Question

# $$f(x) = \left\{\begin{matrix}x^2 \left (\dfrac{e^{1/x}-e^{-1/x}}{e^{1/x} + e^{-1/x}} \right ); & x \neq 0\\ 0; & x=0\end{matrix}\right.$$. Then

A
f(x) is discontinuous at x=0
B
f(x) is continuous but non-differentiable at x=0
C
f(x) is differentiable at x=0
D
f(0)=2

Solution

## The correct option is D $$f(x)$$ is differentiable at $$x = 0$$At $$x = 0$$$$L.H.L = \displaystyle \lim_{ x \rightarrow 0^-} f(x) = \lim_{h \rightarrow 0} f(0-h)$$        $$= \displaystyle \lim_{h \rightarrow 0} h^2 \left ( \frac{e^{-1/h} - e^{1/h}}{e^{-1/h} + e^{1/h}} \right )$$        $$\displaystyle = \lim_{h \rightarrow 0}h^2 \left ( \frac{e^{-2/h} - 1}{e^{-2/h} + 1}\right )$$        $$= 0\displaystyle \left ( \frac{0-1}{0+1} \right ) = 0$$$$R.H.L. = \lim_{x \rightarrow 0^+} f(x) = \lim_{h \rightarrow 0} f(0 + h)$$        $$\displaystyle = \lim_{h \rightarrow 0} h^2 \left ( \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} \right )$$        $$\displaystyle = \lim_{h \rightarrow 0} h^2 \left ( \frac{1 - e^{-2/h}}{1+e^{-2/h}} \right)$$        $$= 0 \displaystyle \left ( \frac{1-0}{1+0} \right ) = 0$$and $$f(0) = 0$$$$\therefore L.H.L. = R.H.L. =f(0)$$Hence, $$f(x)$$ is continuous at $$x = 0$$.Also, $$\displaystyle L.H.D. = \lim_{h \rightarrow 0} \frac{f(0-h) - f(0)}{-h}$$         $$\displaystyle = \lim_{h \rightarrow 0}\frac{h^2 \frac{e^{-1/h} - e^{1/h}}{e^{-1/h} + e^{1/h}} - 0}{-h}$$         $$\displaystyle = -\lim_{h \rightarrow 0}h \frac{e^{-2/h} - 1}{e^{-2/h} + 1} = 0$$and $$R.H.D \displaystyle = \lim_{h \rightarrow 0} \frac{f(0 + h)- f(0)}{h}$$         $$\displaystyle = \lim_{h \rightarrow 0} \frac{h^2 \frac{e^{1/h} - e^{-1/h}}{e^{1/h} + e^{-1/h}} - 0}{-h}$$         $$\displaystyle = -\lim_{h \rightarrow 0} h \frac{1-e^{-2/h}}{1+e^{-2/h}} = 0$$Hence, $$f(x)$$ is differentiable at $$x= 0$$ and $$f'(0) =0$$Mathematics

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