JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because by rewriting the expression and applying L'Hopital's rule, the limit evaluates to 1.

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\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\}

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

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\lim \limits_{{x \rightarrow a}}\left[\left\{\left(\frac{\sqrt{a}+\sqrt{x}}{a^{1 / 4}-x^{1 / 4}}\right)^{-1}-\frac{2(a x)^{1 / 4}}{x^{3 / 4}-a^{1 / 4} x^{1 / 2}+a^{1 / 2} x^{1 / 4}-a^{3 / 4}}\right\}^{-1}\right.\left.-(\sqrt{2})^{\log _4 a}\right]^8