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

Nov 29, 2016

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

#### Explanation:

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

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

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

The above step is valid due to the continuity of $f \left(x\right) = {e}^{x}$

Evaluating the limit in the exponent, we have

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

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

The above step follows from apply L'Hopital's rule to a $\frac{\infty}{\infty}$ indeterminate form

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

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

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

$= \frac{1}{1 + \frac{1}{\infty}}$

$= \frac{1}{1 + 0}$

$= 1$

Substituting this back into the exponent, we get our final result:

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

$= {e}^{1}$

$= e$