Question

The function $f\left(x\right)=\{\begin{array}{cc}\frac{e^{1/x}-1}{e^{1/x}+1}& x\ne 0\\ 0,& x=0\end{array}$ is

• A. continuous at $x=0$
• B. discontinuous at $x=0$
• C. discontinuous at $x=0$ but can be made continuous at $x=0$
• D. None of these

Answer 1

Answer: (B)

discontinuous at $x=0$

L.H.L $=\underset{x\to 0-}{lim}\frac{e^{1/x}-1}{e^{1/x}+1}=\underset{h\to -\infty }{lim}\frac{e^h-1}{e^h+1}=-1$
R.H.L $=\underset{x\to 0+}{lim}\frac{e^{1/x}-1}{e^{1/x}+1}=\underset{h\to +\infty }{lim}\frac{e^h-1}{e^h+1}=\underset{h\to +\infty }{lim}\frac{1-e^{-h}}{1+e^{-h}}=1$
Clearly limit at $x=0$ does not exist. Hence given function is not continuous at $x=0$.