# What is the limit of  (1+2x)^(1/x) as x approaches infinity?

Feb 26, 2017

1

#### Explanation:

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

We can use exponentials and logs here:

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

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

Because $e$ is a continuous function in $\left(- \infty , \infty\right)$, we can also say this:

$= {e}^{{\lim}_{x \to \infty} \textcolor{red}{\frac{\ln \left(1 + 2 x\right)}{x}}}$

....and we can focus on the term in red.

We know straightaway that, because the log will grow more slowly than the $x$ term, then the term in red will go to zero; and so the limit is $1$.

We can use L'Hopital's Rule to see this out as this is in $\frac{\infty}{\infty}$ indeterminate form. One application of L'H takes us here:

$\ln \frac{1 + 2 x}{x} \to \frac{\frac{2}{1 + 2 x}}{1} = \frac{2}{1 + 2 x}$

And so our limit is:

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

$= {e}^{{\lim}_{x \to \infty} 0} = 1$