Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$
To do this I tried 2 approaches:
1: If $\lim_{x \rightarrow 0} \frac{\ln(x+1)}{x} = 1$, $\lim_{x\to1}\frac{\ln(x)}{x-1}=1$ and $\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\frac{\ln(u)}{u-1}$, then I infer that $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, then I have:
$$\sqrt{3-2} \rightarrow 1
\\ 3^2-9 \rightarrow 0
$$
and so 
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = 1$$
But this is wrong. So I tried another method:
2: 
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = \frac{\ln(\sqrt{x-2})}{(x-3)(x+3)} = \lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{(x-3)} \cdot \lim_{x \rightarrow 3} \frac{1}{x+3} = \frac{1}{6}$$
Which is also wrong.
My questions are: What did I do wrong in each method and how do I solve this?
EDIT:  If $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, and $\ln(x) \rightarrow 0^+$, does this mean that $\frac{0^+}{0^+} \rightarrow 1$?

Hint: Let $x=u+3$ and use log rules.
$$\frac{\ln\sqrt{x-2}}{x^2-9}=\frac{\frac12\ln(x-2)}{(x+3)(x-3)}=\frac1{2(u+6)}\frac{\ln(u+1)}u\to\frac1{12}$$
To the first and second try, you assumed that just because the logarithm goes to 1 and the denominator goes to 0 that the limit is 1, but that is only true if the stuff inside the logarithm and denominator both approach the needed values at the same speed of convergence, which is clearly not the case (too much stuff interfering).
