Limit of $\frac{x^a-a^x}{a^x-a^a}$

Question

Limit of $$\lim_{x\to a} \frac{x^a-a^x}{a^x-a^a},$$ where $$a\in (0,\infty), a\neq1.$$ I know to use L'Hopital, but i am confused with the requirments of $a$.

Answer 1

The function that you are computing the limit of, $$f(x)=\frac{x^a-a^x}{a^x-a^a}\,,$$ is not defined for either $a=1$ or $a=0$, as the denominator would be zero for all $x$. Also, we need $a\geq0$, because $a<0$ implies that $a^a$ is a complex number. Those are the reason of the conditions on $a$: $$ a\geq0 \wedge (a\notin\{0,1\}) \Rightarrow a\in(0,\infty)\wedge a\neq1.$$

To compute the limit, use L'Hopital knowing that $\frac{{\rm d}}{{\rm d}x}a^x = a^x\log(a)$ and $\frac{{\rm d}}{{\rm d}x}x^a = ax^{a-1}$.

Answer 2

The way without L'Hospital: $$\lim_{x\rightarrow a}\frac{x^a-a^x}{a^x-a^a}=-1+\lim_{x\rightarrow a}\frac{x^a-a^a}{a^x-a^a}=-1+\lim_{x\rightarrow a}\frac{\frac{x^a-a^a}{x-a}}{\frac{a^x-a^a}{x-a}}=$$ $$=-1+\frac{(x^a)'_{x=a}}{(a^x)'_{x=a}}=-1+\frac{a\cdot a^{a-1}}{a^a\ln{a}}=-1+\frac{1}{\ln{a}}.$$