Question

Show that following function $f\left(x\right)$, does not continuous at $x=0$
$f\left(x\right)=\{\begin{array}{c}\frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}x\ne 0\\ 0x=0\end{array}$

Answer 1

$f\left(x\right)=\{\frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}\begin{array}{c}x\ne 0\\ x=0\end{array}$

For continuity at $x=0$

$f\left(0\right)=f\left(0^{-}\right)=f\left(0^{+}\right)$

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

Expanding $e^{1/x}=1+\frac{1}{x}+\frac{1}{x^{2}2!}+.....$

$f\left(0^{+}\right)\underset{x\to 0^{+}}{lim}\frac{1+\frac{1}{x}+\frac{1}{x^{2}2!}+......}{2+\frac{1}{x}+\frac{1}{x^{2}2!}+....}$

Applying $L-$Hospital Rule

$f\left(0^{+}\right)\underset{x\to 0}{lim}\frac{\frac{-1}{x^2}+\frac{-2}{x^{3}2!}+.....}{-\frac{1}{x^2}+\frac{-2}{x^{3}2!}+.....}=1$

$f\left(0\right)\ne f\left(0^{+}\right)$

Function is not continuous at $x=0$