Question

# Let $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{x}^{\mathrm{n}}\sin\frac{1}{\mathrm{x}},\quad \mathrm{x}\neq 0\\0, \quad \mathrm{x}=0\end{array}\right.$$ , then f(x) is continuous but not differentiable at x=0 if

A
n(0,1]
B
n[1,)
C
n(,0)
D
n=0

Solution

## The correct option is A $$\mathrm{n}\in(0, 1 ]$$Since, $$f\left( x \right)$$ is continuous at $$x=0$$, therefore, $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) =0$$$$\displaystyle\Rightarrow \lim _{ x\rightarrow 0 }{ { x }^{ n } } \sin { \left( \frac { 1 }{ x } \right) } =0\Rightarrow n>0$$$$f(x)$$ is differential at $$x=0$$ if $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { f\left( x \right)-f\left( 0 \right) }{ x-0 } }$$ exists finitely$$\displaystyle \Rightarrow \lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }\sin { \left( \frac { 1 }{ x } \right) } -0 }{ x } }$$ exists finitely$$\displaystyle \Rightarrow { x }^{ n-1 }\sin { \left( \frac { 1 }{ x } \right) }$$ exists finitely.$$\Rightarrow n-1>0 \Rightarrow n>1$$If $$n\le1$$, then $$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ n-1 }\sin { \left( \frac { 1 }{ x } \right) } }$$ does not exists and hence, $$f(x)$$ is not differentiable at $$x=0$$Hence, $$f(x)$$ is continuous but not differentiable at $$x=0$$ for $$0<n\le1$$i.e., $$n\in (0,1]$$Mathematics

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