TITLE: Limit Computation of $(e^x+x)^{1/x}$ as $x$ approaches zero
QUESTION [5 upvotes]: I need help computing the limit of $(e^x+x)^{1/x}$ as $x$ approaches zero. I just need help getting started with the computation. The only way I can think of rearranging the equation is distributing the $1/x$.

REPLY [7 votes]: Without using l'Hopital, you may do this:
$$ \lim_{x\to 0} (e^x + x)^{\frac{1}{x}} = \lim_{x\to 0} (e^x(1 + \frac{x}{e^x}))^{\frac{1}{x}}=$$
$$ = \lim_{x\to 0} e (1 + \frac{x}{e^x})^{\frac{1}{x}}= e \lim_{x\to 0} \big[(1 + \frac{x}{e^x})^{\frac{e^x}{x}}\big]^\frac{1}{e^x}=$$
$$ = e \big[\lim_{x\to 0} (1 + \frac{x}{e^x})^{\frac{e^x}{x}}\big]^{\lim_{x\to 0}\frac{1}{e^x}}= e \cdot e^1 = e^2.$$
Use has been made of the facts that:
If $u(x) \to 0$ as $x \to 0$, then $$(1+u(x))^{\frac{1}{u(x)}} \to e,$$
and $$\lim u(x) ^{v(x)} = \big(\lim u(x) \big)^{\lim v(x)},$$
provided that the individual limits exist.