The value of mathop {lim }limits_{x to 0} fra | vedclass

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

Question

(C) We have $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2 \sin^2

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) We have $\mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2 \sin^2(x^2/2)} }}{{2 \sin^2(x/2)}}$.Since $\sqrt{\sin^2 	heta} = |\sin 	heta|$,and for $x 	o 0$,$\sin(x^2/2) > 0$,we have:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt{2} \sin(x^2/2)}}{{2 \sin^2(x/2)}} = \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \frac{\sin(x^2/2)}{\sin^2(x/2)}$.Multiply and divide by $(x^2/2)$ and $(x/2)^2$:$= \frac{1}{\sqrt{2}} \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{(\sin(x/2)/(x/2))^2 \cdot (x^2/4)} ight)$.$= \frac{1}{\sqrt{2}} \cdot \frac{1/2}{1/4} \cdot \mathop {\lim }\limits_{x 	o 0} \frac{1}{1^2} = \frac{1}{\sqrt{2}} \cdot 2 = \sqrt{2}$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ is

Similar Questions

$\lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} = $

If $x$ is a real number in $[0, 1]$,then the value of $\lim_{m \to \infty} \lim_{n \to \infty} [1 + \cos^{2m}(n! \pi x)]$ is given by

$\lim\limits_{x \rightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{-x / 2}-3^{1-x}}$ is equal to

$\lim _{x \rightarrow 0} \frac{|x|}{|x|+x^2} = $

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{{y^2}}}{x}$,where ${y^2} = ax + b{x^2} + c{x^3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module