Let f(x) = begin{cases} x sin left( frac{1}{x | vedclass

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

Question

(B) To check for continuity at $x = 0$,we evaluate $\lim_{x 	o 0} f(x)$.Given $f(x) = x \sin \left( \frac{1}{x} ight) \sin \left( \frac{1}{x \sin \

Vedclass · Accepted Answer

continuous but not differentiable at $x = 0$

Vedclass · Answer

(B) To check for continuity at $x = 0$,we evaluate $\lim_{x 	o 0} f(x)$.Given $f(x) = x \sin \left( \frac{1}{x} ight) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} ight)} ight)$.Since $|\sin 	heta| \leq 1$,we have $|f(x)| = |x \sin \left( \frac{1}{x} ight) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} ight)} ight)| \leq |x \sin \left( \frac{1}{x} ight)| \leq |x|$.As $x 	o 0$,$|x| 	o 0$,so by the Squeeze Theorem,$\lim_{x 	o 0} f(x) = 0 = f(0)$. Thus,$f(x)$ is continuous at $x = 0$.To check for differentiability,we evaluate $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{h \sin \left( \frac{1}{h} ight) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} ight)} ight)}{h} = \lim_{h 	o 0} \sin \left( \frac{1}{h} ight) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} ight)} ight)$.Let $g(h) = \sin \left( \frac{1}{h} ight) \sin \left( \frac{1}{h \sin \left( \frac{1}{h} ight)} ight)$. As $h 	o 0$,$\sin \left( \frac{1}{h} ight)$ oscillates between $-1$ and $1$. The term $\frac{1}{h \sin \left( \frac{1}{h} ight)}$ also oscillates and takes arbitrarily large values,causing the second sine term to oscillate. Thus,the limit does not exist.Therefore,$f(x)$ is continuous but not differentiable at $x = 0$.

Let $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} \right)} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is:

Similar Questions

Consider the function $f(x) = \begin{cases} \frac{x}{[x]} & \text{if } 1 \leqslant x < 2 \\ 1 & \text{if } x = 2 \\ \sqrt{6-x} & \text{if } 2 < x \leqslant 3 \end{cases}$ where $[x]$ denotes the greatest integer function. At $x = 2$,the function:

Let $a, b \in \mathbb{R}$ $(a \neq 0)$. If the function $f$ is defined as $f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \leq x < 1 \\ a, & 1 \leq x < \sqrt{2} \\ \frac{2b^2-4b}{x}, & \sqrt{2} \leq x < \infty \end{cases}$ is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

Let $f(x) = \frac{1-\tan x}{4x-\pi}$,where $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{\pi}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is:

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then the function $f(x)$ at $x = 0$ is:

If $f(x) = \begin{cases} \frac{\sqrt{a^2-ax+x^2}-\sqrt{x^2+ax+a^2}}{\sqrt{a+x}-\sqrt{a-x}}, & x \neq 0 \\ K, & x=0 \end{cases}$ is continuous at $x=0$,then $K=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module