# what is limit of x*log((1+e^x)/((-1)+e^x))) as x approaches infinity ??

Mar 13, 2018

$0$

#### Explanation:

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

$\log \left(1 + x\right) = x - {x}^{2} / 2 + {x}^{3} / 3 - \ldots \text{ (Taylor series of ln(1+x))}$

$\implies \log \left(\frac{{e}^{x} + 1}{{e}^{x} - 1}\right) \approx 2 {e}^{-} x - 2 {e}^{- 2 x} + \ldots$

$\implies x \log \left(\ldots\right) \approx 2 x {e}^{-} x \to 0 \text{ for } x \to \infty$

Mar 13, 2018

${\lim}_{x \to \infty} \left(x \cdot \log \left[\frac{1 + {e}^{x}}{- 1 + {e}^{x}}\right]\right) = 0$

#### Explanation:

Let's think about it this way:

As $x$ gets really, really, large, ${e}^{x}$ will be so large that numbers like $1$ and $- 1$ will barely matter.

We can, therefore, ignore it.

We are left with ${e}^{x} / {e}^{x}$ Since we are dividing same numbers, we conclude that ${\lim}_{x \to \infty} \frac{1 + {e}^{x}}{- 1 + {e}^{x}} = 1$

We have:

$x \cdot \log \left(1\right)$ Simplify

$\implies x \cdot 0$

$\implies 0$ The answer is $0$.

Just note that this method may not work as you go to harder versions of limit.