# How do you find the lim_(x to oo) (e^x+e^-x)/(e^x-e^-x)?

Dec 4, 2017

1

#### Explanation:

We can manipulate and adjust this via multiplieying both numorator and denominator by ${e}^{x}$

$\left(\frac{{e}^{x} + {e}^{- x}}{{e}^{x} - {e}^{- x}}\right) \left({e}^{x} / {e}^{x}\right)$
$= \frac{{e}^{2 x} + 1}{{e}^{2 x} - 1}$

We know that as $x$ gets large, ${e}^{2 x} + 1 \approx {e}^{2 x}$
and also ${e}^{2 x} - 1 \approx {e}^{2 x}$

So hence limit becomes;

${\lim}_{x \to \infty} {e}^{2 x} / {e}^{2 x}$

$= {\lim}_{x \to \infty} 1$

$= 1$

Dec 4, 2017

${\lim}_{x \to \infty} \frac{{e}^{x} + {e}^{-} x}{{e}^{x} - {e}^{-} x} = 1$

#### Explanation:

Given:

${\lim}_{x \to \infty} \frac{{e}^{x} + {e}^{-} x}{{e}^{x} - {e}^{-} x}$

Add 0 to the numerator in the form $- {e}^{-} x + {e}^{-} x$

${\lim}_{x \to \infty} \frac{{e}^{x} - {e}^{-} x + {e}^{-} x + {e}^{-} x}{{e}^{x} - {e}^{-} x}$

Combine like terms:

${\lim}_{x \to \infty} \frac{{e}^{x} - {e}^{-} x + 2 {e}^{-} x}{{e}^{x} - {e}^{-} x}$

Separate into two fractions:

${\lim}_{x \to \infty} \frac{{e}^{x} - {e}^{-} x}{{e}^{x} - {e}^{-} x} + \frac{2 {e}^{-} x}{{e}^{x} - {e}^{-} x}$

The first fraction becomes 1:

$1 + {\lim}_{x \to \infty} \frac{2 {e}^{-} x}{{e}^{x} - {e}^{-} x}$

Multiply the fraction by 1 in the form of ${e}^{x} / {e}^{x}$

$1 + {\lim}_{x \to \infty} {e}^{x} / {e}^{x} \frac{2 {e}^{-} x}{{e}^{x} - {e}^{-} x}$

Perform the multiplication:

$1 + {\lim}_{x \to \infty} \frac{2}{{e}^{2 x} - 1}$

The limit becomes 0; leaving only the 1.

Dec 4, 2017

For a third alternative, see below.

#### Explanation:

Multiply numerator and denominator by ${e}^{-} x$

$\left(\frac{{e}^{x} + {e}^{- x}}{{e}^{x} - {e}^{- x}}\right) \left({e}^{-} \frac{x}{e} ^ - x\right)$
$= \frac{1 + {e}^{- 2 x}}{1 - {e}^{- 2 x}}$

We know that as $x$ increases without bound, ${e}^{x}$ also increases without bound, so

${e}^{- 2 x} = \frac{1}{e} ^ \left(2 x\right)$ goes to $0$

So the limit becomes;

${\lim}_{x \to \infty} \frac{1 + {e}^{- 2 x}}{1 - {e}^{- 2 x}} = \frac{1 + 0}{1 - 0} = 1$