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

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

Question

(D) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{3^x(1 - \cos x) - (1 - \cos x)}$$= \lim _{x \rightarrow 0} \frac{x^2(2^x -

Vedclass · Accepted Answer

$\frac{2}{\log 3}(\log 2-1)$

Vedclass · Answer

(D) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{3^x(1 - \cos x) - (1 - \cos x)}$$= \lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{(3^x - 1)(1 - \cos x)}$$= \lim _{x \rightarrow 0} \frac{2^x - \sin x - 1}{(3^x - 1) \frac{(1 - \cos x)}{x^2}}$Since $\lim _{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,the expression becomes:$= \lim _{x \rightarrow 0} \frac{2(2^x - \sin x - 1)}{3^x - 1}$Applying $L$'Hospital's Rule:$= \lim _{x \rightarrow 0} \frac{2(2^x \ln 2 - \cos x)}{3^x \ln 3}$$= \frac{2(\ln 2 - 1)}{\ln 3}$$= \frac{2}{\log 3}(\log 2 - 1)$

$\lim _{x \rightarrow 0} \frac{x^2 2^x-x^2 \sin x-x^2}{3^x+\cos x-3^x \cos x-1}=$

Similar Questions

$\mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \frac{\frac{\pi }{2} - \theta}{\cot \theta} =$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x}$ is

Evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x (\tan^{-1} t)^2 dt}}{{\sin x - x}}$

$\lim _{x \rightarrow \pi / 2}(\sec x-\tan x)$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module