If f(x) = begin{cases} x e^{-left( frac{1}{|x | vedclass

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(B) Given $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} ight)}, & x 
e 0 \ 0, & x = 0 \end{cases}$.For continuity at $x = 0$:$R.H

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Continuous for all $x$ but not differentiable at $x = 0$

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(B) Given $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} ight)}, & x 
e 0 \ 0, & x = 0 \end{cases}$.For continuity at $x = 0$:$R.H.L. = \lim_{h 	o 0^+} f(0+h) = \lim_{h 	o 0^+} h e^{-\left( \frac{1}{h} + \frac{1}{h} ight)} = \lim_{h 	o 0^+} h e^{-2/h} = 0$.$L.H.L. = \lim_{h 	o 0^+} f(0-h) = \lim_{h 	o 0^+} (-h) e^{-\left( \frac{1}{h} - \frac{1}{h} ight)} = \lim_{h 	o 0^+} (-h) e^0 = 0$.Since $R.H.L. = L.H.L. = f(0) = 0$,$f(x)$ is continuous at $x = 0$.For differentiability at $x = 0$:$Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h e^{-2/h} - 0}{h} = \lim_{h 	o 0^+} e^{-2/h} = 0$.$Lf'(0) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{-h e^0 - 0}{-h} = 1$.Since $Rf'(0) 
e Lf'(0)$,$f(x)$ is not differentiable at $x = 0$.

If $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} \right)}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

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