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 know that $\mathop {\lim }\limits_{u 	o 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{x 	o 0} \frac{	an x - x}{x^3}

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(C) We know that $\mathop {\lim }\limits_{u 	o 0} \frac{1 - \cos u}{u^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{x 	o 0} \frac{	an x - x}{x^3} = \frac{1}{3}$.Given expression: $L = \mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos (1 - \cos x)}{x(	an x - x)}$.Multiply and divide by $(1 - \cos x)^2$ and $x^4$:$L = \mathop {\lim }\limits_{x 	o 0} \left[ \frac{1 - \cos (1 - \cos x)}{(1 - \cos x)^2} ight] 	imes \left[ \frac{(1 - \cos x)^2}{x^4} ight] 	imes \left[ \frac{x^4}{x(	an x - x)} ight]$.$L = \frac{1}{2} 	imes \left( \mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos x}{x^2} ight)^2 	imes \left( \mathop {\lim }\limits_{x 	o 0} \frac{x^3}{	an x - x} ight)$.$L = \frac{1}{2} 	imes \left( \frac{1}{2} ight)^2 	imes \left( \frac{1}{1/3} ight) = \frac{1}{2} 	imes \frac{1}{4} 	imes 3 = \frac{3}{8}$.

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

Similar Questions

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ is equal to :

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - {x^2}} - \sqrt {1 + {x^2}} }}{{{x^2}}}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}} = $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}} = $

$\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module