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  • <img alt="\mathrm{\text { Let } f(x)=\left\{\begin{array}{ccc} (x-1) \sin \frac{1}{x-1} & \text { if } x \neq 1 \\ 0 & \text { if } x=1 \end{array}\right.}" src="https://entrancecorner.onco

\mathrm{\text { Let } f(x)=\left\{\begin{array}{ccc} (x-1) \sin \frac{1}{x-1} & \text { if } x \neq 1 \\ 0 & \text { if } x=1 \end{array}\right.}

Then which one of the following is true? 

Option: 1

f is differentiable at  x=1 but not at x=0


Option: 2

f  is neither differentiable at x=0 nor at x=1


Option: 3

f is differentiable at x=0 and at x=1


Option: 4

f  is differentiable at  x=0 but not at x=1


Answers (1)

best_answer

:By definition

\mathrm{f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \text {, if the limit exists. }}

\mathrm{\therefore \quad \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}}

\mathrm{=\lim _{h \rightarrow 0} \frac{(1+h-1) \sin \frac{1}{(1+h-1)}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}}

As the limit doesn’t exist.
∴ It is not differentiable at x = 1.

\mathrm{\text { Again, } f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \text {, if the limit exists }}

\mathrm{\therefore \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{(h-1) \sin \frac{1}{h-1}-\sin 1}{h}}

But this limit doesn’t exist. Hence it is not differentiable at x = 0.

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