JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{2^{|x|} e^{|x|}-|x|-|x| \log _e 2-1}{x \tan x}

is equal to

View answer and explanation

Correct option: D

\frac{1}{2}(\ln 2)^2+(\ln 2)+\frac{1}{2}

Option D (

\frac{1}{2}(\ln 2)^2+(\ln 2)+\frac{1}{2}

) is correct because it is obtained by using the Taylor series expansion of

a^{|x|}

and simplifying the limit.

More Questions from LIMITS

Practice All Chapter MCQs →

\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}=

\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\}

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}

f^{\prime}(0)

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}