mathop {lim }limits_{x to 0} frac{{{{left( {1 | vedclass

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

Question

(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{\left( {1 - \cos 2x} ight)}^2}}}{{2x	an x - x	an 2x}}$.Using the identity $1 - \cos 2x = 2\

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{\left( {1 - \cos 2x} ight)}^2}}}{{2x	an x - x	an 2x}}$.Using the identity $1 - \cos 2x = 2\sin^2 x$,we have:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{\left( {2\sin^2 x} ight)}^2}}}{{2x	an x - x	an 2x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4\sin^4 x}}{{2x	an x - x	an 2x}}$.Using the Taylor series expansions $\sin x = x - \frac{x^3}{6} + O(x^5)$ and $	an x = x + \frac{x^3}{3} + O(x^5)$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{4(x - \frac{x^3}{6})^4}}{{2x(x + \frac{x^3}{3}) - x(2x + \frac{(2x)^3}{3})}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4x^4}}{{2x^2 + \frac{2x^4}{3} - 2x^2 - \frac{8x^4}{3}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4x^4}}{-\frac{6x^4}{3}} = \mathop {\lim }\limits_{x 	o 0} \frac{{4x^4}}{-2x^4} = -2$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {1 - \cos 2x} \right)}^2}}}{{2x\tan x - x\tan 2x}}$ is

Similar Questions

$\mathop {\lim }\limits_{x \to 1} [x] = $

For $x \in R$,$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}$ is equal to

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$ is equal to

Let $p = \mathop {\lim }\limits_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$,then $\log p = $ . . .

Assertion $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
Reason $(R)$: As the value of $x$ decreases,the value of $\frac{1}{x}$ increases.

Vedclass Test Series

Exam Paper Generator

Online Exam Module