Question

Show that the function 

\[f\left( x \right) = \begin{cases}x^m \sin\left( \frac{1}{x} \right) & , x \neq 0 \\ 0 & , x = 0\end{cases}\]

(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0

Answer 1

Given: 

\[f(x) = \binom{ x^m \sin\left( \frac{1}{x} \right)}{0}\]      x≠0 , x=0

(i) Let m=2, then the function becomes 

\[f(x) = \binom{ x^2 \sin\left( \frac{1}{x} \right)}{0}\] ,  x≠0, x=0

Differentiability at x=0:

\[\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} x \sin\left( \frac{1}{x} \right) = 0 .\]

\[\lim_{x \to 0} x \sin\left( \frac{1}{x} \right) = 0\]

\[\left| x \sin\frac{1}{x} - 0 \right| = \left| x \sin\frac{1}{x} \right| = \left| x \right| \left| \sin\frac{1}{x} \right| \leq \left| x \right|\]

\[\left| x \sin\frac{1}{x} - 0 \right| = \left| x \sin\frac{1}{x} \right| = \left| x \right| \left| \sin\frac{1}{x} \right| \leq \left| x \right|\]

\[\theta\]) and hence 

\[\left| x \sin\frac{1}{x} \right| < 0 \] when 

\[\left| x - 0 \right| < \epsilon\]

So,  

\[f'(0) = 0\] which means f is differentiable at x=0.
Hence the given function is differentiable at x=0.

(ii) Let 

\[m = \frac{1}{2}, 0 < m < 1\]. Then the function becomes

\[f(x) = \left\{ \begin{array}{l}x^\frac{1}{2} \\ 0\end{array}\sin\left( \frac{1}{x} \right) \right.\]  ,     x≠0 , x=0

Continuity at x=0:
(LHL at x=0) = 

\[\lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} ( - h )^\frac{1}{2} \sin\left( \frac{1}{0 - h} \right) = \lim_{h \to 0} h^\frac{1}{2} \sin\left( \frac{1}{h} \right) = \lim_{h \to 0} h^\frac{3}{2} = 0\]

(RHL at x=0) = 

\[\lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} h^\frac{1}{2} \sin\left( \frac{1}{h} \right) = \lim_{h \to 0} h^\frac{3}{2} = 0\]

and 

\[f(0) = 0\]

LHL at x=0 = RHL at x=0 = 

\[\lim_{x \to 0} f(x)\]

Hence continuous.
Now Differentiabilty at x=0 when 0<m<1.
(LHD at x=0) = 

\[\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0} = \lim_{h \to 0} \frac{( - h )^\frac{1}{2} \sin\left( \frac{1}{- h} \right)}{- h}\]