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I am trying to use L'Hopital's rule to evaluate the following limit but I am not sure I am doing it correctly.

$$\lim_{x \to 0} \left(e^x+x\right)^{\frac{1}{x}}.$$

To use L'Hopitals rule I have been told that the limit has to evaluate to certain in-determinants before it can be applied, in the case above we would get $1^{\infty}$ so it can be used.

Here are my steps to evaluate the limit:

$$\lim_{x \to 0} \left(e^x+x\right)^{\frac{1}{x}}.$$

$$\lim_{x \to 0} \ln\left(\left(e^x+x\right)^{\frac{1}{x}}\right) = \lim_{x \to 0} \frac{1}{x} \cdot \ln\left(e^x+x\right)$$ $$\lim_{x \to 0} \frac{\ln\left(e^x+x\right)}{x} = \lim_{x \to 0} \frac{1}{x^2+xe^x}$$ $$\lim_{x \to 0} \frac{0}{2x+e^x+xe^x} = \frac{0}{1}=0$$

As I took the natural log at the start the evaluated limit would be $e^0=1$.

So I have two questions, firstly is my method correct? I am particularly unsure about what to do when (for example in the 4th line) I get a 0 as my numerator, is $\frac{0}{1}$ considered to be an in determinant? Secondly if the method is correct, is my answer correct? Is there a way to prove what I have done is the limit evaluated?

I am not sure on the definition of an in-determinant as in my lecture notes I have that I can apply L'Hopital's rule on problems with $1^{\infty}$ but I am not sure if this is an in-determinant in the same way that $\frac{0}{0}$ is considered to be undefined. I am not sure my last point is entirely clear so if it is not please let me know and I will try to clarify further.

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Argh! You cannot apply De l'Hospital's rule to limits which are not $[0/0]$ or $[?/\infty]$!!! – Siminore Oct 24 '12 at 14:13
But by taking a log of the limit it can be rewritten in the form $\frac{0}{0}$ and therefore it can be used, can it not, as $\frac{1}{0}$ is not defined? – Aesir Oct 24 '12 at 14:17
up vote 7 down vote accepted

L’Hospital’s rule applies only to limits of the form $0/0$ and $\infty/\infty$. However, there is a standard trick for converting the indeterminate form $1^\infty$ to one of these forms so that l’Hospital’s rule can be applied, and that’s essentially what you’re doing here. Let me write it up in a way that makes a little clearer exactly what is going on.

Let $L=\displaystyle\lim_{x\to 0}\left(e^x+x\right)^{1/x}$. Then $$\ln L=\ln\lim_{x\to 0}\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}\;,$$ since $\ln$ is a continuous function. Thus,

$$\ln L=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\frac1x\ln(e^x+x)=\lim_{x\to 0}\frac{\ln(e^x+x)}x\;.$$

This last limit is a $\frac00$ indeterminate form, so l’Hospital’s rule applies:

$$\ln L=\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{\frac{e^x+1}{e^x+x}}1=\lim_{x\to 0}\frac{e^x+1}{e^x+x}=\frac21=2\;.$$

Finally, then $L=e^{\ln L}=e^2$.

You went astray when you wrote $$\lim_{x \to 0} \frac{\ln\left(e^x+x\right)}{x} = \lim_{x \to 0} \frac{1}{x^2+xe^x}\;:$$ you did not differentiate the numerator correctly, and you did not differentiate the denominator at all.

If $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=1$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac01=0$; there is nothing indeterminate here.

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Thanks this has cleared it up perfectly for me, cheers! – Aesir Oct 24 '12 at 14:29
@Aesir: Glad to help! – Brian M. Scott Oct 24 '12 at 14:32

You start out fine, but there are two mistakes in your third line. You're right that you can use L'Hopital's rule, but you forgot to differentiate the bottom, and futhermore the derivative of $\ln(e^x+x)$ is not $\frac{1}{e^x+x}$.

When you differentiate both top and bottom as required you ought to get $$\frac{D[\ln(e^x+x)]}{D(x)}=\frac{\frac{e^x+1}{e^x+x}}{1}=\frac{e^x+1}{e^x+x}$$ Then take $x$ to $0$ to get the logarithm of your limit is $2$, not $0$.

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