lim _{x rightarrow 0} frac{e^x-e^{sin x}}{2(x | vedclass

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(B) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$By factoring out $e^{\sin x}$ in the numerator,we get:$= \lim _

Vedclass · Accepted Answer

$1/2$

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(B) We evaluate the limit: $\lim _{x ightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$By factoring out $e^{\sin x}$ in the numerator,we get:$= \lim _{x ightarrow 0} \frac{e^{\sin x}(e^{x-\sin x}-1)}{2(x-\sin x)}$Using the standard limit $\lim _{u ightarrow 0} \frac{e^u-1}{u} = 1$,where $u = x - \sin x$:$= \lim _{x ightarrow 0} \left( \frac{e^{\sin x}}{2} \cdot \frac{e^{x-\sin x}-1}{x-\sin x} ight)$As $x ightarrow 0$,$e^{\sin x} ightarrow e^0 = 1$ and $\frac{e^{x-\sin x}-1}{x-\sin x} ightarrow 1$.Therefore,the limit is $\frac{1}{2} 	imes 1 = \frac{1}{2}$.

$\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

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