<s>[INST] Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\left( \frac{x+2}{3-x}-1\right)=x\left(\frac{2x-1}{3-x}\right)=x\left(\frac{2x}{-x}\right)=-2x\rightarrow-\infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
 [/INST] $\frac{x+2}{3-x} = \frac{5}{3-x} - 1 < 0$ when $x\to \infty$.
Hence $\underset{x\to\infty}{\text{lim}}f(x) = \underset{x\to\infty}{\text{lim}} x\log(1 +  \frac{5}{x - 3}) = \underset{x\to\infty}{\text{lim}}x \frac{5}{x - 3} = 5$
where the second to last equality follows from $\log(x + 1) = x + o(x) \text{ for } x\to 0 $ thanks to @gimusi's suggesion.
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