First slide

Introduction to limits

Question

$lim_{x \to 0} \frac{e^{1 / x} - 1}{e^{1 / x} + 1},$ is

Moderate

A

-1
B

1
C

0
D

non-existent

Solution

Let $f (x) = \frac{e^{1 / x} - 1}{e^{1 / x} + 1} .$ Then ,
(LHL of $f (x)$ at $x = 0$ )
$= lim_{x \to 0^{-}} f (x) = lim_{h \to 0^{-}} f (0 - h) = lim_{h \to 0} \frac{e^{- 1 / h} - 1}{e^{- 1 / h} + 1}$
$= lim_{h \to 0} (\frac{\frac{1}{e^{1 / h} - 1}}{\frac{1}{e^{1 / h}} + 1}) = - 1 [e^{1 / h} \to \infty \Rightarrow \frac{1}{e^{1 / h}} \to 0]$
and,
(RHL of $f (x)$ at $x = 0$ )
$= lim_{x \to 0^{+}} f (x) = lim_{h \to 0} f (0 + h) = lim_{h \to 0} \frac{e^{1 / h} - 1}{e^{1 / h} + 1}$
$= lim_{h \to 0} (\frac{1 - \frac{1}{e^{1 / h}}}{1 + e^{1 / h}})$ [Dividing $N r$ and $D r$ by $e^{1 / h}$ ]
$= \frac{1 - 0}{1 + 0} = 1$
Clearly, $lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x)$
Hence, $lim_{x \to 0} f (x)$ does not exist.

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