Question

Show that the function f given by $f\left(x\right)=\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}+1}},\text{if}x\ne 0$
$=0,\text{if}x=0$
is discontinuous at $x=0$.

Answer 1

Given for $x\ne 0$, $f\left(x\right)=\frac{e^{\frac{1}{x}}-1}{e^{1+\frac{1}{x}}}=\frac{1}{e}-e^{-1-\frac{1}{x}}$.

Now we have $\underset{x\to 0}{lim}{e}^{-\frac{1}{x}}$ does not exists.

As $\underset{x\to 0+}{lim}{e}^{-\frac{1}{x}}=0$ but $\underset{x\to 0-}{lim}{e}^{-\frac{1}{x}}=\infty$.

So $\underset{x\to 0}{lim}f\left(x\right)$ does not exist.

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