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  • <img alt="\mathrm{\text { Let } f(x)=\left\{\begin{array}{cc} x^n \sin \frac{1}{x^{\prime}}, & x \neq 0 \\ 0, & x=0 \end{array}\right.}" src="https://entrancecorner.oncodecogs.com/gif.latex

\mathrm{\text { Let } f(x)=\left\{\begin{array}{cc} x^n \sin \frac{1}{x^{\prime}}, & x \neq 0 \\ 0, & x=0 \end{array}\right.}  Then f(x) is continuous but not differentiable at x=0 if

Option: 1

n \in(0,1]


Option: 2

n \in[1, \infty)


Option: 3

\mathrm{n \in(-\infty, 0)}


Option: 4

\mathrm{n=0}


Answers (1)

best_answer

Since f(x) is continuous at x=0, therefore
\mathrm{\lim _{x \rightarrow 0} f(x)=f(0)=0 \Rightarrow \lim _{x \rightarrow 0} x^n \sin \left(\frac{1}{x}\right)=0 \Rightarrow n>0 }
f(x) is differentiable at x=0 if
\mathrm{\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} }  exists finitely
\mathrm{\Rightarrow \lim _{x \rightarrow 0} \frac{x^n \sin \frac{1}{x}-0}{x} }  exists finitely
\mathrm{\Rightarrow \lim _{x \rightarrow 0} x^{n-1} \sin \left(\frac{1}{x}\right) }   exists finitely. \mathrm{\Rightarrow n-1>0 \Rightarrow n>1 }

If  \mathrm{n \leq 1 } , then  \mathrm{\lim _{x \rightarrow 0} x^{n-1} \sin \left(\frac{1}{x}\right) }  does not exist and hence f(x) is not differentiable at x=0.

Hence, f(x) is continuous but not differentiable at x=0 for   \mathrm{0<n \leq 1i.e. n \in(0,1] }

Posted by

seema garhwal

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