If f(x) = begin{cases} frac{x^2 ln cos x}{ln( | vedclass

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

Question

(C) To check for continuity and differentiability at $x = 0$,we first evaluate the limit $\lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{x^2 \ln \cos x}{\

Vedclass · Accepted Answer

differentiable at $0$

Vedclass · Answer

(C) To check for continuity and differentiability at $x = 0$,we first evaluate the limit $\lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{x^2 \ln \cos x}{\ln(1 + x^2)} = \lim_{x 	o 0} \left( \frac{x^2}{\ln(1 + x^2)} ight) 	imes \ln \cos x$.Since $\lim_{x 	o 0} \frac{x^2}{\ln(1 + x^2)} = 1$ and $\lim_{x 	o 0} \ln \cos x = \ln(1) = 0$,the limit is $1 	imes 0 = 0$.Since $\lim_{x 	o 0} f(x) = f(0) = 0$,the function is continuous at $x = 0$.Now,check for differentiability at $x = 0$:$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^2 \ln \cos h}{h \ln(1 + h^2)} = \lim_{h 	o 0} \frac{h \ln \cos h}{\ln(1 + h^2)}$.Using the standard limits $\lim_{h 	o 0} \frac{\ln(1 + h^2)}{h^2} = 1$ and $\lim_{h 	o 0} \frac{\ln \cos h}{h^2} = -\frac{1}{2}$:$f'(0) = \lim_{h 	o 0} \left( \frac{h^2}{\ln(1 + h^2)} ight) 	imes \left( \frac{\ln \cos h}{h^2} ight) 	imes h = 1 	imes (-\frac{1}{2}) 	imes 0 = 0$.Since the limit exists and is finite,$f(x)$ is differentiable at $x = 0$.

If $f(x) = \begin{cases} \frac{x^2 \ln \cos x}{\ln(1 + x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

Similar Questions

If $f(x) = |x|,$ then $f'(0) = $

Let $f(x) = \begin{cases} \sin x, & \text{for } x \ge 0 \\ 1 - \cos x, & \text{for } x \le 0 \end{cases}$ and $g(x) = e^x$. Then $(g \circ f)'(0)$ is

The function represented by the following graph is,

$A$ function $f$ is defined as follows:
$f(x) = \begin{cases} \sin x & \text{if } x \le c \\ ax + b & \text{if } x > c \end{cases}$
where $c$ is a known quantity. If $f$ is derivable at $x = c$,then the values of $a$ and $b$ are . . . . . . and . . . . . . respectively.

If the function $f(x) = \begin{cases} \frac{(e^{kx}-1) \sin kx}{4 \tan x}, & x \neq 0 \\ P, & x=0 \end{cases}$ is differentiable at $x=0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module