<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] 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 :
$\lim_{x \rightarrow \infty} xI =5$.
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