A) \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\ne 0\]
B) \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\ne 2\]
C) \[f(x)\]is discontinuous at \[x=1\]
D) None of these
Correct Answer: C
Solution :
\[\underset{x\to 1+}{\mathop{\lim }}\,f(x)=0\]and \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,h\,\,\frac{{{e}^{1/h}}-{{e}^{-1/h}}}{{{e}^{1/h}}+{{e}^{-1/h}}}\] Hence \[f(x)\] is discontinuous at \[x=1\].
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