Evaluate the following limit: _ x 0 x 1- 2x - Vidyadip

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\tan\text{x}}{1-\cos2\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}\tan\text{x}}{2\sin^2\text{x}}$
$=\lim\limits_{\text{x} \rightarrow0}\frac{\frac{\tan\text{x}}{\text{x}}}{\frac{2\sin^2\text{x}}{\text{x}^2}}$
$=\frac{\lim\limits_{\text{x} \rightarrow0}\frac{\tan\text{x}}{\text{x}}}{2\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$
$=\frac{1}{2\times1}$
$=\frac{1}{2}$

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