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Let us have an expression whose limit I want to evaluate by using Algebra of Limits, which requires the existence of both the limits on the RHS.

Suppose, as an example, I want to use the product rule

$\lim f(x)g(x) = \lim f(x) \cdot \lim g(x)$.

Now what I would do is assume that $ \lim f(x)$ and $\lim g(x)$ exists. After solving if I get a finite value (a non-indeterminate form) I will safely assume that both the limits must have existed and the value I get as the answer is correct. But if i get the answer as an indeterminate form (say $ 0 \times \infty$, or $\infty-\infty $) (or say find $ \lim f(x)$ does not exists), we cannot comment on the nature/value of limit of original expression using the rule, it may exists or may not. We must find another way to so.

Is the above way a correct (or an okayish) way of using the Algebra of Limits or are there certain pitfalls I need to be aware of, or some useful extensions of the mentioned way? Is is also correct to go the above way to apply theorems of limits that require existence of the limit in the first place?

Also my teacher told me that the mentioned way is also correct if we get a (positive finite number) $\times \infty$ form or $ \infty \times \infty$ form, it makes the limit of the original expression $\infty$ (or does not exists, as you may like to treat it). It is also mentioned in A course of Pure Mathematics by G.H. Hardy. Is it true & why if so?

Satwik
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  • It is not quite correct at the beginning of your reasoning: you are not "allowed to" assume that $\lim f$ and $\lim g$ exist. You must prove it (by computing them). – Anne Bauval Nov 25 '22 at 16:49
  • @AnneBauval isn't using the rule and getting a non indeterminate value indirectly proving their existence? – Satwik Nov 25 '22 at 16:52
  • No it is not. To "get a value" (indeterminate or not), you have to prove the existence of the two limits (but if what you mean is "computing each of them proves their existence", then I agree). – Anne Bauval Nov 25 '22 at 17:01
  • To see that the existence of $\lim f(x)g(x)$ ("getting a definite answer") does not imply the existence of $\lim f(x)$ and $\lim g(x)$, consider the behaviour of $f(x) = e^{\sin(1/x)}$ and $g(x) = e^{-\sin(1/x)}$ near $x = 0$. – LSpice Nov 25 '22 at 17:08
  • @LSpice it means that $lim f(x)$ does not exists individually which means we cannot comment on the limit of f(x)*g(x) using the rule as stated in the question. To quote "But if i get the answer as an indeterminate form (say 0×∞, or ∞−∞) (or say find limf(x) does not exists), we cannot comment on the nature/value of limit of original expression using the rule" if I understand your comment clearly. – Satwik Nov 25 '22 at 17:15
  • Then I am not sure what you mean. If you have already computed $\lim f(x)$ and $\lim g(x)$ and shown that they exist, then there is no question of whether they exist. If you have not, then how do you have any values to substitute in the right-hand side of the equation $\lim f(x)g(x) = (\lim f(x))(\lim g(x))$? – LSpice Nov 25 '22 at 17:16
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    @LSpice Yes! you are correct, but in books I have read, the authors will first use the algebra of limits and compute both limits simultaneously rather than computing them separately, showing that they both exist, then use limit rule they intend to use. Thats where my confusion lied. – Satwik Nov 25 '22 at 17:59
  • Re, ah, I see. Then @JairTaylor's answer describes exactly what is going on. – LSpice Nov 25 '22 at 18:00
  • You should have access to a set of limit formulas like $\lim_{x\to 0}\frac{\sin x} {x} =1$ which can help you to recognise if some part (factor/term) of a complicated expression has a limit or not. Then you can use algebra of limits in a slightly extended form to replace them with their limit. This is a fully justified and rigorous process for evaluating limits step by step without trying to prove beforehand that each part of expression has a limit. – Paramanand Singh Nov 25 '22 at 23:50

1 Answers1

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I believe I understand what OP is getting at. A specific example might help. Consider the problem: Find $$\lim_{x\rightarrow \infty} (1+1/x)^{x+1}.$$

You may write

By the product rule for limits,

\begin{align*}\lim_{x\rightarrow \infty} (1+1/x)^{x+1} &= \lim_{x\rightarrow \infty} (1+1/x)^x \cdot (1+1/x) \\&= \lim_{x\rightarrow \infty} (1+1/x)^x \cdot \lim_{x\rightarrow \infty}(1+1/x) \\ &= e \cdot 1 = e.\end{align*}

But technically, breaking the limit into the product of limits is not justified until the end, when we know that both limits exist individually. If we want to be slightly more rigorous, we could write:

Note that $$\lim_{x\rightarrow \infty} (1+1/x)^x = e$$ and that $$\lim_{x\rightarrow \infty}(1+1/x) = 1.$$ Therefore, by the product rule for limits,

\begin{align*}\lim_{x\rightarrow \infty} (1+1/x)^{x+1} &= \lim_{x\rightarrow \infty} (1+1/x)^x \cdot (1+1/x) \\&= \lim_{x\rightarrow \infty} (1+1/x)^x \cdot \lim_{x\rightarrow \infty}(1+1/x) \\ &= e \cdot 1 = e.\end{align*}

This would arguably be more correct. However, I would generally consider the first solution as shorthand for the second solution, and I'd consider it correct since you do eventually show the limits exist. The second version takes up more space and so would probably not be written by a professional mathematician in a paper, since there it is assumed that the reader and author both understand limit laws. For a homework problem, possibly the second is better, but I would accept the first as well.

Jair Taylor
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  • However, it is probably important to note that the literal sentence "After solving if I get a finite value (a non-indeterminate form) I will safely assume that both the limits must have existed and the value I get as the answer is correct" is false; the stated deduction is invalid, even if the demonstration is otherwise correct. – LSpice Nov 25 '22 at 17:09
  • Thank you for the answer, i made an edit at the end, may address that as well if possible. – Satwik Nov 25 '22 at 17:12
  • @LSpice Yes, that's true. But if you actually break it into a product of limits and the individual limits end up not existing, the end result will be nonsense rather than a solution. – Jair Taylor Nov 25 '22 at 17:14
  • That is true, but it starts to sound circular to me: if you already found the values of the limits to use in that law, then the limits exist, because you found them (by whatever means)! But a follow-up comment from @Satwik seems to confirm that you have correctly understood their point, so my caveat is needed, if at all, only for future readers. – LSpice Nov 25 '22 at 17:18
  • @Satwik The situation would basically be the same for a limit of the form $\infty \cdot \infty$ or $c \cdot \infty$ where $c \neq 0$. If you get an indeterminate form such as $0 \cdot \infty$, this approach won't work at all and you'll need a different method. – Jair Taylor Nov 25 '22 at 17:25
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    But be aware that if you write something like $\infty \cdot \infty = \infty$, this is shorthand for applying a particular theorem about limits: that if $\lim_{x\rightarrow a} f(x) = \infty$ and $\lim_{x\rightarrow a} g(x) = \infty$ then $\lim_{x\rightarrow a} f(x)g(x) = \infty$. That theorem is what justifies the "arithmetic" of infinity. If you want to be more rigorous, you can give more detail as in the second solution I wrote above. – Jair Taylor Nov 25 '22 at 17:29
  • @JairTaylor Ah! So the mentioned "way" is also correct if we use $ \lim f(x)g(x) = \lim f(x) \cdot \lim g(x)$ and find $ \lim f(x) = \infty$ and $\lim g(x) = \infty$,it makes the limit of original expression i.e. (f(x)g(x)) = $\infty$ as it is no different from finite$\times$finite as we don't get an indeterminate form. Am I interpreting you correctly? – Satwik Nov 25 '22 at 17:40
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    In the context of this comment, note that "indeterminate form" doesn't just mean "expression involving $\infty$"; although usually not precisely defined, in this context it basically means "an expression of the form $\alpha * \beta$ where there are at least two possibilities for $\lim f(x) * g(x)$ with $\lim f(x) = \alpha$ and $\lim g(x) = \beta$" (where $$ is any binary operation, not necessarily multiplication). In this sense, $\infty\cdot\infty$ is determinate precisely because* of the existence of the corresponding limit law. – LSpice Nov 25 '22 at 17:46
  • @Satwik Yes. It should be fine, unless your instructor does not accept this shorthand. – Jair Taylor Nov 25 '22 at 17:47