Limit of a function, not using L'Hospital's Rule I have encountered this problem and I don't know how to approach it, mainly because we can't use L'Hospital's Rule. The limit is as follows:
$$\lim_{x\to a} {x^x-a^a \over x-a}$$
Thanks for any suggestions.
edit: Solved, thanks for the responses.
 A: Hint 1:
$$
\begin{align}
\frac{e^{x\log(x)}-e^{a\log(a)}}{x-a}
&=e^{a\log(a)}\frac{e^{x\log(x)-a\log(a)}-1}{x-a}\\
&=e^{a\log(a)}\frac{e^{\color{#C00000}{x\log(x)-a\log(a)}}-1}{\color{#C00000}{x\log(x)-a\log(a)}}\frac{x\log(x)-a\log(a)}{x-a}
\end{align}
$$
Hint 2:
$$
\begin{align}
\frac{x\log(x)-a\log(a)}{x-a}
&=\frac{x(\log(x)-\log(a))}{x-a}+\log(a)\\
&=\frac xa\frac{\color{#C00000}{\log(x)-\log(a)}}{e^{\color{#C00000}{\log(x)-\log(a)}}-1}+\log(a)
\end{align}
$$
Apply $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$.

Although you commented that you could use $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$, here is a proof of this fact using the inequality $1+x\le e^x$. This inequality is shown using Bernoulli's Inequality in this answer.
For $|x|\lt1$, $1+x\le e^x$ and $1-x\le e^{-x}$. Therefore,
$$
1+x\le e^x\le\frac1{1-x}
$$
Subtracting $1$ and dividing by $x$ yields
$$
\begin{array}{}
1\le\frac{e^x-1}{x}\le\frac1{1-x}&\text{when }x\gt0\\
1\ge\frac{e^x-1}{x}\ge\frac1{1-x}&\text{when }x\lt0
\end{array}
$$
That is, when $|x|\lt1$,
$$
\frac{e^x-1}{x}\text{ is between }1\text{ and }\frac1{1-x}
$$
By the Squeeze theorem,
$$
\lim_{x\to0}\frac{e^x-1}{x}=1
$$
A: No we can use Lopital's rule so
If $y=x^x$ and $x\gt0$ then $\ln(y)=\ln(x^x)$ 
Use properties of logarithmic functions to expand the right side of the above equation as follows. 
$$
\ln(y)=x\ln(x) 
$$
We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right. 
$$
y'(1/y)=\ln(x) + x(1/x)=\ln(x) + 1 \text{, where }y'=\mathrm{d}y/\mathrm{d}x 
$$
Multiply both sides by $y$ 
$$
y'=(\ln(x) + 1)\,y 
$$
Substitute $y$ by $x^x$ to obtain 
$$
y'=(\ln(x) + 1)\,x^x 
$$
then 
$$
\lim_{x\to a} {x^x-a^a \over x-a}=\lim_{x\to a} {(\ln(x) + 1)\,x^x \over 1}=(\ln(a) + 1)\,a^a
$$
