Given,sin x = sum_{n=1}^{infty} (-1)^{n-1} fr | vedclass

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

Question

(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Using the Taylor series expansion for $\sin x = x -

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ and $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$,we substitute $\sin x$ into the cosine series:$\cos(\sin x) = 1 - \frac{(x - x^3/6)^2}{2} + \frac{(x - x^3/6)^4}{24} = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{5x^4}{24}$.Now,$f(x) = \frac{(1 - x^2/2 + 5x^4/24) - (1 - x^2/2 + x^4/24)}{x^4} = \frac{4x^4/24}{x^4} = \frac{1}{6}$.Thus,$k = \lim_{x 	o 0} f(x) = \frac{1}{6}$.

Given,$\sin x = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}$. If the function $f(x)$ given by $f(x) = \frac{\cos(\sin x) - \cos x}{x^4}$ for $x \neq 0$ and $f(0) = k$ is continuous at $x = 0$,then $k =$

Similar Questions

The number of point / points where the function $f(x) = \frac{1}{x^2 - 5|x| + 6}$ is discontinuous is......

Given $f(x) = b ([x]^2 + [x]) + 1$ for $x \geq -1$ and $f(x) = \sin(\pi(x+a))$ for $x < -1$,where $[x]$ denotes the greatest integer function,for what values of $a$ and $b$ is the function continuous at $x = -1$?

If $f(x) = \begin{cases} \frac{|x - a|}{x - a}, & x \neq a \\ 1, & x = a \end{cases}$,then:

If $f(x) = \frac{\log_e(1 + x^2 \tan x)}{\sin x^3}, x \neq 0$ is to be continuous at $x = 0$,then $f(0)$ must be equal to

If the function $f(x) = \frac{1-\sin 2x + \cos 2x}{1+\sin 2x + \cos 2x}$ for $x \neq \frac{\pi}{2}$ and $f(x) = k$ for $x = \frac{\pi}{2}$ is continuous at $x = \frac{\pi}{2}$,then $k = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module