Answer
$\lim\limits_{x \to \infty}(1+\frac{2}{x})^{x}=e^{2}=7.389$
Work Step by Step
We can look at a graph of $f(x)=(1+\frac{2}{x})^{x}$ and see that as $x$ gets larger and larger, $f(x)$ gets closer to 7.389.
$\lim\limits_{x \to \infty}(1+\frac{2}{x})^{x}=(1+\frac{2}{\infty})^{\infty}=1^{\infty}$
We have an indeterminate form of $1^{\infty}$, we can rewrite this in terms of natural logs.
$\ln((1+\frac{2}{x})^{x})=\ln y=x\ln(1+\frac{2}{x})$
$\lim\limits_{x \to \infty} x\ln(1+\frac{2}{x})=\infty*\ln(1+\frac{2}{\infty})=\infty*\ln1=\infty*0$
We now have an indeterminate form of $\infty*0$.
We can rewrite this into $\lim\limits_{x \to \infty} \frac{ln(1+\frac{2}{x})}{\frac{1}{x}}$ and apply L'Hospital's rule.
$\lim\limits_{x \to \infty} \frac{ln(1+\frac{2}{x})}{\frac{1}{x}}=\lim\limits_{x \to \infty} \frac{-2x^{2}}{-x^{2}-2x}$
$\lim\limits_{x \to \infty} \frac{-2x^{2}}{-x^{2}-2x}=\lim\limits_{x \to \infty} \frac{-4x}{-2x-2}$
$\lim\limits_{x \to \infty} \frac{-4x}{-2x-2}=\lim\limits_{x \to \infty} \frac{-4}{-2}$
$\lim\limits_{x \to \infty} \frac{-4}{-2}=\lim\limits_{x \to \infty} 2$
Remember, that this is in terms of logs, in order to cancel out the logs by raising $e$ to the power of the answer
$\lim\limits_{x \to \infty} 2=\lim\limits_{x \to \infty}\ln y= \lim\limits_{x \to \infty}e^{ln y}=\lim\limits_{x \to \infty} e^{2}$
$e^{2}=7.389$
