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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Show that the function f given by:
$f(x) = \begin{matrix} \frac{e^{1/x} - 1}{e^{1/x} + 1} & ,\text{if }x \neq 0 \\ -1 & ,\text{if }x = 0 \\ \end{matrix} $
is discontinuous at $x = 0$
✓

Answer

$\text{f(x)} = \begin{matrix} \frac{e^{\frac{1}{\text{x}}} - 1}{e^{\frac{1}{\text{x}}} + 1} &\text{x} \neq 0 \\ -1 & \text{x} = 0 \\ \end{matrix} $
$\text{LHL: } \lim\limits_{ x \rightarrow\overline{0}} \frac{\text{e}^{\frac{1}{\text{x}}} - 1}{\text{e}^{\frac{1}{\text{x}}} + 1}$
$= \lim\limits_{ h \rightarrow{0}} \frac{\text{e}^{\frac{1}{\text{h}}} - 1}{\text{e}^{\frac{1}{\text{h}}} + 1} = \frac{0 - 1}{0 + 1} = -1$
$\text{RHL: } \lim\limits_{ h \rightarrow{0}} \frac{\text{e}^{\frac{1}{\text{h}}} - 1}{\text{e}^{\frac{1}{\text{h}}} + 1} = \lim\limits_{ h \rightarrow{0}} \frac{\text{1 - e}^{\frac{1}{\text{h}}}} {\text{1 + e}^{\frac{1}{\text{h}}}} = 1 $
$\text{LHL}\neq\text{RHL}$
$\therefore$ f(x) is discontinuous at x = 0

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