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Tipe Soal Limit Trigonometri

3 HAL YANG HARUS DIINGAT

1. \displaystyle \lim_{x \rightarrow 0}\frac{\sin x}{x}=\lim_{x \rightarrow 0}\frac{x}{\sin x}=1

Karena menurut aturan L’hopital :

\displaystyle \lim_{x \rightarrow 0}\frac{\sin x}{x}=\lim_{x \rightarrow 0}\frac{\cos x}{1}=1

2. \displaystyle \lim_{x \rightarrow 0}\frac{1-\cos x}{x}=0

Karena menurut aturan L’hopital :

\displaystyle \lim_{x \rightarrow 0}\frac{1-\cos x}{x}=\lim_{x \rightarrow 0}\frac{\sin x}{1}=0

3. \displaystyle \lim_{x \rightarrow 0}\frac{\tan x}{x}=\lim_{x \rightarrow 0}\frac{x}{\tan x}=1

Karena menurut aturan L’hopital :

\displaystyle \lim_{x \rightarrow 0}\frac{\tan x}{x}=\lim_{x \rightarrow 0}\frac{\sec^2 x}{1}=1


5 TIPE SOAL LIMIT TRIGONOMETRI

1. \displaystyle \lim_{x \rightarrow \pi}\frac{\sin (x-\pi)}{(x-\pi)}=1

Kita analogikan soal di atas seperti

\displaystyle \lim_{x \rightarrow 0}\frac{\sin x}{x}=1

2. \displaystyle \lim_{x \rightarrow 0}\frac{\sin 2x}{1-\sqrt{1-x}}=

Pakai aturan L’hopital

\displaystyle \lim_{x \rightarrow 0}\frac{\sin 2x}{1-\sqrt{1-x}}= \lim_{x \rightarrow 0}\frac{(\cos 2x)(2)}{-\frac{1}{2}(1-x)^{-1/2}}=4

3. \displaystyle \lim_{x \rightarrow 2}\frac{1-\cos^2(x-2)}{x^2-4x+4}}=

Pakai identitas trigonometri :

\displaystyle \lim_{x \rightarrow 2}\frac{1-\cos^2(x-2)}{x^2-4x+4}}=\lim_{x \rightarrow 2}\frac{\sin^2(x-2)}{(x-2)^2}}=1

4. \displaystyle \lim_{x \rightarrow 0} \frac{\sin 2x \tan^2 3x+6x^3}{2x^2\sin 3x \cos 2x}

Pakai manipulasi bentuk sehingga bisa dicoret :

\displaystyle \lim_{x \rightarrow 0} \frac{\sin 2x \tan^2 3x+6x^3}{2x^2\sin 3x \cos 2x}=\frac{2.3^2}{2.3}+\frac{6}{2.3}=4

5. \displaystyle \lim_{x \rightarrow \frac {\pi}{2}}\frac{\cos^2x}{(\frac{\pi}{2}-x)^2 \sin x}

Pakai kesamaan bentuk trigonometri :

\displaystyle \lim_{x \rightarrow \frac {\pi}{2}}\frac{\cos^2x}{(\frac{\pi}{2}-x)^2 \sin x}=\lim_{x \rightarrow \frac {\pi}{2}}\frac{\sin^2(\frac{\pi}{2}-x)}{(\frac{\pi}{2}-x)^2 \sin x}=1

Sumber : pinterest

Belajarlah agama dengan tekun, niscaya akan menjadi pelita di saat gelap.

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