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  4. If f(x)=(e1/x/1+e1/x) for x≠ 0 and f(0)=0, then at x=0 the function f(x) is

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Q. If $ f(x)=\frac{{{e}^{1/x}}}{1+{{e}^{1/x}}} $ for $ x\ne 0 $ and $ f(0)=0, $ then at $ x=0 $ the function $ f(x) $ is

J & K CETJ & K CET 2007Continuity and Differentiability

Solution:

Given, $ f(x)=\frac{{{e}^{1/x}}}{1+{{e}^{1/x}}},\,\,f(0)=0 $
$ LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h) $
$ =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{-1/h}}}{1+{{e}^{-1/h}}}=0 $
$ RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h) $
$ =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{1/h}}}{1+{{e}^{1/h}}}=1 $
Since, $ LHL\ne RHL $
So, $ f(x) $ is discontinuous at $ x=0. $