Question

evaluate limit X tends to zero sin 2x / sin x ​

Answer 1

Given,

$\displaystyle \lim_{x\to0}\dfrac {\sin (2x)}{\sin x}$

On taking x directly as 0, we get the indeterminate form.

$\dfrac {\sin(2\times 0)}{\sin 0}=\dfrac {\sin0}{\sin 0}=\dfrac {0}{0}$

But, we know that,

$\sin (2x)=2\sin x\cos x$

Then,

$\displaystyle \lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=\lim_{x\to0}\dfrac {2\sin x\cos x}{\sin x}\\\\\\\lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=\lim_{x\to0}2\cos x\\\\\\\lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=2\cos 0\\\\\\\lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=\mathbf {2}$

Or we can use L'hospital's Rule.

$\displaystyle \lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=\lim_{x\to0}\dfrac {(\sin (2x))'}{(\sin x)'}\\\\\\\lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=\lim_{x\to0}\dfrac {2\cos (2x)}{\cos x}\\\\\\\lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=\dfrac {2\cos (2\times0)}{\cos 0}\\\\\\\lim_{x\to0}\dfrac {\sin (2x)}{\sin x}=\mathbf {2}$

where $f'$ is the first derivative of $f$ wrt $x$.

However 2 is the answer.

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