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I am not sure about what happens with this property when the limits are not real numbers, i.e., they exist, but are $\infty$. Say $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = -\infty$. In this case, can we apply the property? Is it true that $\lim_{x \to a} f(x) + \lim_{x \to a} g(x) = \lim_{x \to a} \left( f(x)+g(x) \right)$? If it is not, is there any way to deal with such a sum of limits? I'll leave an example from where I asked myself about this. $$\lim_{x \to 0} \left( \frac{\pi}{2} - x \right) \ln^2 {x} - \frac{\pi}{2} \lim_{x \to 0} \frac{1}{x^2}$$ I think the result would be $- \infty$ if I could apply this property and bring them into a single limit, but I wasn't sure about that as they both are $\infty$ so that's why I'm asking.

user267998
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  • What does $L \to+ \infty$ mean ?? Do you mean that $\lim_{x \rightarrow a} f(x) = +\infty$ ? – TheSilverDoe Sep 04 '20 at 10:58
  • @TheSilverDoe The limit approaches $\infty$, something like a vertical assymptote – user267998 Sep 04 '20 at 10:59
  • Limits are fixed things. They approach nothing. – José Carlos Santos Sep 04 '20 at 11:00
  • Ok, so you mean that the limit is $+\infty$. – TheSilverDoe Sep 04 '20 at 11:00
  • Sorry for bad use of words and concepts. I updated the question, I hope it's better – user267998 Sep 04 '20 at 11:03
  • $\lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) = \lim_{x \to a} \left( f(x)\pm g(x) \right)$ is a property of limits. There is no exception. Check https://www.math24.net/properties-limits/#:~:text=This%20rule%20states%20that%20the,%E2%86%92ag(x). – DatBoi Sep 04 '20 at 11:05
  • @DatBoi But I don't know if the property applies when both limits are $\infty$, not real numbers – user267998 Sep 04 '20 at 11:05
  • Well, the sum of a quantity that tends to $+\infty$ and a quantity that tends to $-\infty$, can have a finite limit, or an infinite limit, or no limit at all. It is what we call an indeterminate form. Therefore you are not allowed to apply a rule such "the sum of the limit is the limit of the sum". – TheSilverDoe Sep 04 '20 at 11:05
  • @DatBoi To write such an equality, you have to be sure that all the limits exist ! But the fact that the RHS has a limit is not always true, given that the limits of the LHS exist. – TheSilverDoe Sep 04 '20 at 11:14
  • @TheSilverDoe Of course!! Both limits must be defined! – DatBoi Sep 04 '20 at 11:16
  • @DatBoi If you apply the rule with infinite limits, you can conclude things like $\infty - \infty = 2$ or $\infty - \infty = 3$. It is really sloppy. – TheSilverDoe Sep 04 '20 at 11:18

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If$$\lim_{x\to a}f(x)=\infty\quad\text{and}\quad\lim_{x\to a}g(x)=-\infty,$$then $\displaystyle\lim_{x\to a}\bigl(f(x)+g(x)\bigr)$ can be anything. For instance, if $b\in\Bbb R$, if $f(x)=\frac1{(x-a)^2}+b$ and if $g(x)=-\frac1{(x-a)^2}$, then$$\lim_{x\to a}f(x)=\infty,\ \lim_{x\to a}g(x)=-\infty\text{ and }\lim_{x\to a}\bigl(f(x)+g(x)\bigr)=b.$$

José Carlos Santos
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  • But you did apply the limit sum property in order to conclude that, I mean you did $\lim_{x \to a} \left( \frac{1}{(x-a)^2} + b \right) + \lim_{x \to a} - \frac{1}{(x-a)^2} = \lim_{x \to a} \left( \frac{1}{(x-a)^2} + b - \frac{1}{(x-a)^2} \right) = \lim_{x \to a} b = b$ – user267998 Sep 04 '20 at 11:15
  • No, I did not! Where, in my answer, do you see the first equality of your comment? – José Carlos Santos Sep 04 '20 at 11:18
  • Sorry I wrongly assumed you were answering my question. I wondered if I could do something with the first part that I wrote, the sum of the two limits – user267998 Sep 04 '20 at 11:19
  • But I *did answer your question. I told you that, from the knowledge that $\lim_{x\to a}f(x)=\infty$ and that $\lim_{x\to a}g(x)=-\infty$, you get no information whatsoever about $\lim_{x\to a}\bigl(f(x)+g(x)\bigr)$. – José Carlos Santos Sep 04 '20 at 11:22
  • The thing I didn't understand is, do we just let $\lim_{x \to a} \left( \frac{1}{(x-a)^2} + b \right) + \lim_{x \to a} - \frac{1}{(x-a)^2}$ undefined then? I'm sorry if I'm not understanding correctly – user267998 Sep 04 '20 at 11:27
  • Yes, $\infty-\infty$ is undefined. – José Carlos Santos Sep 04 '20 at 11:30
  • It's hard for me to grasp it. I got to the example I wrote in the question from an integral, and it turns out that this sum of limits should be zero in order for my integral to make sense. What can I do about it? Do I just look for a different technique which does not involve ending up with such undefined expressions? – user267998 Sep 04 '20 at 11:34
  • The expression$$\lim_{x \to 0} \left( \frac{\pi}{2} - x \right) \ln^2 {x} - \frac{\pi}{2} \lim_{x \to 0} \frac{1}{x^2}$$doesn't make sense. Where did you get it from? – José Carlos Santos Sep 04 '20 at 11:38
  • I got it after doing some substitutions. I originally wrote another question about the integral but it seems I didn't explain my problem well because I only got answers on how to solve the integral, which wasn't what I wanted. You can check the process here https://math.stackexchange.com/q/3812962/804320 – user267998 Sep 04 '20 at 11:43
  • It's the same problem all over again. The expression$$\left.x \log^2\left({\cos{x}} \right) \right|_0^{\pi/2} +2 \int_0^{\pi/2} x \tan x \log\left(\cos x\right),\mathrm dx$$doesn't make sense, since it is $\infty-\infty$. – José Carlos Santos Sep 04 '20 at 11:56
  • So a simple integration by parts fails because I ended up writing a number (convergent integral) as $\infty - \infty$ which doesn't make sense. Then, am I only allowed to do this if the result of the integration by parts is a sum of two real numbers? – user267998 Sep 04 '20 at 12:02
  • Or if you get $\infty+\infty$. Or $-\infty-\infty$. Or if one of them is $\pm\infty$ and the other one is a real number. – José Carlos Santos Sep 04 '20 at 12:03
  • I get it now. Thank you very much for your time. Now it makes sense. Have a nice day! – user267998 Sep 04 '20 at 12:04
  • If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos Sep 04 '20 at 12:06
  • Sorry I forgot about that option, I'm new. It was useful, I marked it now – user267998 Sep 04 '20 at 12:07