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

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(2^\text{x}-1\big)}{1-\cos\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(2^\text{x}-1\big)}{1-\cos\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(2^\text{x}-1\big)}{2\sin^2\big(\frac{\text{x}}{2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(2^\text{x}-1\big)}{\text{x}}\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{\begin{pmatrix}\frac{\sin\big(\frac{\text{x}}{2}\big)}{\frac{\text{x}}{2}}\end{pmatrix}^2\times\frac{\text{x}^2}{2}}$
$=\text{2log2}$
$=\text{log4}$

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