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  4. displaystyle lim x arrow 0 (ex-e-x-2 x/x- sin x) is equal to

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Q. $\displaystyle\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}$ is equal to

Limits and Derivatives

Solution:

$\displaystyle\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{1-\cos x}$(Using L'Hospital's Rule)
$=\displaystyle\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin x}$(Using L’Hospital’s Rule)
$=\displaystyle\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}}{\cos x}$ (Using L’Hospital’s Rule)
$=2$