JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\left( {1 - {e^x}} \right)\sin x}}{{{x^2} + {x^3}}}

is equal to

View answer and explanation

Correct option: A

- 1

Option A (

-1

) is correct because the limit can be evaluated by using Taylor series expansions for

e^x

and

\sin x

or by applying L'Hopital's rule.

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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\}

\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}