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

Let $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$,then for all values of $x$

For $a \neq 0$ and $b \neq 0$,if the real valued function $f(x) = \frac{\sqrt[5]{a(625+x)} - 5}{\sqrt[4]{625+bx} - 5}$ is continuous at $x = 0$,then $f(0) =$

If $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then at $x = 0$ the function $f(x)$ is

If $f(x) = |x - 2|$,then

If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module