Tipe Soal Limit Trigonometri

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

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

Pakai manipulasi bentuk sehingga bisa dicoret :

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

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

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