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$$\lim_{x \to 0}\frac{x+ x\cos(x)}{\sin(x)\cos(x)} $$

This question is very easy by just using L'Hopital's, but I'm trying to do the question without it but I'm stuck on what to do. Any help would be greatly appreciated.

Hanul Jeon
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3 Answers3

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$$\lim_{x\to 0}\frac{x+x\cos x}{\sin x\cos x} = \lim_{x\to 0}\frac {x}{\sin x}\cdot\frac{1 +\cos x}{\cos x} = 1\cdot\frac 21$$

player3236
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$\lim\limits_{x\to0}\dfrac{x+x\cos x}{\sin x\cos x}=\lim\limits_{x\to0}\left(\dfrac x{\sin x}\dfrac1{\cos x}+ \dfrac x{\sin x}\right)=\lim\limits_{x\to0}\dfrac x{\sin x}\lim\limits_{x\to0}\dfrac1{\cos x}+\lim\limits_{x\to0}\dfrac x{\sin x}$

$=1+1=2$

J. W. Tanner
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Another way:

$$\frac{x+x\cos x}{\sin x\cos x}=\frac{x(1+\cos x)}{\frac12\sin 2x}=\frac{2x}{\sin 2x}(1+\cos x)\xrightarrow[x\to0]{}1\cdot(1+1)=2$$

DonAntonio
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