Question

Show that given function doesn't exist
$\sf{ \lim x \rightarrow \: 0} \: \: \displaystyle{\sf{ \frac{ {e}^{ \frac{1}{x} } - 1 }{ {e}^{ \frac{1}{x} } + 1 } }}$
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Answer 1

$\LARGE{\underline{\underline{\red{\sf{Answer :}}}}}$

We are given,

$\large \displaystyle{\boxed {\sf{ \displaystyle \lim_ {x \to 0} \: \frac{ e^{\frac{1}{x}} \: - \: 1}{e^{\frac{1}{x}} \: + \: 1}}}}$

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Take L.H.L (Left Hand Limit)

As we know for left hand side,

⇒ x = 0 - h

$\Large {\sf{x \: \to \: 0^{-} \: \: , \: \: h \: \to \: 0}}$

Now Substute these values,

$\Large \displaystyle {\sf{\lim_ {h \to 0} \: \frac{e^{\frac{-1}{h}} \: - \: 1}{e^{\frac{-1}{h}} \: + \: 1}}}$

Which is also equal to,

$\displaystyle \sf{ \lim_ {x \to 0}} \: \: \frac{ {e}^{ \frac{1}{h} } \: - \: 1}{ {e}^{ \frac{1}{h} } \: + \: 1 }$

${\displaystyle {\sf {(1) \: \begin{cases}h \to 0 \\ \frac{1}{h} \to \infty \\ {e}^{ \frac{1}{h} } \to \infty \\ \frac{1}{ {e}^{ \frac{1}{h} } } \to0 \end{cases}}}}$

By using these values we get,

$\Large{\sf{L.H.L \: = \: \frac{0 \: - \: 1}{0 \: + \: 1}}}$

Thus L.H.L is,

$\Large \implies {\underline{\boxed{\sf{L.H.L \: = \: -1}}}}$

$\rule{200}{2}$

$\rule{200}{2}$

Now Take R.H.L (Right Hand Limit)

See Attached Figure for R.H.L

______________________

Now,

This Function is Wrong Because,

L.H.L ≠ R.H.L

✌ ✌ ✌ ✌

Answer 2

Answer:

Refer to the attachment