Question-39

Consider the function

\[ \begin{equation*} f( x) =\begin{cases} x\sin\frac{1}{x} , & x\neq 0\\ 0, & x=0 \end{cases} \end{equation*} \]

• \(f\) is continuous at \(x = 0\)
• \(f\) is not continuous at \(x = 0\)
• \(f\) is differentiable at \(x = 0\)
• \(f\) is not differentiable at \(x = 0\)

Hint

Use first principles approach.

Answer

• \(f\) is continuous at \(x = 0\)
• \(f\) is not continuous at \(x = 0\)
• \(f\) is differentiable at \(x = 0\)
• \(f\) is not differentiable at \(x = 0\)

Solution

We note that:

\[ \begin{equation*} 0\leqslant x\sin\frac{1}{x} \leqslant x \end{equation*} \]

Using sandwich theorem, we see that the limit as \(\displaystyle x\rightarrow 0\) exists and is equal to \(\displaystyle 0\). Since \(\displaystyle f( 0) =0\), the function is continuous. For differentiability at \(\displaystyle x=0\), we can approach this from first principles:

\[ \begin{equation*} \begin{aligned} \lim\limits _{h\rightarrow 0}\frac{f( h) -f( 0)}{h} & =\lim\limits _{h\rightarrow 0} \ \frac{h\sin\frac{1}{h} -0}{h}\\ & \\ & =\lim\limits _{h\rightarrow 0} \ \sin\frac{1}{h} \end{aligned} \end{equation*} \]

We see that this limit does not exist. Hence, \(\displaystyle f\) is not differentiable at \(\displaystyle x=0\).