# What is the limit of (1+4/x)^x as x approaches infinity?

Apr 17, 2016

${\lim}_{x \to \infty} {\left(1 + \frac{4}{x}\right)}^{x} = {e}^{4}$

#### Explanation:

Notice that

${\left(1 + \frac{4}{x}\right)}^{x} = {e}^{x \ln \left(1 + \frac{4}{x}\right)}$

and if the limit exists,

${\lim}_{x \to \infty} \left({e}^{x \ln \left(1 + \frac{4}{x}\right)}\right) = {e}^{{\lim}_{x \to \infty} \left(x \ln \left(1 + \frac{4}{x}\right)\right)}$

as the exponential function is continuous everywhere.

To evaluate the limit at the exponent, we first write it as

$x \ln \left(1 + \frac{4}{x}\right) = \frac{\ln \left(1 + \frac{4}{x}\right)}{\frac{1}{x}}$

Since the form is indeterminate $\frac{0}{0}$, use the L'hospital rule.

${\lim}_{x \to \infty} \left(\ln \frac{1 + \frac{4}{x}}{\frac{1}{x}}\right) = {\lim}_{x \to \infty} \left(\frac{\frac{d}{\mathrm{dx}} \left(\ln \left(1 + \frac{4}{x}\right)\right)}{\frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)}\right)$

$= {\lim}_{x \to \infty} \left(\frac{\frac{- \frac{4}{x} ^ 2}{\left(1 + \frac{4}{x}\right)}}{- \frac{1}{x} ^ 2}\right)$

$= {\lim}_{x \to \infty} \left(\frac{4}{1 + \frac{4}{x}}\right)$

$= \frac{4}{1 + 0}$

$= 4$

Therefore, the limit is ${e}^{4}$.

Apr 17, 2016

If you are familiar with the sometimes definition of $e$, as ${\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u} = e$, then we don't need l"Hospital.

#### Explanation:

${\lim}_{x \rightarrow \infty} {\left(1 + \frac{4}{x}\right)}^{x} = {\lim}_{x \rightarrow \infty} {\left(1 + \frac{1}{\frac{x}{4}}\right)}^{4 \left(\frac{x}{4}\right)}$

$= {\lim}_{x \rightarrow \infty} {\left({\left(1 + \frac{1}{\frac{x}{4}}\right)}^{\frac{x}{4}}\right)}^{4}$

Now, with $u = \frac{x}{4}$, we have

$= {\lim}_{u \rightarrow \infty} {\left({\left(1 + \frac{1}{u}\right)}^{u}\right)}^{4}$

$= {\left({\lim}_{u \rightarrow \infty} {\left(1 + \frac{1}{u}\right)}^{u}\right)}^{4} = {e}^{4}$