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

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(B) To check continuity at $x = 0$:$f(0 + 0) = \lim_{h 	o 0} f(h) = \lim_{h 	o 0} h \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} = \lim_{h 	o 0} h

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$f$ is continuous at every point but is not differentiable

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(B) To check continuity at $x = 0$:$f(0 + 0) = \lim_{h 	o 0} f(h) = \lim_{h 	o 0} h \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} = \lim_{h 	o 0} h \frac{1 - e^{-2/h}}{1 + e^{-2/h}} = 0 	imes 1 = 0$.$f(0 - 0) = \lim_{h 	o 0} f(-h) = \lim_{h 	o 0} (-h) \frac{e^{-1/h} - e^{1/h}}{e^{-1/h} + e^{1/h}} = \lim_{h 	o 0} (-h) \frac{e^{-2/h} - 1}{e^{-2/h} + 1} = 0 	imes (-1) = 0$.Since $f(0 + 0) = f(0 - 0) = f(0) = 0$,$f$ is continuous at $x = 0$. For $x 
e 0$,$f$ is a composition of continuous functions,hence continuous.To check differentiability at $x = 0$:$L f'(0) = \lim_{h 	o 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h 	o 0} \frac{-h \frac{e^{-1/h} - e^{1/h}}{e^{-1/h} + e^{1/h}} - 0}{-h} = \lim_{h 	o 0} \frac{e^{-1/h} - e^{1/h}}{e^{-1/h} + e^{1/h}} = \frac{0 - 1}{0 + 1} = -1$.$R f'(0) = \lim_{h 	o 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h 	o 0} \frac{h \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} - 0}{h} = \lim_{h 	o 0} \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} = \lim_{h 	o 0} \frac{1 - e^{-2/h}}{1 + e^{-2/h}} = \frac{1 - 0}{1 + 0} = 1$.Since $L f'(0) 
e R f'(0)$,$f$ is not differentiable at $x = 0$. Thus,$f$ is continuous everywhere but not differentiable at $x = 0$.

If $f(x) = \begin{cases} x \frac{e^{(1/x)} - e^{(-1/x)}}{e^{(1/x)} + e^{(-1/x)}}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ then which of the following is true?

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