How do I find the following limits with L'Hospital? I need to find the limits at $0$ and infinity of the following function: 

$$f(x):=\ln(3^x-1)-\ln(2^x-1)$$

where $x$ is bigger than $0$. I have to use L'Hospital Rule but I am just getting a harder limit when I apply it. Can you say what to do?
 A: $$\lim _{ x\rightarrow 0 }{ \ln { \left( \frac { { 3 }^{ x }-1 }{ { 2 }^{ x }-1 }  \right)  }  } =\lim _{ x\rightarrow 0 }{ \ln { \left( \frac { { 3 }^{ x }\ln { 3 }  }{ { 2 }^{ x }\ln { 2 }  }  \right)  }  } =\ln { \left( \log _{ 2 }{ 3 }  \right)  } $$
A: Exponential functions grow fast.
When $x$ is large $a^x$ grows faster than any polynomial function.  At least it does if $a>1$
As $x$ gets to be big, $2^x - 1$ increasingly starts to resemble $2^x$ and the constant term becomes a triviality.
$\ln(3^x -1) - \ln (2^x - 1)$ starts to look like $\ln 3^x - \ln 2^x = x(\ln 3 - \ln 2)$
And as $x$ goes to infinity $x(\ln 3 - \ln 2)$ goes to infinity, too.
Now this may feel a bit hand-wavy.  If we wanted to be more rigorous we would show that as $x$ gets to be large $|(\ln (3^x - 1) - \ln (2^x-1)) - x (\ln 3 - \ln 2)|$ is small.
Or, for any $\epsilon > 0,$ there exists and $N > 0$ such that $x> N$ implies
$|(\ln (3^x - 1) - \ln (2^x-1)) - x (\ln 3 - \ln 2)|<\epsilon$
But, I don't think that that is really necessary.
A: Write,
$$e^{f(x)}=\frac{3^x-1}{2^x-1}$$
Now use L'hospitals  rule on the right hand side. To get,
$$e^{f(x)} \to \frac{\ln 3}{\ln 2}$$
So,
$$e^{\lim_{x \to 0} f(x)}=\frac{\ln 3}{\ln 2}$$
$$\lim_{x \to 0} f(x) = \ln (\frac{\ln 3}{\ln 2})$$
