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Evaluate $$\lim_{x\to 0}x^x$$

I tried by writing $x$ in terms of exponentials:

$x=e^{\ln x}$ so $x^x=e^{x\ln x}$

$\lim_{x \to 0}x \ln x=\lim_{x \to 0}(\ln x +1) =-\infty$

Thus $\lim_{x\to 0}x^x=0$

Is this correct?

Martin Sleziak
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Rescy_
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  • No, $x\ln x\ne\ln x+1$. –  Jul 30 '15 at 01:42
  • I get L'Hopital's rule wrong... Thanks everybody for your help. It is also interesting to see so many version of the rule, 'L Hospital', 'L Hopital' and 'L Hopitale'... – Rescy_ Jul 30 '15 at 01:57
  • You need to restrict to $x\to 0^+$, since the function is not defined for $x<0$. – Thomas Andrews Jul 30 '15 at 02:14
  • @ThomasAndrews Technically the function is defined at $x<0$. It can exist in $\left{\left.x=-\frac{m}{2n+1}\right|m,n \in \mathbb{N}\right}$. – Arbuja Sep 11 '16 at 01:03

4 Answers4

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It is correct to first calculate $\lim_{x \to 0}x\ln (x)=L$ and then, since $\exp$ is continuous the desire result will be $e^L$, but why do you think $x \ln (x) = \ln(x)+1$? That it is not true. Better use that $$ x \ln (x) = \frac{\ln(x)}{1/x} $$ Now apply L´Hopital's rule to get that $ \lim_{x \to 0}x \ln (x)= -0 $, and conclude that $\lim_{x \to 0}x^x =1$ .

Martin Sleziak
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Alonso Delfín
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2

Apply L'hopitale rule:

$x\ln x = \dfrac{\ln x}{\dfrac{1}{x}}\to \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^2}}=-x \to 0^{-}\Rightarrow x^x \to e^{0}=1$

DeepSea
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2

Here is a way that does not involve L'Hopital's rule. Consider the logarithm $f(x)=x\ln x$. Write $x=\mathrm e^{-y}$, so that $x\to0$ means $y\to\infty$. Then $$f(x)=-\frac{y}{\mathrm e^{y}}.$$Just from the third term $\frac12y^2$ of the exponential series, we can see that this tends to $0$ (on the negative side) as $y\to\infty$. So the limit of the original expression $x^x=\exp f(x)$ is $1$.

John Bentin
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  • Note: Here you use $e^x=\sum\limits_{k=0}^\infty \frac{x^k}{k!}$ instead of $e^x=1+\sum\limits_{k=1}^\infty \frac{x^k}{k!}$ which is the right form. You get $0^0=1$ by (a senseful) definition. – user90369 Sep 09 '16 at 18:51
  • $0^0$ isn't defined. – PinkyWay May 20 '20 at 17:58
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    @Cheesecake That depends on whether one defines it or not. Where I come from, it standardly is. And before the age of the internet I've never encountered anyone who preferred to leave it undefined. – Daniel Fischer May 20 '20 at 18:10
  • Nice solution (+1)upvote. –  Jan 05 '21 at 09:57
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Put $y = x^x$. Then $\log(y) = x\log(x)$. As $x\downarrow 0$, we have a $0\cdot \infty$ situation. So we rearrange to get $$\log(y) = {\log(x)\over 1/x}$$ As $x\to 0$ we have $\infty/\infty$ so we can invoke L'hospital to get $$\lim_{x\downarrow 0} y = {1/x\over -1/x^2} = \lim_{x\downarrow 0} -x = 0.$$

Therefore the limit is 1. Caveat: You must take the limit as $x\downarrow 0$, or you have domain problems.

ncmathsadist
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