# How do you find the limit of (e^x + x)^(1/x) as x approaches 0?

Feb 23, 2016

${\lim}_{x \setminus \to 0} h \left(x\right) = {e}^{2}$

#### Explanation:

Given $h \left(x\right) = {\left({e}^{x} + x\right)}^{\frac{1}{x}}$

First, we'll check how the base comes out if we apply the limit $x \setminus \to 0$. It seems that if we do so, then ${e}^{x} + x = {e}^{0} + 0 = 1$
But, check the power, it seems that then the equation would become ${1}^{\setminus} \infty$. We could leave it at that but since this is limits, we need to be careful here.

So, to solve this we have an equation, that is if $h \left(x\right) = g {\left(x\right)}^{f} \left(x\right)$ and as $x \setminus \to a$ such that $g \left(x\right) \setminus \to 0$ and $f \left(x\right) \setminus \to \infty$, then we have a solution which is
lim_{x\toa}h(x)=e^(h(x)*(g(x)-1)

So, substituting $g \left(x\right) = {e}^{x} + x$ and $f \left(x\right) = \frac{1}{x}$, we get
lim_{x\to0}h(x)=e^(1/x(e^x+x-1)

Let's now just take care of the powers, so we get
${\lim}_{x \setminus \to 0} \frac{{e}^{x} + x - 1}{x} = {\lim}_{h \setminus \to 0} \frac{{e}^{x} - 1}{x} + \frac{x}{x}$
We know about the $\frac{{e}^{x} - 1}{x}$ at $x$ tends to $0$ limit equaling $1$, so we get it in the end as $1 + 1 = 2$

So, that is why there's an $e$ in the answer.