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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 $1/x$.

Julian Kuelshammer
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Lizi
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4 Answers4

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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.

Brian M. Scott
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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 07:25
  • @Lizi: Yes. He said that nicely. – Mikasa Dec 20 '12 at 07:27
  • @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 07:27
  • @BrianM.Scott Should the first line be $\ln \lim\ldots$, not $\ln\lim\ln\ldots$? – Daryl Dec 20 '12 at 07:30
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    @Daryl: It should indeed; that was a copy-paste error during editing. Thanks for catching it. – Brian M. Scott Dec 20 '12 at 07:36
  • Very frequently, L'Hospital's rule is overkill and may be replaced by the more elementary recognition and calculation of some derivative. This is the case here: setting $$f(x)=\ln(e^x+x),$$ we find immediately $$\lim_{x\rightarrow 0} \frac{\ln(e^x+x)}x=f'(0)=2.$$ – Anne Bauval Jun 01 '23 at 13:59
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    @AnneBauval: Perfectly true. In practice, however, it’s likely to be quicker to apply the mechanical procedure, especially in the likely event that one took the log with that procedure in mind. (It’s a useful point, though, and even at this late date you might consider posting it as another answer; if you did so and left me a note, I’d certainly upvote it.) – Brian M. Scott Jun 01 '23 at 18:03
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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.

Teddy
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You might start by taking the logarithm.

Robert Israel
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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 $e^2$.

Joe
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