# What is the limit of (x/(x+1))^x  as x approaches infinity?

Mar 21, 2016

${\lim}_{x \to \infty} {\left(\frac{x}{x + 1}\right)}^{x} = \frac{1}{e}$

#### Explanation:

First, we will use the following:

• ${e}^{\ln} \left(x\right) = x$
• Because ${e}^{x}$ is continuous on $\left(- \infty , \infty\right)$, we have ${\lim}_{x \to \infty} {e}^{f} \left(x\right) = {e}^{{\lim}_{x \to \infty} f \left(x\right)}$

With these:

${\lim}_{x \to \infty} {\left(\frac{x}{x + 1}\right)}^{x} = {\lim}_{x \to \infty} {e}^{\ln \left({\left(\frac{x}{x + 1}\right)}^{x}\right)}$

=lim_(x->oo)e^(xln(x/(x+1))

$= {e}^{{\lim}_{x \to \infty} x \ln \left(\frac{x}{x + 1}\right)}$

Next, we will use L'Hopital's rule:

${\lim}_{x \to \infty} x \ln \left(\frac{x}{x + 1}\right) = {\lim}_{x \to \infty} \ln \frac{\frac{x}{x + 1}}{\frac{1}{x}}$

$= {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(\frac{x}{x + 1}\right)}{\frac{d}{\mathrm{dx}} \frac{1}{x}}$

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

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

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

$= - 1$

Putting them together, we get our final result.

${\lim}_{x \to \infty} {\left(\frac{x}{x + 1}\right)}^{x} = {e}^{{\lim}_{x \to \infty} x \ln \left(\frac{x}{x + 1}\right)} = {e}^{-} 1 = \frac{1}{e}$