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  • For a whole number n ,if <img alt="\mathrm{f(x)=x^{n-1} \sin (1 / x), x \neq 0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3Dx%5E%7Bn-1%7D%20%5Csin%20%281%20/%20x%29

For a whole number n ,if \mathrm{f(x)=x^{n-1} \sin (1 / x), x \neq 0} and \mathrm{f(0)=0} then in order that \mathrm{f} is differentiable at all \mathrm{x}, the value of \mathrm{n} can be

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

0


Answers (1)

best_answer

\mathrm{f} is clearly differentiable except possibly \mathrm{x=0}.

\mathrm{\begin{aligned} f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{h^{n-1} \sin (1 / h)}{h} \\ & =\lim _{h \rightarrow 0} h^{n-2} \sin (1 / h) \end{aligned}}

The last limit exists and is equal to 0 only if \mathrm{n-2 \geq 1, i.e. n \geq 3}. [Using Sandwich Theorem]

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manish

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