4
$\begingroup$

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

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

  • 1
    To the edit, yes. – 2017-02-19
  • 1
    Use the fact that $\log \sqrt{x-2} = \frac{1}{2} \log (x-2)$. Check the computation in part (2). – 2017-02-19

4 Answers 4

4

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).

  • 0
    Does that mean that for this rule to work the derivatives of x and y must be the same? – 2017-02-19
  • 0
    Yes. It's like saying that since $\lim_{x\to0^+}\frac xx=1$, then all things of the indeterminate form $\frac00$ are equal to one. – 2017-02-19
  • 0
    Does that mean that $\frac{\infty}{\infty} = 1$ ? Or is that only also when you can prove that the derivative is the same on both the numerator and the denominator? – 2017-02-19
  • 0
    @MarkRead Each indeterminate form must be worked out on its own, since it depends how it approaches the indeterminate form. $\frac\infty\infty$ is not always equal to one. – 2017-02-19
  • 0
    My point was $$\lim_{x\to a}\frac{\ln(\dots)}{\dots}=\frac{\ln(1)}0\ne\text{ always equal to one}$$ – 2017-02-19
  • 0
    Right and in this case you could have also done $\frac{\ln(x-2)}{x-3} \cdot \frac{1}{2x+6}$ on the 3rd step, as $x-2 \rightarrow 1$, $x-3 \rightarrow 0$ and their derivatives are the same. – 2017-02-19
  • 0
    @MarkRead Yup. I just like to do the substitution to make the limit 'click' better. – 2017-02-19
1

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.

1

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).

  • 0
    But if in order for this rule to work $y$ had to be $1+x$, then their derivatives are still the same. In other words, the derivative of $x$ and $y$ have to be equal for this rule to work. Is this correct? – 2017-02-19
  • 0
    @MarkRead: Why do you need to think of derivatives? The limit formula says that the argument of $\log$ must be one more than the denominator and that the denominator must tend to $0$ and only in this case the ratio tends to $1$. There is no talk of derivatives here and neither do we need to think like that. Limits have nothing to do with derivatives, but derivatives are defined in terms of limits. – 2017-02-19
0

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