Indeterminate Forms 0^0, 1^Infinity and Infinity^0

The forms $0^0,\ 1^{\infty},\ \infty^0$ all arise from an expression $g(x)^{h(x)}$ where the base and exponent separately approach values that make the combination indeterminate. Each is resolved by the same three-step logarithm procedure:

1. Let $A=\displaystyle\lim_{x\to a}g(x)^{h(x)}$; assume $A>0$ (to keep $\log$ continuous) and take logarithms: $\log A=\displaystyle\lim_{x\to a}\log\!\big(g(x)^{h(x)}\big)=\displaystyle\lim_{x\to a}h(x)\log g(x)$.
2. The right-hand limit is now a $0\times\infty$ form; rewrite it as a $\tfrac00$ or $\tfrac{\infty}{\infty}$ ratio and apply l'Hôpital's Rule (possibly more than once) to evaluate it.
3. If that evaluated limit is $\alpha$, then $\log A=\alpha$, so the original limit is $A=e^{\alpha}$ (exponentiating undoes the logarithm taken in step 1). …