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In the calculus textbooks I've come across, the limit laws are given on the condition that both individual limits exist.

Is it safe to weaken that condition by saying that they are valid as long as they do not lead to indeterminate forms, and then abusing notation a bit for determinate forms involving $\infty$?

For example, I would say that: $$ \lim\limits_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x\rightarrow a} f(x)}{\lim\limits_{x\rightarrow a} g(x)} $$ and then say that for this purpose $\frac{\infty}{x} = \infty$ and $\frac{x}{\infty}=0$ where $x\in\mathbb R$.

This seems to work for me, but I know calculus and not analysis. So I'm wondering if there are any pitfalls.

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  • $\begingroup$ Sorta kinda. Do you accept non-standard analysis? $\endgroup$ – Brevan Ellefsen Dec 15 '15 at 4:42
  • $\begingroup$ I'm not talking about non-standard analysis here (I only have a very vague idea of what it is, so I can't say if I "accept" it). I'm just saying that the notation still works. $\endgroup$ – MGA Dec 15 '15 at 4:46
  • $\begingroup$ Well, technically $\frac{\infty}{x}$ is undefined in the real number system. You can naively use the notation now, but it might come back to bite you later... I wouldn't recommend it. Non-standard analysis does allow you to extend this a bit. The problem is that using this might cause nonsensical answers if it isn't the last step, allowing you to show that $1=0$. Further, this might lead you to believe that similar notations are valid, and doesn't allow for indeterminate forms except by exception (such as when $x=0$). I would say limits are the safer way to go $\endgroup$ – Brevan Ellefsen Dec 15 '15 at 4:49
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The following answer assumes that a function $f(x)$ defined in a certain deleted neighborhood of $x = a$ has the following five options:

  1. $f(x) \to L$ where $L$ is a real number as $x \to a$. This is what I mean when I say that $\lim_{x \to a}f(x)$ exists.
  2. $f(x) \to \infty$ as $x \to a$.
  3. $f(x) \to -\infty$ as $x \to a$.
  4. $f(x)$ oscillates finitely as $x \to a$.
  5. $f(x)$ oscillates infinitely as $x \to a$.

Limit laws related to "algebra of limits" are normally presented in the following fashion:

If $\lim_{x \to a}f(x), \lim_{x \to a}g(x)$ exist then

  • $\lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$
  • $\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$
  • $\lim_{x \to a}\frac{f(x)}{g(x)} = \dfrac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}$ provided that $\lim_{x \to a}g(x) \neq 0$

Out of these the first law can be easily modified so that the existence of only one of the limits $\lim_{x \to a}f(x), \lim_{x \to a}g(x)$ is required. Thus we have the following result:

If $\lim_{x \to a}f(x)$ exists the nature of $\lim_{x \to a}\{f(x) \pm g(x)\}$ is same as the nature of $\lim_{x \to a}g(x)$. This means that

  • if $\lim_{x \to a}g(x)$ exists then $\lim_{x \to a}\{f(x) \pm g(x)\}$ also exists and $\lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$
  • if $\lim_{x \to a}g(x)$ is $\pm \infty$ then $\lim_{x \to a}\{f(x) \pm g(x)\}$ is also $\pm \infty$.
  • if $g(x)$ oscillates finitely (or infinitely) then $f(x) \pm g(x)$ also oscillates finitely (or infinitely).

The laws relating to product and quotient of $f(x), g(x)$ can also be extended in a similar fashion where we require the existence of the limit of only one of the functions $f(x), g(x)$.

However in case of products is it important that the limit of one of the functions (say $f(x)$) must be non-zero and then the behavior of the product is same as the behavior of the other function (namely $g(x)$).

The case of quotients is bit difficult to handle and it is better to think of $f(x)/g(x)$ as $f(x)\times (1/g(x))$ and formulate the rules for $1/g(x)$ and use the laws of products mentioned above. For $1/g(x)$ the idea is simple. If $g(x) \to \pm\infty$ then $1/g(x) \to 0$. If $g(x) \to 0$ and $g(x)$ is positive (or negative) then $1/g(x) \to \infty$ (or $1/g(x) \to -\infty$). If $g(x) \to L \neq 0$ then $1/g(x) \to 1/L$. In other cases $1/g(x)$ oscillates.

Note that these rules suffice to handle all cases except the case of quotients when both the functions tend to $0$ or both of them tend to $\pm\infty$. This is where we need some more work (involving algebraic manipulation, standard limits, L'Hospital's Rule, Taylor series, or Squeeze Theorem).

Thus what you think intuitively is almost correct, however it is better not to abuse notation and start doing arithmetical operations with symbol $\infty$.

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    $\begingroup$ @user21820: i should have been explicit, normally i assume existence means that the limit is a real number and not an extended real number. i will put this in the answer. $\endgroup$ – Paramanand Singh Dec 15 '15 at 9:42
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    $\begingroup$ @LarsSmith: with multiplication or division you must note that the one limit which you want to "take into" must be non-zero. Also this operation has to be done step by step. Take one limit at a time $\endgroup$ – Paramanand Singh Jan 5 at 3:10
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    $\begingroup$ @LarsSmith: well you can't combine those two. The split and combine work the same way. For product you need one of the factors with non-zero limit. In your example you can do a combine if the first factor is $x-2$ instead of $x-1$ and limit of this factor is $-1$ (non-zero). $\endgroup$ – Paramanand Singh Jan 5 at 13:46
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    $\begingroup$ @LarsSmith: the expression $\lim_{x\to 1}(x-1)\times\lim_{x\to 1}\dfrac{1}{x-1}$ is undefined/indeterminate. At the same time $\lim_{x\to 1}(x-1)\cdot\dfrac{1}{x-1}$ is $1$. $\endgroup$ – Paramanand Singh Jan 6 at 9:34
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    $\begingroup$ @LarsSmith: yes. You are right. $\endgroup$ – Paramanand Singh Jan 6 at 12:01
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No this is false.

$\lim_{x \to 0} \dfrac{2+\sin(\frac{1}{x})}{2+\sin(\frac{1}{x})} = 1$ but $\lim_{x \to 0} ( 2+\sin(\frac{1}{x}) )$ does not even exist, neither in the affinely extended real line nor the projective real line, intuitively because it 'oscillates and never goes anywhere'.

Also $\lim_{x \to 0} \dfrac{x^2}{\sin(x)} = 0$ but it is certainly not $\dfrac{ \lim_{x \to 0} x^2 }{ \lim_{x \to 0} \sin(x) }$ because the latter is meaningless in the first place since both numerator and denominator are zero.

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  • $\begingroup$ The example you have chosen does not have a limit according to most definitions of limit given in introductory texts. The definition requires that if we are discussing about $\lim_{x \to a}f(x)$ then we must ensure that $f(x)$ is defined in a certain deleted neighborhood of $a$. However some advanced texts (perhaps your source) only require that $a$ be a limit point of the domain of $f$. No big deal here but better to be explicit. $\endgroup$ – Paramanand Singh Dec 15 '15 at 8:22
  • $\begingroup$ @ParamanandSingh: Okay fine just add two to both numerator and denominator! $\endgroup$ – user21820 Dec 15 '15 at 8:39
  • $\begingroup$ @ParamanandSingh: I also added another example where the separate limits exist but the claimed limit law still fails. $\endgroup$ – user21820 Dec 15 '15 at 8:46
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    $\begingroup$ fully agree with your examples and your comments, just wanted to highlight a small detail. my +1 $\endgroup$ – Paramanand Singh Dec 15 '15 at 9:02
  • $\begingroup$ @ParamanandSingh: Yes I'm grateful for your comment, and it spurred me to create the second example to show that it's not just one way that can fail. $\endgroup$ – user21820 Dec 15 '15 at 9:06

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