Limit Computation of $(e^x+x)^{1/x}$ as $x$ approaches zero 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 $1/x$.
 A: 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.
A: You might start by taking the logarithm.
A: This limit is a $1^\infty$ indeterminate form, and there’s a standard method for attacking such limits. Let $$L=\lim_{x\to 0}\left(e^x+x\right)^{1/x}\;.$$ The log function is continuous, so
$$\begin{align*}
\ln L&=\ln\lim_{x\to 0}\left(e^x+x\right)^{1/x}\\
&=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}\\
&=\lim_{x\to 0}\frac{\ln\left(e^x+x\right)}x\;.
\end{align*}$$
Here the numerator and denominator both tend to $0$ as $x\to 0$, so you can apply l’Hospital’s rule.
A: Hint:
Take the logarithm and rewrite the limit to use L'Hopital. The result ends up being $e^2$.
What I mean is writing it as
$$\exp\left(\lim_{x\to 0} \frac{\log(x+e^x)}{x} \right)$$
It is now of $0/0$, so you can use L'Hopital.
Alternative hint:
Possibly rewrite it as a series and notice the higher order terms go to $0$. Thus, you can approximate the limit as just the first term of the Taylor Series, which is just $e^2$.
