I need to do this limit without using L'Hopital rule $$ \lim_{x\to a} \frac{x^x-a^a}{x-a} $$
I tried to make a substitution and to write $x^x$ as $e^{x\ln x}$ and to approach it using the fact that $$ \lim_{h\to 0} \frac{a^h-1}{h}= \ln{a} $$ but that doesn't seem to go anywhere. Any suggestion would help.
I thought it might be instructive to present an approach that relies on pre-calculus methodologies only. Proceeding, we can write
$$\begin{align} \lim_{x\to a}\frac{x^x-a^a}{x-a}&=\lim_{x\to a}\frac{(x^x-x^a)+(x^a-a^a)}{x-a}\\\\ &=\lim_{x\to a}\frac{e^{x\log(x)}-e^{a\log(x)}}{x-a}+\lim_{x\to a}\frac{e^{a\log(x)}-e^{a\log(a)}}{x-a}\\\\ &=\lim_{x\to a}e^{a\log(x)}\left(\frac{e^{(x-a)\log(x)}-1}{x-a}\right)\tag1\\\\ &+\lim_{x\to a}a^a\left(\frac{e^{a\log(x/a)}-1}{x-a}\right)\\\\ &=a^a\log(a)+a^a\tag2 \end{align}$$
where in going from $(1)$ to $(2)$, we used the Squeeze Theorem along with the estimates $$1+x<e^x<\frac1{1-x}$$ for $x<1$ and $$\frac{x-1}{x}<\log(x)<x-1$$ that I derived in THIS ANSWER, which were based on the limit definition of the exponential function and Bernouli's inequality.
TOOLS USED:
The limit defintion of the exponential function, Bernoulli's Inequality, and basic algebra to obtain estimates for $e^x$ and $\log(x)$, and the Squeeze Theorem
