# Find whether f(x) is continuous or discontinuous at the indicated point $f(x)=\left \{\begin{array}{1 1}\large\frac{e^ {\Large\frac{1}{x}}}{1+e^ {\Large\frac{1}{x}}},\normalsize & if\;x\neq 0\\0, & if\;x=0\end{array}\right.$ at $x=0$.

Toolbox:
• A function is discontinuous at a point if both the LHL and the RHL does not exist.
• It is also discontinuous if LHL $\neq$ RHL.
Step 1:
Given : $f(x)=\left \{\begin{array}{1 1}\large\frac{e^ {\Large\frac{1}{x}}}{1+e^ {\Large\frac{1}{x}}},\normalsize & if\;x\neq 0\\0, & if\;x=0\end{array}\right.$ at $x=0$.
We have $f(x)=\large\frac{e^{\Large\frac{1}{x}}}{1+e^{\Large\frac{1}{x}}}$, $x\neq 0$
$f(0)=0$
The function is defined on both side of $x=0$,by the same relation,but it is still discontinuous .Let us see that.
Step 2:
The left hand limit :
$\lim\limits_{\large x\to 0^-}f(x)=\lim\limits_{\large h\to 0}f(0-h)$
$\qquad\qquad=\lim\limits_{\large h\to 0}f(-h)$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{e^{-\Large\frac{1}{h}}}{1+e^{-\Large\frac{1}{h}}}$
As $h\to 0\Rightarrow \large\frac{1}{h}$$\to \infty$
But $e^{-\infty}=0$
$\qquad\qquad=\large\frac{e^{-\infty}}{1+e^{-\infty}}$
$\qquad\qquad=\large\frac{0}{1+0}$
$\qquad\qquad=0$
Step 3:
The right hand limit :
$\lim\limits_{\large x\to 0^+}f(x)=\lim\limits_{\large h\to 0}f(0+h)$
$\qquad\qquad=\lim\limits_{\large h\to 0}\large\frac{e^{\Large\frac{1}{h}}}{1+e^{\Large\frac{1}{h}}}$
Divide the numerator and the denominator by $e^{\Large\frac{1}{h}}$
$\lim\limits_{\large h\to 0}f(h)=\large\frac{1}{1+\large\frac{1}{e^{\Large h}}}$
On applying limits
$\qquad\qquad=\large\frac{1}{1+0}$
$\qquad\qquad=1$
Therefore the function is discontinuous at $x=0$
edited Jun 26, 2013