7
$\begingroup$

I'm having a hard time understanding why this limits equals to 0.

By simply using logarithm identity I get that this limit equals

$ \lim_{x\to 0} \ e^{x\ln x^{x}}=\lim_{x\to 0}\ e^{x^{2}\ln x}$ and by L'Hopital we get that the limit of the exponent is 0 and because $f(a)=e^{a}$ is continuous I get that the final limit should be 1.

I'd love your help with understanding what I did wrong.

Thank you.

  • 0
    @Joel Cohen: A good way of remembering. My way is "its the really really fast one."2011-06-21

2 Answers 2

9

You considered $(x^x)^x$ instead of $x^{(x^x)}$.

  • 1
    Note that, in any case, the limit should be one-sided, since $x^x$ is not defined on the negative real axis.2011-06-21
12

The mistake you made is parenthesis in the exponent. In any case, here is another way to solve this problem:

Since $\lim_{x\rightarrow 0^+} x^x =1$, there is an interval $(0,\delta)$, $0<\delta<1$ such that $\frac{1}{2}. Then on this interval, $x^{\frac{3}{2}}\leq x^{x^x} \leq x^{\frac{1}{2}}$ so we see the limit is $0$ by the Squeeze Theorem.

Hope that helps,