The value of lim _{x rightarrow 0} frac{cos ( | vedclass

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

Question

(B) Let $L = \lim _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1 - \frac{	heta^2}{2!}

Vedclass · Accepted Answer

$1/6$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1 - \frac{	heta^2}{2!} + \frac{	heta^4}{4!} - O(	heta^6)$,we have:$\sin x = x - \frac{x^3}{6} + O(x^5)$$\cos(\sin x) = 1 - \frac{(x - x^3/6)^2}{2} + \frac{(x - x^3/6)^4}{24} = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{5x^4}{24}$.$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24}$.Substituting these into the limit:$L = \lim _{x ightarrow 0} \frac{(1 - \frac{x^2}{2} + \frac{5x^4}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24})}{x^4}$.$L = \lim _{x ightarrow 0} \frac{\frac{5x^4}{24} - \frac{x^4}{24}}{x^4} = \lim _{x ightarrow 0} \frac{4x^4/24}{x^4} = \frac{4}{24} = \frac{1}{6}$.

The value of $\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to

Similar Questions

$\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = $

If $\lim _{x \rightarrow 0}\left(\frac{1+cx}{1-cx}\right)^{1/x}=4$,then $\lim _{x \rightarrow 0}\left(\frac{1+2cx}{1-2cx}\right)^{1/x}$ is

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$

$\mathop {\lim }\limits_{x \to 0} {(\cos mx)^{n/{x^2}}}$ equals

Find $\mathop {\lim }\limits_{x \to 1} f(x)$,where $f(x) = \begin{cases} x^{2}-1, & x \leq 1 \\ -x-1, & x > 1 \end{cases}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module