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I've been casually rereading some Calculus when I've come across an argument I feel is circular.

Consider a polynomial

$$f(x) = \sum_{k=0}^{n}a_{k} x^{k} = x^{n}\left(\sum_{k=0}^{n}a_{k}x^{k-n}\right).$$

Now, as $\lvert x\rvert$ grows large, the second term tends to $a_{n}$ so that $f(x)\sim a_{n}x^{n}$. Therefore,

$$\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow\infty} a_{n}x^{n}.$$

I can't help but think that even in a first course in Calculus, i.e., without the rigor of Real Analysis, this is a pedagogically unsatisfying result since the usual limit of the product is the product of the limits law cannot be applied here, but that is seemingly brushed aside.

In a similar example, one sees

$$\lim_{x\rightarrow \infty} \frac{x^{2}}{\sqrt{x^{3}+1}} = \lim_{x\rightarrow \infty} \frac{x^{2}}{x^{3/2} \sqrt{ 1 + x^{-3}}} = \lim_{x\rightarrow \infty} \frac{x^{2}}{x^{3/2}}$$

and as before, I find this dissatisfying because the quotient law would not apply to be able to justify that one can simply set $\sqrt{1+x^{-3}} = 1$ under the limit.

I've found this old stackexchange post asking a similar question with a surprising number of downvotes.

The only answer has a large number of downvotes, but no comments. Without dwelling on the overall quality of the answer, the answerer claims that you can apply the limit laws provided an indeterminant form does not arise sounds reasonable to me and would have helped resolve infinite limits like the above.

My question is essentially, how would you resolve this if a first year calculus student raised the alarm like I have here? And is the author of the other posts claim correct?

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  • $\begingroup$ Since $\lim_{x\to\infty}x^n=\infty$ and $\lim_{x\to\infty}\sum_{k=0}^na_kx^{k-n}=a_n\ne0$, then$$\lim_{x\to\infty}f(x)=\infty\times a_n=\pm\infty.$$What's wrong here? Why do you claim that “the usual limit of the product is the product of the limits law cannot be applied here”? $\endgroup$ Sep 17 at 10:38
  • $\begingroup$ @JoséCarlosSantos I'm sorry, that is my mistake. The limit law I am referring to is that $\lim_{x\rightarrow c} f(x)g(x) = \lim_{x\rightarrow c}f(x) \lim_{x\rightarrow c}g(x)$ provided that $\lim_{x\rightarrow c} f(x)$ and $\lim_{x\rightarrow c} g(x)$ exists with the convention that we say that a limit that diverges to $\pm\infty$ does not exist. $\endgroup$
    – JessicaK
    Sep 17 at 10:41
  • $\begingroup$ But the law is valid also when one of the limits (or both) is equal to $\pm\infty$, as long as the other limit is not $0$. $\endgroup$ Sep 17 at 10:45
  • $\begingroup$ @JoséCarlosSantos Of course, I do not disagree with that. For this reason, I chose to bring up the downvoted answer in the other post that claimed that the laws are valid at infinity provided an indeterminant form does not arise. Is that claim true for the usual sum, difference, product, and quotient laws with the other laws defined in an analogous way to how the product one was defined in my previous comment? $\endgroup$
    – JessicaK
    Sep 17 at 10:48
  • $\begingroup$ Yes, it is valid in that context. $\endgroup$ Sep 17 at 10:54
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since the usual limit of the product is the product of the limits law cannot be applied here

Sure, but there is another rule, saying that if $\lim_{x\to\infty} f(x)=\infty$ and there exists some $c>0$ and some $M>0$ such that, for all $x>M$, you have $g(x)>c$, then $$\lim_{x\to\infty} f(x)\cdot g(x) = \infty.$$

Note that if the limit $\lim_{x\to\infty} g(x)$ exists and is positive, then the condition about $c$ and $M$ above is also satisfied.

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You are not correct when you claim that “the usual limit of the product is the product of the limits law cannot be applied here”. They do apply. For instance, the assertion “if both limits $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}g(x)$ exist, then $\lim_{x\to\infty}f(x)g(x)=\left(\lim_{x\to\infty}f(x)\right)\left(\lim_{x\to\infty}g(x)\right)$” holds when one of the limits (or both) is equal to $\pm\infty$, as long as the other limit is not $0$.

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