Show that the function f(x) x^m (1 x ), &x 0 0 ,& x=0 Differentia… - Vidyadip

Question

Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Differential at x = 0, if m > 1

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Answer

Let m = 2, then the function $\text{f(x)}=\begin{Bmatrix}\text{x}^2\sin(\frac{1}{\text{x}}) &\text{x}\neq0 \\0 & \text{x}=0 \end{Bmatrix}$
Differentiability at x = 0:
$\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}-\text{f(0)}}{\text{x}-0}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}}=\lim_\limits{\text{x}\rightarrow0}\text{x}\sin\Big(\frac{1}{\text{x}}\Big)=0.$
$\big[\therefore\lim_\limits{\text{x}\rightarrow0}\text{x}\sin\Big(\frac{1}{\text{x}}\Big)\text{x}=0,$ as
$\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}}-0\end{vmatrix}=\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}} \end{vmatrix}=\begin{vmatrix}\text{x} \end{vmatrix}\begin{vmatrix}\sin\frac{1}{\text{x}} \end{vmatrix} \leq\begin{vmatrix}\text{x}\end{vmatrix}$
$\because\ |\sin\theta|\leq1\ \text{for all }\theta$
Hence, $\begin{vmatrix}\text{x}\sin\frac{1}{\text{x}} \end{vmatrix}<0\ \text{when}\ |\text{x-0}|<\in|\text{x}-0|<\in\big]$
Therefore, f'(x) = 0, which means f is differentiable at x = 0.
Hence the given function is differentiable at x = 0.

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