The function f (x) = e−|x| is

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continuous everywhere but not differentiable at x = 0 Given: `f(x) = e^(-|x|) = {(e^x , xge0),(e^(-x) ,x< 0):}` 	ext{ Continuity }: \lim_{x 	o 0^-} f(x) = \lim_{h 	o 0} f(0 - h) = \lim_{h 	o 0} e^{- (0 - h)} = \lim_{h 	o 0} e^h = 1 RHL at x = 0 \lim_{x 	o 0^+} f(x) = \lim_{h 	o 0} f(0 + h) = \lim_{h 	o 0} e^{(0 + h)} = 1 and f(0) = f(0) = e^0 = 1 Thus, \lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f Hence, function is continuous at x = 0 Differentiability at x = 0 (LHD at x = 0) \lim_{x 	o 0^-} f(x) - f(0)/x - 0 = \lim_{h 	o 0} \frac{f(0 - h) - f(0)}{0 - h - 0} = \lim_{h 	o 0} \frac{e^{- (0 - h)} - 1}{- h} = \lim_{h 	o 0} e^h/h = \infty Therefore, left hand derivative does not exist. Hence, the function is not differentiable at x = 0.

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