mathop {Limit}limits_{x to 0} {(cos 2x)^{3/x^ | vedclass

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

Question

(D) Let $L = \mathop {Limit}\limits_{x 	o 0} (\cos 2x)^{3/x^2}$.This is of the form $1^\infty$,so we use the formula $\mathop {Limit}\limits_{x 	o a

Vedclass · Accepted Answer

$e^{-6}$

Vedclass · Answer

(D) Let $L = \mathop {Limit}\limits_{x 	o 0} (\cos 2x)^{3/x^2}$.This is of the form $1^\infty$,so we use the formula $\mathop {Limit}\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {Limit}\limits_{x 	o a} g(x)(f(x)-1)}$.$L = e^{\mathop {Limit}\limits_{x 	o 0} \frac{3}{x^2}(\cos 2x - 1)}$.Using the identity $\cos 2x - 1 = -2\sin^2 x$,we get:$L = e^{\mathop {Limit}\limits_{x 	o 0} \frac{3}{x^2}(-2\sin^2 x)}$.$L = e^{-6 \mathop {Limit}\limits_{x 	o 0} (\frac{\sin x}{x})^2}$.Since $\mathop {Limit}\limits_{x 	o 0} \frac{\sin x}{x} = 1$,we have $L = e^{-6(1)^2} = e^{-6}$.

$\mathop {Limit}\limits_{x \to 0} {(\cos 2x)^{3/x^2}}$ has the value equal to . . . . . . .

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x} = $

Let $f(2) = 4$ and $f'(2) = 4$,then $\mathop {\lim }\limits_{x \to 2} \,\frac{{xf(2) - 2f(x)}}{{x - 2}}$ equals

Let $f: R \rightarrow R$ be a function such that $f(2)=4$ and $f^{\prime}(2)=1$. Then,the value of $\lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$ is equal to:

The value of the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$ is

$\lim _{x \rightarrow 0} \left( \frac{x \cdot 10^x - x}{1 - \cos x} \right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module