Evaluate the limit.
![\[ \lim_{x \to 0^+} x^{\left( x^x - 1 \right)}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-319e927fac61cd591e37f440ca960a09_l3.png "Rendered by QuickLaTeX.com")
To evaluate this we’ll use the definition of a real number raised to a real power (i.e., 
), and use the continuity of the exponential function. Then we apply L’Hopital’s rule twice. (We also use the notation 
to avoid huge expressions in the exponent.)
![\begin{align*} \lim_{x \to 0^+} x^{\left( x^x - 1 \right)} &= \lim_{x \to 0^+} e^{\left(x^x -1 \right) \log x} \\[9pt] &= \exp \left( \lim_{x \to 0^+} \left((x^x-1) \log x \right)\right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ \log x}{\frac{1}{x^x-1}} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ \frac{1}{x}}{ \frac{x^x (\log x + 1)}{(x^x-1)^2}} \right) &(\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{(x^x-1)^2}{x^x(x \log x + x)} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{2(x^x-1) x^x (\log x + 1)}{x^x (x \log x + x)(\log x + 1) + x^x(\log x+ 2)}\right) &\text{L'Hopital} \\[9pt] &= \exp \left( \lim_{ x \to 0^+} \frac{2(x^x-1)(\log x + 1)}{(x \log x + x)(\log x + 1) + \log x + 2} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{2(x^x -1)}{x(\log x + 1) + 1 + \frac{1}{\log x + 1}} \right) \\[9pt] &= e^0 \\ &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4e851db3e01d29c9f82f1104b7cbb1a5_l3.png "Rendered by QuickLaTeX.com")
The second to last line follows since 
by Example #3 on page 302 of Apostol, which means the numerator is going to 0 as 
. However, in the denominator both the terms 
and 
go to 0 as , but then the 
in the denominator means the denominator is going to 1 as . Hence, the whole expression is going to 0.
Looks like an error in the 2nd application of L’Hopital, in the denominator; (xlogx+x)’ is logx+2, not logx+1. Works out to the same answer though.
Yeah, you’re right. It did still get to the right answer, but wasn’t as simple as when those
terms cancelled everywhere. Anyway, I think it’s fixed now.