The value of lim _{x rightarrow 0} frac{(1-co | vedclass

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

Question

(A) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this in the limit,we get: $\lim _{x ightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x 	an 4x}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) We know that $1 - \cos 2x = 2 \sin^2 x$. Substituting this in the limit,we get: $\lim _{x ightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x 	an 4x}$ $= \lim _{x}$ ${ightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{x}{	an 4x} \cdot (3 + \cos x) ight]$ $= \lim _{x}$ ${ightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{1}{4 \cdot \frac{	an 4x}{4x}} \cdot (3 + \cos x) ight]$ Using the standard limits $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$ and $\lim_{	heta ightarrow 0} \frac{	an 	heta}{	heta} = 1$: $= 2 \cdot (1)^2 \cdot \frac{1}{4 \cdot 1} \cdot (3 + \cos 0)$ $= 2 \cdot \frac{1}{4} \cdot (3 + 1)$ $= \frac{2}{4} \cdot 4 = 2$

The value of $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ is

Similar Questions

If $a > 0$ and $b < 0$,then $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$ is equal to:

$\mathop {\lim }\limits_{x \to 0} \frac{{{x^2} - \tan 2x}}{{\tan x}} = $

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)+\cos (2020 x)]}$

$\mathop {\lim }\limits_{\theta \to 0} \frac{{5\theta \cos \theta - 2\sin \theta }}{{3\theta + \tan \theta }} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^{\sin x}} - 1}}{{{b^{\sin x}} - 1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module