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

Nov 14, 2016

$e$

#### Explanation:

${\left({e}^{x} + x\right)}^{\frac{1}{x}} = e {\left(1 + \frac{x}{e} ^ x\right)}^{\frac{1}{x}}$ now making $y = \frac{x}{e} ^ x$ we have
$e {\left(1 + y\right)}^{{e}^{- x} / y} = e {\left({\left(1 + y\right)}^{\frac{1}{y}}\right)}^{{e}^{- x}}$ here
$\left\{\begin{matrix}x \to \infty \\ y \left(x\right) \to 0\end{matrix}\right.$ then

${\lim}_{x \to \infty} {\left({e}^{x} + x\right)}^{\frac{1}{x}} = e {\lim}_{x \to \infty} {\left({\lim}_{y \to 0} {\left(1 + y\right)}^{\frac{1}{y}}\right)}^{{e}^{- x}} = e \cdot {e}^{0} = e$

Nov 14, 2016

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

#### Explanation:

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

A helpful tool in finding limits with exponential functions like this is to use the equivalence $a = {e}^{\ln} a$. Thus

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

This can be rewritten using the log rule $\log \left({a}^{b}\right) = b \log a$ as

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

We can move the limit inside of the exponential function:

L=e^(lim_(xrarroo)(ln(e^x+x)/x)

Notice that ${\lim}_{x \rightarrow \infty} \ln \frac{{e}^{x} + x}{x}$ is in the indeterminate form $\frac{\infty}{\infty}$, so L'Hopital's rule applies. We can take the derivatives of the numerator and denominator:

L=e^(lim_(xrarroo)((d/dxln(e^x+x))/(d/dxx))

L=e^(lim_(xrarroo)((e^x+1)/(e^x+x))/1

L=e^(lim_(xrarroo)(e^x+1)/(e^x+x)

You may immediately recognize that this limit is $1$, since the ${e}^{x}$ terms in the numerator and denominator will overpower the other terms, so as $x$ approaches infinity, $\frac{{e}^{x} + 1}{{e}^{x} + x} \approx {e}^{x} / {e}^{x} = 1$, but in case this isn't enough for you we can do L'Hopital's again since we are in the $\frac{\infty}{\infty}$ form.

L=e^(lim_(xrarroo)(e^x/(e^x+1))

And L'Hopital's once more, as we're still in $\frac{\infty}{\infty}$:

L=e^(lim_(xrarroo)(e^x/e^x)

$L = {e}^{{\lim}_{x \rightarrow \infty} 1}$

$L = {e}^{1}$

$L = e$