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 $\lim\limits_{x\to 1} a = a$
 A: Hint
Make life simpler using $x=1+y$ which makes
$$\lim\limits_{x\to 1} \frac{x^a -1}{x-1}=\lim\limits_{y\to 0} \frac{(1+y)^a -1}{y}$$
Now, use the binomial theorem or Taylor expansion.
A: Do the substitution $y=a\log x$ (natural logarithm). Then $x=e^{y/a}$, so the limit becomes
$$
\lim_{y\to0}\frac{e^y-1}{e^{y/a}-1}=
\lim_{y\to0}a\frac{e^y-1}{y}\frac{y/a}{e^{y/a}-1}
$$
Since
$$
\lim_{y\to0}\frac{e^y-1}{y}=1
$$
we can conclude that
$$
\lim_{x\to1}\frac{x^a-1}{x-1}=a
$$
A: This is literally the definition of the derivative of $x^a$ at $x=1$, so L'Hôpital's rule would have been circular anyway. We'll have to differentiate, viz. $$y=x^a\implies \ln y=a\ln x\implies \frac{y'}{y}=\frac{a}{x}\implies y'=ax^{a-1}\implies y'\left.\right|_{x=1}=a.$$
A: mIf $f(x)=x^a$, $\;\lim\limits_{x\to 1}\dfrac{x^a-1^a}{x-1}\;$ is the definition of $f'(1)$. So the limit  is 
$$a\,x^{a-1}\biggm|_{x=1}=\color{red}a.$$
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.
A: You can use Bernoulli's ineq (which can be proved by am-gm and by continuity argument, so minimal calculus knowledge i think) and the squeeze theorem.  You can assume  wlog that $a\geq 1$, or otherwise, if $0<a<1$, take $X=x^a$, so
$$\lim_{x\to 1}\frac{x^a-1}{x-1}=\lim_{X\to 1}\frac{X-1}{X^{\frac1a}-1}=\frac{1}{\lim_{X\to 1}\frac{X^{\frac1a}-1}{X-1}}.$$
So, if $\lim_{x\to 1}\frac{x^a-1}{x-1}=a$ for $a\geq 1$ holds, we also have $\lim_{x\to 1}\frac{x^a-1}{x-1}=a$ for $0<a<1$ too.  
Suppose that $a\geq 1$.  Let $x=1+y$.  Wlog, we may say that $y>-1$ is a small real number.  For $a\geq 1$, Bernoulli's ineq gives $$x^a-1=(1+y)^a-1 \geq ay=a(x-1).$$  So, $\frac{x^a-1}{x-1}\geq a$ if $y>0$ (or $x>1$).  (The ineq is reversed for $y<0$, or $x<1$.)
Now write $x=\frac{1}{1-z}$ for some small real number $z<\frac1a$ (that is, $z=y/(1+y)$).  By Bernoulli's ineq, $(1-z)^a\geq 1-az$, so $$x^a-1=\frac{1}{(1-z)^a}-1\leq \frac{1}{1-az}-1=\frac{az}{1-az}.$$
That is, if $y>0$ (i.e. $x>1$), we get $z>0$ and
$$\frac{x^a-1}{x-1}=\frac{\frac1{(1-z)^a}-1}{\frac1{1-z}-1}\leq \frac{az}{1-az}\left(\frac{1-z}{z}\right)=\frac{a(1-z)}{1-az}.$$
The ineq is reversed if $y<0$ (i.e., $x<1$ and $z<0$).
Therefore,
$$a\leq \frac{x^a-1}{x-1}\leq \frac{a(1-z)}{1-az}=\frac{a}{x}\left(\frac{1}{1-a\left(1-\frac1x\right)}\right)=\frac{a}{x-a(x-1)}$$
if $x>1$.  
Similarly, for $a\geq1$ and $0<x<1$, we have
$$\frac{a}{x-a(x-1)}\leq \frac{x^a-1}{x-1}\leq a.$$
Taking $x\to 1$ and using the squeeze thm, we have $\lim_{x\to 1}\frac{x^a-1}{x-1}=a$ for $a\geq 1$.
