\tan bx}=\frac{a}{b}$
3. $\lim\limits_{x \to 0}\frac{\sin ax}{\sin bx}=\lim\limits_{x \to 0}\frac{\tan ax}{\tan bx}=\frac{a}{b}$
4. $\lim\limits_{x \to 0}\frac{\sin ax}{\tan bx}=\lim\limits_{x \to 0}\frac{\tan ax}{\sin bx}=\frac{a}{b}$

Rumus Pengembangan

1. $\lim\limits_{x \to k}\frac{\sin a(x-k)}{b(x-k)}=\lim\limits_{x \to k}\frac{a(x-k)}{\sin b(x-k)}=\frac{a}{b}$
2. $\lim\limits_{x \to k}\frac{\tan a(x-k)}{b(x-k)}=\lim\limits_{x \to k}\frac{a(x-k)}{\tan b(x-k)}=\frac{a}{b}$
3. $\lim\limits_{x \to k}\frac{\sin a(x-k)}{\sin b(x-k)}=\lim\limits_{x \to k}\frac{\tan a(x-k)}{\tan b(x-k)}=\frac{a}{b}$
4. $\lim\limits_{x \to k}\frac{\sin a(x-k)}{\tan b(x-k)}=\lim\limits_{x \to k}\frac{\tan a(x-k)}{\sin b(x-k)}=\frac{a}{b}$

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Contoh 1.
Tentukan nilai $\lim\limits_{x \to 0}\frac{\sin 3x+\tan 4x}{6x+\sin 8x}$.
Penyelesaian:
$\begin{align}\lim\limits_{x \to 0}\frac{\sin 3x+\tan 4x}{6x+\sin 8x} &= \lim\limits_{x \to 0}\frac{\frac{\sin 3x}{x}+\frac{\tan 4x}{x}}{\frac{6x}{x}+\frac{\sin 8x}{x}} \\ &= \frac{\frac{3}{1}+\frac{4}{1}}{\frac{6}{1}+\frac{8}{1}} \\ &= \frac{7}{14} \\ &= \frac{1}{2} \end{align}$

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Contoh 2.
Tentukan nilai $\lim\limits_{x \to 0}\frac{x^2}{\cos 2x-\cos 4x}$.
Penyelesaian:
$\lim\limits_{x \to 0}\frac{x^2}{\cos 2x-\cos 4x}$
= $\lim\limits_{x \to 0}\frac{x^2}{-2\sin \left( \frac{2x+4x}{2} \right)\cdot \sin \left( \frac{2x-4x}{2} \right)}$
= $\lim\limits_{x \to 0}\frac{x^2}{-2\sin 3x\cdot \sin (-x)}$
= $\lim\limits_{x \to 0}\frac{1}{-2}\cdot \frac{x}{\sin 3x}\cdot \frac{x}{\sin (-x)}$
= $\frac{1}{-2}\cdot \frac{1}{3}\cdot \frac{1}{-1}$
= $\frac{1}{6}$

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Contoh 3.
Tentukan nilai $\lim\limits_{x \to 0}\frac{\sin 3x-\sin 3x\cos 8x}{10x^2\sin 6x}$.
Penyelesaian:
$\lim\limits_{x \to 0}\frac{\sin 3x-\sin 3x\cos 8x}{10x^2\sin 6x}$
= $\lim\limits_{x \to 0}\frac{\sin 3x(1-\cos 8x)}{10x^2\sin 6x}$
= $\lim\limits_{x \to 0}\frac{\sin 3x\cdot 2\sin^2\left( \frac{8x}{2} \right)}{10x^2\sin 6x}$
= $\lim\limits_{x \to 0}\frac{\sin 3x\cdot \sin^24x}{5x^2\sin 6x}$
= $\lim\limits_{x \to 0}\frac{\sin 3x}{5x}\cdot \frac{\sin 4x}{x}\cdot \frac{\sin 4x}{\sin 6x}$
= $\frac{3}{5}\cdot \frac{4}{1}\cdot \frac{4}{6}$
= $\frac{8}{5}$

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Contoh 4.
Tentukan nilai $\lim\limits_{x \to 1}\frac{1-\cos^2(3x-3)}{4(x^2-2x+1)}$.
Penyelesaian:
$\lim\limits_{x \to 1}\frac{1-\cos^2(3x-3)}{4(x^2-2x+1)}$
= $\lim\limits_{x \to 1}\frac{\sin^2(3x-3)}{4(x^2-2x+1)}$
= $\lim\limits_{x \to 1}\frac{\sin^23(x-1)}{4(x-1)(x-1)}$
= $\lim\limits_{x \to 1}\frac{\sin 3(x-1)}{4(x-1)}\cdot \frac{\sin 3(x-1)}{(x-1)}$
= $\frac{3}{4}\cdot \frac{3}{1}$
= $\frac{9}{4}$

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Contoh 5.
Tentukan nilai $\lim\limits_{x \to 3}\frac{\sin (4x-12)}{2(x-3)+\tan (5x-15)}$.
Penyelesaian:
$\lim\limits_{x \to 3}\frac{\sin (4x-12)}{2(x-3)+\tan (5x-15)}$
= $\lim\limits_{x \to 3}\frac{\sin 4(x-3)}{2(x-3)+\tan 5(x-3)}$
= $\lim\limits_{x \to 3}\frac{\frac{\sin 4(x-3)}{(x-3)}}{\frac{2(x-3)}{(x-3)}+\frac{\tan 5(x-3)}{(x-3)}}$
= $\frac{\frac{4}{1}}{\frac{2}{1}+\frac{5}{1}}$
= $\frac{4}{7}$

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D. Soal Latihan

1. Tentukan nilai $\lim\limits_{x \to \frac{\pi }{3}}\frac{2\tan x-\sin x}{\sin \frac{1}{2}x}$.
2. Tentukan nilai $\lim\limits_{x \to 0}\frac{5x^2}{\sin^24x}$.
3. Tentukan nilai $\lim\limits_{x \to \frac{\pi }{4}}\frac{\sin x-\cos x}{1-\tan x}$.
4. Tentukan nilai $\lim\limits_{x \to 0}\left( \frac{2}{x^2}-\frac{\sin 2x}{x^2\tan x} \right)$.
5. Tentukan nilai $\lim\limits_{x \to 2}\frac{1-\cos^2(x-2)}{3x^2-12x+12}$.

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