1. $\lim_{{x}\to{0}} \frac{3^x - 2^x}{x^2 - x}$

I'm trying to solve this limit. I'm not sure how to go about it. I can try multiplying the numerator and denominator by either $\frac{1}{x^2}$, $\frac{1}{x}$, $\frac{1}{3^x}$, $\frac{1}{2^x}$, but that doesn't lead anywhere.

2. We are given to evaluate:

$\displaystyle L=\lim_{x\to0}\left(\frac{3^x-2^x}{x^2-x}\right)$

We have the indeterminate form 0/0, so we may apply L'Hôpital's Rule:

$\displaystyle L=\lim_{x\to0}\left(\frac{\ln(3)3^x-\ln(2)2^x}{2x-1}\right)=$?

3. What about series? You've posted in Pre-Calculus, so I'm guessing neither of these methods are available.

4. Oops...I didn't notice this wasn't posted in the Calculus forum. So L'Hôpital is out.

Okay, let's consider:

$\displaystyle \frac{3^x-2^x}{x^2-x}=\frac{2^x}{x-1}\cdot\frac{\left(\dfrac{3}{2}\right)^x-1}{x}$

Next, let's consider the limit:

$\displaystyle L_a=\lim_{x\to0}\left(\frac{a^x-1}{x}\right)$ where $0<a$

Let:

$\displaystyle a^x=y+1\implies x=\frac{\ln(y+1)}{\ln(a)}$

And we observe that as $x\to0$ we have $y\to0$, hence:

$\displaystyle L_a=\lim_{y\to0}\left(\frac{\ln(a)}{\ln\left((y+1)^{\frac{1}{y}}\right)}\right)$

Now, using $\displaystyle \lim_{u\to0}\left((1+u)^{\frac{1}{u}}\right)=e$ we conclude:

$\displaystyle L_a=\ln(a)$

And so:

$\displaystyle L=\lim_{x\to0}\left(\frac{2^x}{x-1}\right)\cdot\lim_{x\to0}\left(\frac{\left(\dfrac{3}{2}\right)^x-1}{x}\right)=(-1)\left(\ln\left(\frac{3}{2}\right)\right)=\ln\left(\frac{2}{3}\right)$