limlimits _{x rightarrow 0} frac{cos (sin x)- | vedclass

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

Question

(C) We evaluate the limit: $\lim\limits _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(C) We evaluate the limit: $\lim\limits _{x ightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.Using the Taylor series expansion for $\cos 	heta = 1 - \frac{	heta^2}{2!} + \frac{	heta^4}{4!} - \dots$:$\sin x = x - \frac{x^3}{6} + O(x^5)$$\cos(\sin x) = 1 - \frac{(x - \frac{x^3}{6})^2}{2} + \frac{(x)^4}{24} + O(x^6) = 1 - \frac{x^2 - \frac{x^4}{3}}{2} + \frac{x^4}{24} + O(x^6) = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} + O(x^6) = 1 - \frac{x^2}{2} + \frac{5x^4}{24} + O(x^6)$.$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$.Subtracting these:$\cos(\sin x) - \cos x = (1 - \frac{x^2}{2} + \frac{5x^4}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24}) + O(x^6) = \frac{4x^4}{24} = \frac{x^4}{6}$.Therefore,$\lim\limits _{x ightarrow 0} \frac{\frac{x^4}{6}}{x^4} = \frac{1}{6}$.

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

Similar Questions

$\mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}}$ is equal to

$\lim _{x \rightarrow-\infty} \log _e(\cosh x)+x=$

$\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !-n !} = $

If $\lim _{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{q x}=e^9$ where $p, q \in \mathbb{N}$,then $p+q=$

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\frac{1}{x}}}}}{{{e^{\left( {\frac{1}{x} + 1} \right)}}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module