Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule

Question

$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$

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 $\frac{0^+}{0^+} \rightarrow 1$?

Answer 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).

Answer 2

Just in case you want to see another approach.

Consider $$A=\frac{\ln(\sqrt{x-2})}{x^2-9}=\frac 12\frac{\ln({x-2})}{x^2-9}$$ Now, as Simply Beautiful Art answered, let $x=u+3$ which makes $$A=\frac{\log (1+u)}{2 u (u+6)}$$ Now, use Taylor series around $u=0$ $$\log(1+u)=u-\frac{u^2}{2}+O\left(u^3\right)$$ which makes $$A=\frac{u-\frac{u^2}{2}+O\left(u^3\right)}{2 u (u+6)}=\frac{1-\frac{u}{2}+O\left(u^2\right)}{2 (u+6)}$$ Now, long division $$A=\frac{1}{12}-\frac{u}{18}+O\left(u^2\right)$$ which shows the limit and how it is approached.

Answer 3

$$\lim_{x\to3}\frac{\ln \sqrt {x-2}}{x^2-9} = \lim_{x\to3}\frac{\ln (1+\sqrt{x-2}-1)}{\sqrt {x-2}-1}\cdot\frac{\sqrt {x-2}-1}{x^2-9} = \lim_{x\to3}\frac{x-3}{1+\sqrt {x-2}}\cdot\frac{1}{x^2-9} = \frac{1}{12}$$

Answer 4

I think you started off with the right approach and moved into a blind alley. Note that $$\lim_{x\to 0}\dfrac{\log(1+x)}{x}=1\tag{1}$$ cannot lead to the fact that $(\log y) /x\to 1$ when $x\to 0,y\to 1$ independently. This holds only when $x, y$ are related by $y=1+x$ and this is the only meaning of formula $(1)$. You are trying to infer more than what is possible to infer from equation $(1)$. Trying to use intuition when dealing with limits is a sure way to go astray (why? because limits are not an intuitive concept). The right way to deal with them is to learn rules of limits and use them properly. Thus we can see that $$\lim_{x\to 3}\frac{\log\sqrt{x-2}} {x^{2}-9}=\frac{1}{2}\lim_{x\to 3}\frac{\log(x-2)}{x-3}\cdot\frac{1}{x+3}=\frac{1}{12}\lim_{u\to 1}\frac{\log u}{u-1}=\frac{1}{12}$$ where $u=x-2$. This is almost the same as your second approach but without the mistake you made (I hope you are able to see your mistake in the second approach).