Evaluate: $\lim_{x \to 2}\frac{\ln(3-x)+3\ln(x-1)}{\ln(2x-3)+3\ln(5-2x)}$

Question

Evaluate:

$$\lim_{x \to 2}\frac{\ln(3-x)+3\ln(x-1)}{\ln(2x-3)+3\ln(5-2x)}$$

I found this limit in my calculus book. The answer is supposed to be $-\frac1 3$. I used the Taylor expansion about $x=2$ for each term and got $-\frac1 2$ as the final answer.

The question also specifies not to use L'Hôpital's rule.

Answer 1

Your answer is correct, the book is wrong. (Assuming you copied the statement accurately.)

Set $h=2-x$. (I tend to find Taylor expansions around $0$ much simpler, at least for me.) Then $$\begin{align} \frac{\ln(3-x)+3\ln(x-1)}{\ln(2x-3)+3\ln(5-2x)} &= \frac{\ln(1+h)+3\ln(1-h)}{\ln(1-2h)+3\ln(1+2h)} = \frac{h-3h+o(h)}{-2h+6h+o(h)} \\&= \frac{-2h+o(h)}{4h+o(h)} = \frac{-1+o(1)}{2+o(1)} \xrightarrow[h\to0]{} -\frac{1}{2} \end{align}$$

where we used the first-order Taylor expansion (i.e., equivalently, the asymptotic equivalent) $\ln(1+u)=u+o(u)$ around $0$.

Answer 2

This is a possible approach based on the substitution $t=x-2$ and the fact that $\displaystyle \lim_{t\to 0}\frac{\ln(1+at)}{t}=a$ (this limit should be in your notes): $$\lim_{x \to 2}\frac{\ln(3-x)+3\ln(x-1)}{\ln(2x-3)+3\ln(5-2x)}= \lim_{t \to 0}\frac{\ln(1-t)+3\ln(1+t)}{\ln(1+2t)+3\ln(1-2t)}\\=\lim_{t \to 0}\frac{\frac{\ln(1-t)}{t}+3\frac{\ln(1+t)}{t}}{\frac{\ln(1+2t)}{t}+3\frac{\ln(1-2t)}{t}}=\frac{-1+3}{2-3\cdot 2}=-\frac{1}{2}.$$ So your final answer is right!!