Limits - Calculating $\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$, where $a \gt 0$, without using L'Hospital's rule 
Calculate  $\displaystyle\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$, where $a \gt 0$, without using L'Hospital's rule.

I'm messing around with this limit. I've tried using substitution for $x^a -1$, but it didn't work out for me.
I also know that $(x-1)$ is a factor of  $x^a -1$, but I don't know where to go from here. 
EDIT: Solved it, posting here for future generations :)
a) We can write $x^a$ as $e^{a\ln x}$ ending up with $\lim\limits_{x\to 1} \frac{e^{a\ln x} -1}{x-1}$
b) Multiplying by $\frac{a\ln x}{a\ln x}$ we end up with: $\lim\limits_{x\to 1} \frac{e^{a\ln x} -1}{a\ln x} \cdot \frac{\ln x}{x-1} \cdot a$
c) Now we just have to show that the first 2 limits are equal 1, and $\lim\limits_{x\to 1} a = a$

For $a$ rational, let $a=\dfrac pq$. We set $x=t^q$, and
$$\frac{x^{p/q}-1}{x-1}=\frac{t^p-1}{t^q-1}=\frac{\dfrac{t^p-1}{t-1}}{\dfrac{t^q-1}{t-1}}$$
which tends to $\dfrac pq$.

For irrational $a$, the result could be extended using continuity, but this is more technical and depends on your definition of the powers.
