lim _{x rightarrow 0} frac{tan x - sin x}{x^3 | vedclass

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

Question

(C) We evaluate the limit $L = \lim _{x ightarrow 0} \frac{	an x - \sin x}{x^3}$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'Hosp

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) We evaluate the limit $L = \lim _{x ightarrow 0} \frac{	an x - \sin x}{x^3}$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'Hospital's Rule:$L = \lim _{x ightarrow 0} \frac{\sec^2 x - \cos x}{3x^2}$.This is still a $\frac{0}{0}$ form. Applying $L$'Hospital's Rule again:$L = \lim _{x ightarrow 0} \frac{2 \sec^2 x 	an x + \sin x}{6x}$.Using the standard limits $\lim _{x ightarrow 0} \frac{	an x}{x} = 1$ and $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$:$L = \lim _{x}$ ${ightarrow 0} \left( \frac{2 \sec^2 x \cdot 	an x}{6x} + \frac{\sin x}{6x} ight) = \frac{2(1)(1)}{6} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

$\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$ is equal to

Similar Questions

$\lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1} = $

$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ is

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\frac{\pi }{2}}^x t \,dt}}{{\sin (2x - \pi )}}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{\frac{1}{3}}}} - 3}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$ equals.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module