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

Apr 27, 2016

$y = {e}^{4}$

#### Explanation:

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

This is an indeterminate type so we use l'Hopital's Rule. That is, find the limit of the derivative of the top divided by the derivative of the bottom.

Let $y = {\left(1 + \frac{4}{x}\right)}^{x}$

$\ln y = \ln {\left(1 + \frac{4}{x}\right)}^{x}$-> Take $\ln$ of both sides

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

$\ln y = \ln \frac{1 + \frac{4}{x}}{x} ^ - 1$

${\lim}_{x \to \infty} \ln y = {\lim}_{x \to \infty} \ln \frac{1 + \frac{4}{x}}{x} ^ - 1$

${\lim}_{x \to \infty} \ln y = {\lim}_{x \to \infty} \frac{\frac{1}{1 + \frac{4}{x}} \cdot - \frac{4}{x} ^ 2}{- \frac{1}{x} ^ 2}$

${\lim}_{x \to \infty} \ln y = {\lim}_{x \to \infty} \frac{1}{1 + \frac{4}{x}} \cdot - \frac{4}{x} ^ 2 \times - {x}^{2}$

${\lim}_{x \to \infty} \ln y = {\lim}_{x \to \infty} \frac{4}{1 + \frac{4}{x}}$

$\ln y = \frac{4}{1 + \frac{4}{\infty}}$ Note that the limit of ln y is just ln y since it is a constant

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

${e}^{\ln} y = {e}^{4}$

$y = {e}^{4}$

Apr 27, 2016

${e}^{4}$

#### Explanation:

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

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

As $x \rightarrow \infty$, the ration $\frac{x}{4} \rightarrow \infty$ so the inside expression goes to $e$.

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

$= {\left(e\right)}^{4} = {e}^{4}$