JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 0}}(\cos a x)^{\operatorname{cosec}^2 b x}

View answer and explanation

Correct option: C

e^{\left(-\frac{a^2}{2 b^2}\right)}

Option C (

e^{\left(-\frac{a^2}{2 b^2}\right)}

) is correct because the limit can be solved by taking the natural logarithm of the expression and applying L'Hopital's rule or using trigonometric identities.

More Questions from LIMITS

Practice All Chapter MCQs →

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

is equal to

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

The value of

\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)

is equal to If

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

, then the value of

f^{\prime}(0)

is equal to

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

equals