# Problem 18 **

Strictly speaking, we look at x \to 0^+ to avoid dealing with complex exponentiation.

Note that:

\displaystyle x \ln x = \frac {\ln x} {\frac 1 x}

As x \to 0^+, \ln x \to -\infty and \frac 1 x \to \infty. We also have:

\displaystyle \lim_{x \to 0^+} \frac {(\ln x)'} {\left(\frac 1 x\right)'} = \lim_{x \to 0^+} \frac 1 {x \left(-\frac 1 {x^2}\right)} = -\lim_{x \to 0^+} x = 0

Importantly, this limit is existent so we can apply L’Hopital’s rule to conclude that:

\displaystyle \lim_{x \to 0^+} x \ln x = 0

Since the exponential function is continuous we therefore have:

\begin{align*} \lim_{x \to 0^+} x^x & = \lim_{x \to 0^+} e^{x \ln x} \\ & = e^{\lim_{x \to 0^+} (x \ln x)} \\ & = e^0 \\ & = 1\end{align*}

For x^{x^x}, write x^{x^x} = e^{x^x \log x}. As x \to 0^+, x^x \to 1 but \log x \to -\infty, so x^x \log x \to -\infty as x \to 0^+, (ie. because for small positive x, x^x \approx 1 but \log x will be a very large negative number) meaning that \lim_{x \to 0^+} e^{x^x \log x} = 0 since e^x \to 0 as x \to -\infty.

Therefore:

\begin{align*}\lim_{x \to 0^+} \left(x^{x^x} - x^x\right) & = \lim_{x \to 0^+} x^{x^x} - \lim_{x \to 0^+} x^x \\ & = 0 - 1 \\ & = -1\end{align*}