# 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 $1/x$.

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But you can’t distribute the $1/x$: $(a+b)^c\ne a^c+b^c$. And you really ought to read this on accepting answers. – Brian M. Scott Dec 20 '12 at 7:11

## 4 Answers

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.

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So then it would be the limit of (e^x+1)/(e^x+x) as x goes to zero? – Lizi Dec 20 '12 at 7:25
@Lizi: Yes. He said that nicely. – Babak S. Dec 20 '12 at 7:27
@Lizi: As Babak says, yes; and that limit is a nice easy one. Oh, and thank you for going back and accepting some of the old answers. – Brian M. Scott Dec 20 '12 at 7:27
@BrianM.Scott Should the first line be $\ln \lim\ldots$, not $\ln\lim\ln\ldots$? – Daryl Dec 20 '12 at 7:30
@Daryl: It should indeed; that was a copy-paste error during editing. Thanks for catching it. – Brian M. Scott Dec 20 '12 at 7:36

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.

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You might start by taking the logarithm.

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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 $e^2$.

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