lim _{x rightarrow 0} frac{x tan 4x - 2x tan | vedclass

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

Question

(B) We use the Taylor series expansion for $	an 	heta = 	heta + \frac{	heta^3}{3} + \frac{2	heta^5}{15} + \dots$ and $1 - \cos 	heta = 2 \sin^2(

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(B) We use the Taylor series expansion for $	an 	heta = 	heta + \frac{	heta^3}{3} + \frac{2	heta^5}{15} + \dots$ and $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2}) \approx 2(\frac{	heta}{2})^2 = \frac{	heta^2}{2}$.The denominator is $(1 - \cos 4x)^2 \approx (\frac{(4x)^2}{2})^2 = (8x^2)^2 = 64x^4$.The numerator is $x(4x + \frac{(4x)^3}{3} + \dots) - 2x(2x + \frac{(2x)^3}{3} + \dots)$.$= (4x^2 + \frac{64x^4}{3}) - (4x^2 + \frac{16x^4}{3}) = \frac{48x^4}{3} = 16x^4$.Thus,the limit is $\lim _{x ightarrow 0} \frac{16x^4}{64x^4} = \frac{16}{64} = \frac{1}{4}$.

$\lim _{x \rightarrow 0} \frac{x \tan 4x - 2x \tan 2x}{(1 - \cos 4x)^2} = $

Similar Questions

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{\tan ^2(\pi \sec ^4 x)}{\pi^2 x^4}$

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

The value of $ \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} $ is

Evaluate the given limit: $\mathop {\lim }\limits_{x \to \pi } \frac{\sin (\pi-x)}{\pi(\pi-x)}$

$\lim _{t}$ ${\rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots +n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ is equal to $.......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module