Question

If $f\left(x\right)=\frac{e^{1/x}-1}{e^{1/x}+1},x\ne 0$ and $f\left(0\right)=0$, then $f\left(x\right)$ is

• A. Continuous at $0$
• B. Right continuous at $0$
• C. Discontinuous at $0$
• D. Left continuous at $0$

Answer 1

The correct option is C Discontinuous at $0$

$f\left(x\right)=\frac{e^{1/x}-1}{e^{1/x}+1}f{\left(x\right)}_{x\to 0^{-}}={lim}_{h\to 0}\frac{e^{-1/h}-1}{e^{-1/h}+1}f{\left(x\right)}_{x\to 0^{-}}={lim}_{h\to 0}\frac{\frac{1}{e^{1/h}}-1}{\frac{1}{e^{1/h}}+1}=\frac{0-1}{0+1}=-1f{\left(x\right)}_{x\to 0^{+}}={lim}_{h\to 0}\frac{e^{1/h}-1}{e^{1/h}+1}f{\left(x\right)}_{x\to 0^{+}}={lim}_{h\to 0}\frac{1-\frac{1}{e^{1/h}}}{1-\frac{1}{e^{1/h}}}=\frac{1-0}{1+0}=1f{\left(x\right)}_{x\to 0^{-}}\ne f{\left(x\right)}_{x\to 0^{+}}$

L.H.L is not equal to R.H.L at $x=0$.

So, function is not continuous at $x=0$.

Option C is correct.