5. Calculate
$$
\lim _{x \rightarrow 0^{+}}\left(x^{x^{x}}-x^{x}\right) .
$$
Solution: $\square$
We first calculate $\lim _{x \rightarrow 0^{+}} x^{x}$ : it is just $\exp \left(\lim _{x \rightarrow 0^{+}} x \ln x\right)$. But
$$
\lim _{x \rightarrow 0^{+}} x \ln x=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{1 / x}=\lim _{x \rightarrow 0^{+}} \frac{1 / x}{-1 / x^{2}}=\lim _{x \rightarrow 0^{+}}-x=0
$$
by L'Hôpital's Rule. Therefore $\lim _{x \rightarrow 0^{+}} x^{x}=1$. Then $\lim _{x \rightarrow 0^{+}} x^{x^{x}}=0^{1}=0$, so $\lim _{x \rightarrow 0^{+}}\left(x^{x^{x}}-x^{x}\right)=-1$