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Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to 0^+} x^{\left( x^x - 1 \right)}. \]

To evaluate this we’ll use the definition of a real number raised to a real power (i.e., a^x = e^{x \log a}), and use the continuity of the exponential function. Then we apply L’Hopital’s rule twice. (We also use the notation e^x = \exp x 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*}

The second to last line follows since \lim_{x \to 0} x^x = 1 by Example #3 on page 302 of Apostol, which means the numerator is going to 0 as x \to 0^+. However, in the denominator both the terms x (\log x+ 1) and \frac{1}{\log x + 1} go to 0 as x \to 0^+, but then the +1 in the denominator means the denominator is going to 1 as x \to 0^+. Hence, the whole expression is going to 0.


  1. tom says:

    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.

    • RoRi says:

      Yeah, you’re right. It did still get to the right answer, but wasn’t as simple as when those \log x + 1 terms cancelled everywhere. Anyway, I think it’s fixed now.

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