# How do you find the Limit of ln(1+e^x)/(5x)  as x approaches infinity?

##### 1 Answer
May 21, 2016

$\frac{1}{5}$

#### Explanation:

We can rewrite the argument of the natural logarithm function:

${\lim}_{x \rightarrow \infty} \ln \frac{1 + {e}^{x}}{5 x} = {\lim}_{x \rightarrow \infty} \ln \frac{{e}^{x} \left({e}^{-} x + 1\right)}{5 x}$

Use the logarithm rule $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$ to split up the logarithm in the numerator:

$= {\lim}_{x \rightarrow \infty} \frac{\ln \left({e}^{x}\right) + \ln \left({e}^{-} x + 1\right)}{5 x} = {\lim}_{x \rightarrow \infty} \frac{x + \ln \left(\frac{1}{e} ^ x + 1\right)}{5 x}$

Split the fraction:

$= {\lim}_{x \rightarrow \infty} \left(\frac{1}{5} + \ln \frac{\frac{1}{e} ^ x + 1}{5 x}\right) = \frac{1}{5} + {\lim}_{x \rightarrow \infty} \ln \frac{\frac{1}{e} ^ x + 1}{5 x}$

Note that as $x$ approaches infinity in this limit, $\frac{1}{e} ^ \infty$ goes to $0$:

$= \frac{1}{5} + \ln \frac{\frac{1}{e} ^ \infty + 1}{5 \cdot \infty} = \frac{1}{5} + \ln \frac{1}{\infty} = \frac{1}{5}$

Recall that $\ln \left(1\right) = 0$.