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Home  >>  CBSE XII  >>  Math
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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.

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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$

 

answered Mar 11, 2013 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 

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