Let g(x) = x cdot f(x),where f(x) = begin{cas | vedclass

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(A) Given $f(x) = \begin{cases} x \sin \frac{1}{x}, & x 
e 0 \ 0, & x = 0 \end{cases}$.Then $g(x) = x \cdot f(x) = \begin{cases} x^2 \sin \frac{1}{x

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$g$ is differentiable but $g'$ is not continuous

Vedclass · Answer

(A) Given $f(x) = \begin{cases} x \sin \frac{1}{x}, & x 
e 0 \ 0, & x = 0 \end{cases}$.Then $g(x) = x \cdot f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x 
e 0 \ 0, & x = 0 \end{cases}$.To check the differentiability of $g(x)$ at $x = 0$:$g'(0) = \lim_{h 	o 0} \frac{g(0+h) - g(0)}{h} = \lim_{h 	o 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h 	o 0} h \sin(1/h)$.Since $\lim_{h 	o 0} h = 0$ and $\sin(1/h)$ is bounded in $[-1, 1]$,by the squeeze theorem,$g'(0) = 0$.Thus,$g(x)$ is differentiable at $x = 0$.Now,for $x 
e 0$,$g'(x) = \frac{d}{dx} (x^2 \sin \frac{1}{x}) = 2x \sin \frac{1}{x} + x^2 \cos \frac{1}{x} \cdot (-\frac{1}{x^2}) = 2x \sin \frac{1}{x} - \cos \frac{1}{x}$.As $x 	o 0$,$2x \sin \frac{1}{x} 	o 0$,but $\lim_{x 	o 0} \cos \frac{1}{x}$ does not exist.Therefore,$\lim_{x 	o 0} g'(x)$ does not exist,which means $g'(x)$ is not continuous at $x = 0$.

Let $g(x) = x \cdot f(x)$,where $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Discuss the differentiability of $g$ at $x = 0$.

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