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) First, we check for continuity at $x=0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x^2 \ln \cos x}{\ln (1+x^2)}$.Using standard limits $\lim_{u

Vedclass · Accepted Answer

differentiable at zero

Vedclass · Answer

(C) First, we check for continuity at $x=0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{x^2 \ln \cos x}{\ln (1+x^2)}$.Using standard limits $\lim_{u 	o 0} \frac{\ln(1+u)}{u} = 1$, we have $\lim_{x 	o 0} \frac{\ln(1+x^2)}{x^2} = 1$.Thus, $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left( \frac{x^2}{\ln(1+x^2)} ight) \cdot \ln \cos x = 1 \cdot \ln(1) = 0$.Since $\lim_{x 	o 0} f(x) = f(0) = 0$, the function is continuous at $x=0$.Now, we check for differentiability at $x=0$ using the definition $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h^2 \ln \cos h}{h \ln(1+h^2)} = \lim_{h 	o 0} \frac{h \ln \cos h}{\ln(1+h^2)}$.Dividing numerator and denominator by $h^2$: $\lim_{h 	o 0} \frac{\frac{\ln \cos h}{h}}{\frac{\ln(1+h^2)}{h^2}} = \lim_{h 	o 0} \frac{\frac{\ln \cos h}{h}}{1}$.Using $L$'Hospital's rule on $\frac{\ln \cos h}{h}$: $\lim_{h 	o 0} \frac{-	an h}{1} = 0$.Since the limit exists and is equal to $0$, $f(x)$ 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

The derivative of $y = 1 - |x|$ at $x = 0$ is

$f(x) = |\log_e |x||$ is differentiable at

Let $f : (-1, 1) \to \mathbb{R}$ be a function defined by $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$. If $K$ is the set of all points at which $f$ is not differentiable,then $K$ has exactly

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module