Evaluate $\lim_{x\to 1}\frac{x^{x^a}-x^{x^b}}{\ln^2 x}$ 
Evaluate $$\lim_{x\to 1}\frac{x^{x^a}-x^{x^b}}{\ln^2 x}$$

I've tried to use fundamental limits to solve it and conclude that the limit is $a-b$ as follows:
\begin{align*}
\lim_{x\to 1}\frac{e^{x^a\ln x}-e^{x^b\ln x}}{\ln^2 x}&=\lim_{x\to 1}\frac{e^{x^a\ln x}-1-(e^{x^b\ln x}-1)}{\ln^2 x}\\
&=\lim_{x\to 1}\left(\frac{e^{x^a\ln x}-1}{\ln^2 x}-\frac{e^{x^b\ln x}-1}{\ln^2 x}\right)\\
&=\lim_{x\to 1}\frac{x^a-x^b}{\ln x}=\lim_{x\to 1}\left(\frac{a(e^{a\ln x}-1)}{a\ln x}-\frac{b(e^{b\ln x}-1)}{b\ln x}\right)\\
&= a-b 
\end{align*}
I uses the fact that the $\lim_{x\to a}\dfrac{e^{u(x)}-1}{u(x)}=1$ where  $\lim_{x\to a} u(x)=0$

It's not clear to me what happens in the step from blue to red:

$$\begin{align*}
\lim_{x\to 1}\frac{e^{x^a\ln x}-e^{x^b\ln x}}{\ln^2 x}&=\lim_{x\to 1}\frac{e^{x^a\ln x}-1-(e^{x^b\ln x}-1)}{\ln^2 x}\\
&=\color{blue}{\lim_{x\to 1}\left(\frac{e^{x^a\ln x}-1}{\ln^2 x}-\frac{e^{x^b\ln x}-1}{\ln^2 x}\right)}\\
& = \color{red}{\lim_{x\to 1}\frac{x^a-x^b}{\ln x}}=\lim_{x\to 1}\left(\frac{a(e^{a\ln x}-1)}{a\ln x}-\frac{b(e^{b\ln x}-1)}{b\ln x}\right)\\
&= a-b 
\end{align*}$$

If you take $x = e^u$, then:
$$\lim_{x\to 1}\frac{e^{x^a\ln x}-e^{x^b\ln x}}{\ln^2 x}
=\lim_{u\to 0}\frac{e^{ue^{au}}-e^{ue^{bu}}}{u^2}$$
Now if $u \to 0$, then $ue^{au} \to 0$ so you can use a Taylor expansion on $e^{ue^{au}}$ and $e^{ue^{bu}}$.
