Pessimist Singularitarian

#4
December 11th, 2016,
00:24
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)$