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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Similar Questions

$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :

$\lim _{x \rightarrow 3} \frac{(84-x)^{\frac{1}{4}}-3}{x-3}$ is

If $f$ is a strictly increasing function,then $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}$ is equal to

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x)\tan x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module