TITLE: Compute a limit without L'Hopital's rule $\lim_{x \to a} \frac{a^x-x^a}{x-a}$
QUESTION [7 upvotes]: I would want to know how we can compute the following limit by using only fundamental limits. 
$$\lim_{x \to a} \dfrac{a^x-x^a}{x-a},$$ where $a$ is a positive real number. 
My idea was to use a substitution: $y=x-a$. We get
$$\lim\limits_{y \to 0} \dfrac{a^aa^y-(y+a)^a}{y}
=a^a\left[ \lim\limits_{y \to 0} \dfrac{a^y-1+1-(\frac{y+a}{a})^a}{y} \right]
=a^a\left[ \ln a+\lim\limits_{y \to 0}\frac{1-(\frac{y+a}{a})^a}{y} \right].
$$
I am looking forward to read any tips on how I can continue from this point. Any help is appreciated.

REPLY [13 votes]: $$\lim_{x \to a} \dfrac{a^x-x^a}{x-a}=\lim_{x \to a} \dfrac{a^x+a^a-a^a-x^a}{x-a}$$$$=\lim_{x \to a} \dfrac{a^x-a^a}{x-a}-\lim_{x \to a} \dfrac{x^a-a^a}{x-a}=f'(a)-g'(a)$$
where $f(x)=a^x$ and $g(x)=x^a$