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# Show that the function :$f(x)= \left\{ \begin{array}{1 1} \Large\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}, & \quad when\;x\neq0 \\ 0, & \quad when\; x=0 \end{array} \right.$ is discontinuous at x = 0.

Toolbox:
• Check for $LHL = RHL$ or not
• Use LHL definition and RHL definition.
• LHL = $\lim\limits_{h \to 0} f(a-h)$ and $RHL = \lim\limits_{h \to 0}f(a+h)$
LHL
$\lim\limits_{x \to 0^-} \large\frac{e^{\large\frac{1}{x}}-1}{e^{\large\frac{1}{x}}+1}$
$= \lim\limits_{h \to 0} \large\frac{e^{\large\frac{1}{0-h}}-1}{e^{\large\frac{1}{0-h}}+1}$
$= \lim\limits_{h \to 0} \large\frac{e^{\large\frac{1}{\large\frac{1}{h}}}-1}{e^{\large\frac{1}{\large\frac{1}{h}}}+1} = -1$

RHL
$\lim\limits_{x \to 0^+} \large\frac{e^{\large\frac{1}{x}}-1}{e^{\large\frac{1}{x}}+1}$
$= \lim\limits_{h \to 0} \large\frac{e^{\large\frac{1}{\large\frac{1}{h}}}-1}{e^{\large\frac{1}{\large\frac{1}{h}}}+1}$
$= \lim\limits_{h \to 0} \large\frac{1-e^{\large\frac{1}{\large\frac{1}{h}}}}{1+e^{\large\frac{1}{\large\frac{1}{h}}}} = 1$

$LHL \neq RHL \Rightarrow f$ is discontinuous at $x = 0$

edited Mar 26, 2013