Evaluate the limit: lim _{x rightarrow 0} fra | vedclass

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(B) Given the limit: $L = \lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$We can rewrite the expression as: $L = \lim _{x \rightarrow 0} \f

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(B) Given the limit: $L = \lim _{x ightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$We can rewrite the expression as: $L = \lim _{x ightarrow 0} \frac{e^{\sin x}(e^{x-\sin x}-1)}{2(x-\sin x)}$Using the standard limit $\lim _{u ightarrow 0} \frac{e^u-1}{u} = 1$,where $u = x - \sin x$:As $x ightarrow 0$,$u = x - \sin x ightarrow 0$.Therefore,the limit becomes: $L = \lim _{x ightarrow 0} \frac{e^{\sin x}}{2} 	imes \lim _{u ightarrow 0} \frac{e^u-1}{u}$$L = \frac{e^0}{2} 	imes 1 = \frac{1}{2} 	imes 1 = \frac{1}{2}$

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

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