Let $f(x) = \left\{ \begin{array}{r} x^{P}\sin{\frac{1}{x},\ x
eq 0} \ 0,\ \ \ \ \ \ \ \ \ x = 0 \end{array} ight.\$ then $f(x)$ is continuous but not differentiable at $x = 0$, if

Question

Let $f(x) = \left\{ \begin{array}{r} x^{P}\sin{\frac{1}{x},\ x 
eq 0} \ 0,\ \ \ \ \ \ \ \ \ x = 0 \end{array} ight.\$ then $f(x)$ is continuous but not differentiable at $x = 0$, if

Dr. Abubacker · Accepted Answer

(a)
Given, $f(x)$ is continuous at $x = 0$
$	herefore$ Limit must exist
$ie,\ \ \lim_{x ightarrow 0}x^{p}\sin{\frac{1}{x} = (0)^{p}\sin{\infty = 0}}$, when, $0 < p < \infty$ ...(i)
Now, RHD$= \lim_{h ightarrow 0}{\frac{h^{p}\sin{\frac{1}{h} - 0}}{h} =}\lim_{h ightarrow 0}\ h^{p - 1}\sin\frac{1}{h}$
LHD$= \lim_{h ightarrow 0}\ \frac{( - h)^{p}\sin\left( - \frac{1}{h} ight) - 0}{- h}$
$= \lim_{h ightarrow 0}\ ( - 1)^{p}h^{p - 1}\sin\frac{1}{h}$
Since, $f(x)$ is not differentiable at $x = 0$
$	herefore\ \ \ p \leq 1$ ...(ii)
From Eqs.(i) and (iii), $0 < p \leq 1$

Q. Let $f(x) = \left\{ \begin{array}{r} x^{P}\sin{\frac{1}{x},\ x \neq 0} \\ 0,\ \ \ \ \ \ \ \ \ x = 0 \end{array} \right.\$ then $f(x)$ is continuous but not differentiable at $x = 0$, if

Explanation

Reviewed by Dr. Abubacker

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