For the function f(x) = l e^ 1/x - 1 e^ 1/x + 1 , , ,x 0 0 , , , … - Vidyadip

MCQ

For the function $f(x) = \left\{ \begin{array}{l}\frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}},\,\,x \ne 0\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,x = 0\end{array} \right.$, which of the following is correct

A
$\mathop {\lim }\limits_{x \to 0} f(x)$ does not exist
B
$f(x)$ is continuous at $x = 0$
C
$\mathop {\lim }\limits_{x \to 0} f(x) = 1$
✓
$\mathop {\lim }\limits_{x \to 0} f(x)$ exists but $f(x)$ is not continuous at $x = 0$

✓

Answer

Correct option: D.

$\mathop {\lim }\limits_{x \to 0} f(x)$ exists but $f(x)$ is not continuous at $x = 0$

d
(d) Given $f(x) = \left\{ \begin{array}{l}\frac{{{e^{\frac{1}{x}}} - 1}}{{{e^{\frac{1}{x}}} + 1}}\,\,,\,\,x \ne 0\\0\,\,\,\,\,\,\,\,\,\,\,\,\,,\,x = 0\end{array} \right.$

==> $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^{\frac{1}{x}}} - 1}}{{{e^{\frac{1}{x}}} + 1}} = \frac{{{e^\infty } - 1}}{{{e^\infty } + 1}} = - 1$

==> $\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{\frac{1}{x}}} - 1}}{{{e^{\frac{1}{x}}} + 1}} = \frac{{1 - {e^{ - \frac{1}{x}}}}}{{1 + {e^{\frac{1}{x}}}}} = \frac{{1 - {e^{ - \infty }}}}{{1 + {e^\infty }}} = 1$

So, $\mathop {\lim }\limits_{x \to 0} f(x)$ exists at $x = 0$, but at $x = 0$ it is not continuous.

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