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What is $$\lim_{x\to -1^-}\frac{x}{-x^2+2+x}?$$

It should be equal to:

$\frac{-1}{-(-1)^2+2-1} = \frac{-1}{-1+2-1} = \frac{-1^+}{0^-} = -\infty$

Or do you ignore the signs in the numerator and then it's $+\infty$?

user21820
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SDWayne
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2 Answers2

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You want to approach $-1$ from the negative side. So consider $-1-d$ where $d$ approaches zero from the positive side.

As $x \to -1^-$:

  Let $d = -1 - x$, and hence $x = -1 - d$.

  Then $d \to 0^+$.

  $\frac{x}{-x^2+2+x} = \frac{-1-d}{-(-1-d)^2+2+(-1-d)} = \frac{-1-d}{-3d-d^2} \to\ ?$. [You should be able to figure out now.]

user21820
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  • What is $d$? I don't get it! Why can't I just replace $x$ with $-1$ like they tought us in class? – SDWayne Aug 08 '15 at 15:53
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    @SDWayne: If you were taught to "just replace", then your teacher is simply totally wrong. A limit is defined as the value that is approached, not at the point itself, where the expression may be undefined. The limit of $\frac{x}{x}$ as $x \to 0$ is $1$ but the expression is simply meaningless for $x = 0$. – user21820 Aug 08 '15 at 16:05
  • @SDWayne: Read "$x \to -1^-$" as "$x$ eventually stays close to $-1$ on the negative side". $d = -1 - x$ is then the difference of $-1$ and $x$. We use $d$ because we do not want to deal with things going to $-1$, as it is easier to see the behaviour if we have things going to $0$. If $x$ goes to $-1$, then the difference of $-1$ and $x$ goes to zero. – user21820 Aug 08 '15 at 16:08
  • I know... I didn't express myself correctly. Thank you for your asnwer – SDWayne Aug 08 '15 at 16:09
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It's $+\infty$ because you do not "ignore the signs". You have a negative number dividing a negative number, tending to $0$, therefore the fraction will be positive and will diverge in absolute value.

Note that $$\lim_{x\to+\infty}\frac{1}{x}=\lim_{x\to-\infty}\frac{1}{x}$$ because what it counts is that the expression approaches the same value $0$ regardless of the sign of $x$, not, indeed, the values attained in the process.

Vincenzo Oliva
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  • Oh because $0^-$ is negative. That's all I needed. Thank you – SDWayne Aug 08 '15 at 15:57
  • @SDWayne You're very welcome. – Vincenzo Oliva Aug 08 '15 at 16:08
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    @SDWayne: You are wrong. It is not at all what you need. Your method was completely wrong because you cannot identify correctly how fast the quantities approach the values you claim in the denominator of your result. So you do not know whether it is $0^-$ or $0^+$. – user21820 Aug 08 '15 at 16:11
  • @user21820 Yes we do, using your small $d$. I did not address that matter because it seemed the OP had already got it. – Vincenzo Oliva Aug 08 '15 at 16:30
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    @VincenzoOliva: You do, but the asker probably didn't. And you should not edit a question to change its content (so you should only fix minor things like spelling or grammar or mathjax). – user21820 Aug 08 '15 at 16:37
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    @VincenzoOliva: Besides, what happens when you get something like $1^+ + 1^-$? The point is that that notation is intrinsically wrong from the beginning and should not be taught. – user21820 Aug 08 '15 at 16:38
  • @user21820 Limits are covered by real analysis, and I added something for clarity and coherence (like $-1^-$). Anyway it's true I shouldn't have because he did mean not to write like I did, as the new question of his shows the same mistake. – Vincenzo Oliva Aug 08 '15 at 16:42
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    @user21820 In that case of course one can do nothing about it. I obviously agree that in general it's impracticable, and it should be made clear when taught, if taught. However in this particular limit there are no similar issues, as clarified by my previous edit, so I had made it to expand what I believed the OP's thoughts were. But I was wrong. – Vincenzo Oliva Aug 08 '15 at 16:44