_ x 1 f(x) is equal to, where f(x) = * 20 c e^ 1 x - 1 - 2 e^ 1 x… - Vidyadip

MCQ

$\mathop {\lim }\limits_{x \to 1} f(x)$ is equal to, where

$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{e^{\frac{1}{{x - 1}}}} - 2}}{{{e^{\frac{1}{{x - 1}}}} + 2}}}&{x \ne 1}\\
{1\,\,\,\,\,\,\,\,\,\,\,\,\,}&{x = 1}
\end{array}} \right.$

A
$-1$
B
$1$
C
$0$
✓
does not exist

✓

Answer

Correct option: D.

does not exist

d
$L.H.L$ $ = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{e^{\frac{1}{{x - 1}}}} - 2}}{{{e^{\frac{1}{{x - 1}}}} + 2}} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{{e^{ - \frac{1}{h}}} - 2}}{{{e^{ - \frac{1}{h}}} + 2}} = - 1$

$R.H.L$ $ = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{e^{\frac{1}{{x - 1}}}} - 2}}{{{e^{\frac{1}{{x - 1}}}} + 2}} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{{e^{\frac{1}{h}}} - 2}}{{{e^{\frac{1}{h}}} + 2}}$

$ = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{1 - 2{e^{\frac{1}{h}}}}}{{1 + 2{e^{ - \frac{1}{h}}}}} = 1$

as $\mathrm{L} . \mathrm{H} . \mathrm{L} \neq \mathrm{R} . \mathrm{H} . \mathrm{L}.$

limit does not exist.

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