# Evaluating $\lim\limits_{x \to a} \frac{x^a - a^x}{x^a - a^a}$ without L'Hopital

I'm trying to solve this limit without using L'Hopital rule. I already tried multiplying up and down by $x^a+a^a$, finding bounds for squeeze theorem, substitution of variables, but got nothing...

$a$ is a positive real number different than $1$.

Any help would be appreciated.

• Is $a \in \mathbb N$? – Staki42 Jan 21 '18 at 18:12
• Oh, sorry, forgot to mention. $a \in \mathbb{R} , a>0, a \neq 1$ – user286485 Jan 21 '18 at 18:15
• @Lorenzo \displaystyle (and similarly \dfrac, \dbinom, etc.) should not be used in question titles. For more details, see: Guidelines for good use of $\LaTeX$ in question titles – Martin Sleziak Jan 30 '18 at 4:14
• @MartinSleziak Thank you, I'll be more careful in the future – Lorenzo B. Jan 30 '18 at 15:56

By definition of the derivative we obtain: $$\lim_{x\rightarrow a}\frac{x^a-a^x}{x^a-a^a}=\lim_{x\rightarrow a}\frac{\frac{x^a-a^a}{x-a}-\frac{a^x-a^a}{x-a}}{\frac{x^a-a^a}{x-a}}=\frac{a\cdot a^{a-1}-a^a\ln{a}}{a\cdot a^{a-1}}=1-\ln{a}.$$
$$=1-\dfrac{a^x-a^a}{x^a-a^a}$$
Now $$\dfrac{a^x-a^a}{x-a}=a^a\cdot\dfrac{a^{x-a}-1}{x-a}$$
Use $\lim_{h\to0}\dfrac{a^h-1}h=\ln a$ for $a>0$
Similarly for $a^{a-1}\cdot\dfrac{(x/a)^a-1}{(x/a)-1}$