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?
 A: We have that for $x>3$
$$\log{\left|\frac{x+2}{3-x}\right|}=\log{\left(\frac{x+2}{x-3}\right)}=\log{\left(1+\frac{5}{x-3}\right)}$$
and therefore
$$x\log{\left|\frac{x+2}{3-x}\right|}=\frac{5x}{x-3}\frac{\log{\left(1+\frac{5}{x-3}\right)}}{\frac{5}{x-3}}\to 5\cdot 1=5$$
A: $\frac{x+2}{3-x} = \frac{5}{3-x} - 1 < 0$ 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.
A: Let $x>4$;
$I:= \log\dfrac {|x+2|}{|x-3|}=  {\displaystyle \int_{x-3}^{x+2}}(1/t)dt.$
MVT:
$I = (1/r){\displaystyle  \int_{x-3}^{x+2}}dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r \in [x-3,x+2]$.
Hence:
$5\dfrac {x}{x+2} \le xI \le 5\dfrac{x}{x-3}$.
Squeeze :
$\lim_{x \rightarrow \infty} xI =5$.
A: We have for big $x > 0$
$$
\ln \left|\frac{x+2}{3-x}\right|^x = \ln \left(\frac{x+2}{x-3}\right)^x
$$
and also
$$
\lim_{x\to \infty}\left(\frac{x+2}{x-3}\right)^x = e^5
$$
hence
$$
\lim_{x\to\infty}x\ln\left|\frac{x+2}{3-x}\right| = 5
$$
