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 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} = $

If $[ \cdot ]$ denotes the greatest integer function,then evaluate the limit: $\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{[\sin x]-[\cos x]+1}{2}$

If $x$ is a real number in $[0, 1]$,then the value of $\lim_{m \to \infty} \lim_{n \to \infty} [1 + \cos^{2m}(n! \pi x)]$ is given by

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x)$ is equal to

$\mathop {\lim }\limits_{x \to {a^ + }} \left( \frac{{|x{|^3}}}{a} - {\left[ {\frac{x}{a}} \right]^3} \right) \,(a > 0)$ is equal to :- (where $[x]$ is the greatest integer function and $|x|$ is the modulus function)

Vedclass Test Series

Exam Paper Generator

Online Exam Module