Question

without L'hopitals rule
\[\lim_{x\to a}\frac{a^x-x^a}{a^ax-a^{a+1}}\]

Answer 1

simplification shows that
\[\frac{1}{a^{a}}\lim_{x\to a}\frac{a^x-x^a}{x-a}\]
so a question can be asked as
\[\lim_{x\to a}\frac{a^x-x^a}{x-a}\]

Answer 2

\[ \lim_{x\to a}\frac{a^x-a^a+a^a-x^a}{x-a}=\lim_{x\to a}\frac{a^x-a^a}{x-a}-\lim_{x\to a}\frac{x^a-a^a}{x-a}\]

if we let \[h=x-a\] we get
\[\lim_{h\to0}\frac{(a)^{a+h}-a^a}{h}-\lim_{h\to0}\frac{(a+h)^{a}-a^a}{h}\]

now consider two functions \[f(x)=a^x\\f'(x)=\lim_{h\to0}\frac{(a)^{a+h}-a^x}{h}=a^x\ln a\\g(x)=x^a\\g'(x) =\lim_{h\to0}\frac{(a+h)^{a}-x^a}{h}=ax^{a-1}\]
so our limit is the derivative evaluated at \[x=a\]
hence
\[\frac{1}{a^a}\lim_{x\to a}\frac{(a)^{x}-x^a}{x-a}=\frac{1}{a^a}(a^a\ln a-aa^{a-1})=\ln a-1\]