5
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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?

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  • $\begingroup$ You actually can't remove the absolute value since $3-x \leqslant 0$ when $x \rightarrow \infty$ ! $\endgroup$
    – tmaths
    Aug 20 '18 at 9:42
  • $\begingroup$ Your error comes from the fact that $\log u\sim u-1$ is valid for $u\to\infty$, which is not the case of $\dfrac{x+2}{3-x}$. $\endgroup$
    – Bernard
    Aug 20 '18 at 9:44
  • $\begingroup$ @Bernard my textbook says that if $u\rightarrow1$, then $\log{u}\sim u -1$ for $x\rightarrow x_{0}$ $\endgroup$
    – Cesare
    Aug 20 '18 at 9:48
  • $\begingroup$ @Cesare;: That's right. Sorry for the *lapsus calami *! However, your fraction tends to $-1$, not $1$. You should have taken its absolute value. $\endgroup$
    – Bernard
    Aug 20 '18 at 9:57
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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

$$x\log{\left|\frac{x+2}{3-x}\right|}=\frac{5x}{x-3}\frac{\log{\left(1+\frac{5}{x-3}\right)}}{\frac{5}{x-3}}\to 5\cdot 1=5$$

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5
  • $\begingroup$ Thanks, this is indeed correct. I don't grasp how we "flip" $3-x$ so that it comes $x-3$. Is it because $|x| = -x$ when $x <= 0$? $\endgroup$
    – Cesare
    Aug 20 '18 at 9:50
  • $\begingroup$ @Cesare Yes that's the key point to solve it! I've used standard limit to conclude but of course you can also use other methods. You are welcome. Bye $\endgroup$
    – user
    Aug 20 '18 at 9:54
  • $\begingroup$ Gimusi. Very nice:)) $\endgroup$ Aug 20 '18 at 11:05
  • $\begingroup$ @PeterSzilas Thanks, not so difficult ;) $\endgroup$
    – user
    Aug 20 '18 at 11:44
  • $\begingroup$ Gimusi.Well, well:)) $\endgroup$ Aug 20 '18 at 12:04
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$\frac{x+2}{3-x} = \frac{5}{3-x} - 1 < 0$ 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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  • $\begingroup$ The step $\underset{x\to\infty}{\text{lim}} x\log(1 + \frac{5}{x - 3}) = \underset{x\to\infty}{\text{lim}}x \frac{5}{x - 3}$ is true but we need to justify that, for example by $\log(1 + \frac{5}{x - 3})=\frac{5}{x - 3}+o(1/x)$. $\endgroup$
    – user
    Aug 20 '18 at 9:58
  • $\begingroup$ Yes, I forgot to write it explicitly because I assumed it. Thanks! $\endgroup$
    – kevin
    Aug 20 '18 at 10:00
  • $\begingroup$ maybe you should revise it $\endgroup$
    – user
    Aug 20 '18 at 10:32
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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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1
  • $\begingroup$ Very nice method Peter! $\endgroup$
    – user
    Aug 20 '18 at 13:36
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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

$$ \lim_{x\to\infty}x\ln\left|\frac{x+2}{3-x}\right| = 5 $$

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