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I have formulated an old question but in that case I have committed many errors in limit calculus and I understand my doubt was not absolutely clear so I have decided to remove it (it had not answer) and now I try to formulate it better.

If I have a limit as $x$ goes to $\infty$ or $x\to 0$ I can multiply and divide by $x$ in order for instance to regain a notable limit. For instance $$\lim_{x\to 0}\frac{\sin^2{x}}{x}=\lim_{x\to 0}x\frac{\sin^2{x}}{x^2}=0$$ Now: if I have for istance a function $f(x)$ s.t $\lim_{x\to \infty}\frac{f(x)}{x}=1$ and I have to compute $\lim_{x\to \infty}f(x)-x$, if I apply the previous argument ONLY on the term $f(x)$ I will obtain: $\lim_{x\to \infty}x-x=0$.

$\textbf{My doubt is: }$is it allowed what i have done in the last?

Sisi
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4 Answers4

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Others can cover counterexamples, but here I'm focusing on the computational mechanism.

Here's what you need: assuming $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exist (and in the case of division, assuming $\lim_{x \to a}g(x) \neq 0$) and are finite: $$\lim_{x \to a}\left[f(x) \mathrel{\substack{+ \\- \\\cdot \\/}} g(x) \right] = \lim_{x \to a}f(x) \mathrel{\substack{+ \\- \\\cdot \\/}} \lim_{x \to a}g(x)\text{.}$$

In the first case, you can show that $$\lim_{x \to 0}\dfrac{\sin^2(x)}{x^2} = 1$$ thus this limit exists, and we also know that $$\lim_{x \to 0}x = 0$$ thus this limit exists, hence $$\lim_{x \to 0}x \cdot \dfrac{\sin^2(x)}{x^2} = 0 \cdot 1 = 0\text{.}$$ What needs to be emphasized here is that the limits exist and are finite for the computation to work out.

Now, suppose you have $\lim_{x \to \infty}\dfrac{f(x)}{x} = 1$ and you want to compute $\lim_{x \to \infty}[f(x) - x]$.

Let's consider your proposed idea: $$\lim_{x \to \infty}\left[x \cdot \dfrac{f(x)}{x} - x\right]\text{.}$$ What you had implicitly done was turned the $\dfrac{f(x)}{x}$ into $1$ by way of the limit, but the problem with that is that you're basically trying to boil it down to $$\lim_{x \to \infty}x \cdot \dfrac{f(x)}{x} = \lim_{x \to \infty}x \cdot \lim_{x \to \infty}\dfrac{f(x)}{x} = \lim_{x \to \infty}x$$ which is NOT TRUE because $\lim_{x \to \infty}x$ is NOT finite.

Clarinetist
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  • So when I have the limit as $x\to \infty$ i can't divide and multiply by $x$? – Sisi Apr 28 '21 at 20:30
  • @Sisi You're missing the point slightly. The point here is that you can do operations and distribute the limits only if the corresponding limits are finite. Multiplication and division by $x$ and then splitting the limits here does not work because the limits are not all finite. More specifically, this step: $$\lim_{x \to \infty}x \cdot \dfrac{f(x)}{x} = \lim_{x \to \infty}x \cdot \lim_{x \to \infty}\dfrac{f(x)}{x}$$ is wrong because only one of those two limits is finite. – Clarinetist Apr 28 '21 at 20:33
  • @Sisi To be more direct, you can do whatever algebra you want to attempt to make it tractable to compute a limit. But the underlying problem is this: you cannot split a limit into multiple limits unless you know that each specific limit exists and is finite. – Clarinetist Apr 28 '21 at 20:40
  • Do you mean tha: 1)t if I have $h$ and $g$ divergent I can't say that $\lim_{x\to \infty}(h-g)=\lim_{x\to \infty}h-\lim_{x\to \infty}g$ and moreover if it is not allowed to say that $\lim_{x\to \infty}f(x)=\lim_{x\to \infty}x\frac{f(x)}{x}$ and this last since $f$ and $x$ are divergent and really i need to do this that $f$ is convergent and also $x$? 2) So I can conclude that when I have $x\to\infty$ it is not allowed to divide ans multiply by $x$, right? – Sisi Apr 28 '21 at 20:48
  • @Sisi Correct on $(h - g)$, for the second part: it IS true that $\lim_{x \to \infty}f(x) = \lim_{x \to \infty}x \cdot \dfrac{f(x)}{x}$ but you cannot split the limit into a product of two separate limits since you do not have that the two corresponding limits exist. Regarding 2, there may be some cases where you might have to do that to solve limits at $\infty$, but I can't think of any examples at the moment. – Clarinetist Apr 28 '21 at 20:52
  • why "you do not have that the two corresponding limits exist"? by "exist" do you mean that the limit must be finite? or that the limit does not exists (so indeterminate) as in case of $\lim_{x\to \infty (-x1)^x}$...sorry but I see your exaplanation is very clear and think you can really clarify my doubts (that I have fo my fault) – Sisi Apr 28 '21 at 21:00
  • @Sisi You have in that example that $\lim_{x \to \infty}\dfrac{f(x)}{x}$ exists and is finite, and $\lim_{x \to \infty}x$ is not finite. – Clarinetist Apr 28 '21 at 21:01
  • @Sisi I really should have clarified that to mean "exist and are finite". – Clarinetist Apr 28 '21 at 21:02
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I am not completely sure what you're asking, but if you're wondering whether $$\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1$$ implies that $$\lim_{x\to\infty} \left[f(x)-g(x)\right]=0,$$ then the answer is no. Consider as a counterexample $f(x) = x$ and $g(x) = x+1$.

What is true is that if $\lim_{x\to\infty} f(x)$ and $\lim_{x\to\infty} g(x)$ both individually exist, then $$\lim_{x\to\infty} f(x)g(x) = \left(\lim_{x\to\infty} f(x)\right)\left(\lim_{x\to\infty} g(x)\right),$$ and likewise for addition, subtraction, division (by non-zero), etc.

You cannot use this rule in the first example above because the individual limits do not exist ($f(x)$ and $g(x)$ diverge to infinity).

user7530
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$x/x = 1$ so it is not a problem. Technically $x/x$ is undefined for $x=0$ but the limit does not depend on what happens at $0$, only what happens near $0$.

As for your second example: It sounds like you are trying to argue something like this:

\begin{align*} \lim_{x\rightarrow \infty} f(x) - x &= \lim_{x\rightarrow \infty} x \frac{f(x)}{x} - x \\&= \lim_{x\rightarrow \infty} x \frac{f(x)}{x} - \lim_{x\rightarrow \infty} x \\&= (\lim_{x\rightarrow \infty} x) (\lim_{x\rightarrow \infty} \frac{f(x)}{x}) - \lim_{x\rightarrow \infty} x\\ &= \lim_{x\rightarrow \infty} x - \lim_{x\rightarrow \infty} x\\ &= \lim_{x\rightarrow \infty} x - x \\ &= 0. \end{align*}

However the step $$\lim_{x\rightarrow \infty} x \frac{f(x)}{x} - x = \lim_{x\rightarrow \infty} x \frac{f(x)}{x} - \lim_{x\rightarrow \infty} x $$ is not justified. In general the rule $$\lim_{x\rightarrow a} f(x) + g(x) = \lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x)$$ is only a valid inference if both $\lim_{x\rightarrow a} f(x)$ and $\lim_{x\rightarrow a} g(x)$ exist (and are finite).

Jair Taylor
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  • Yes it is exactly what I wanted to do...vut I can't understand where I am wrong...can you tell me where i am wrong? – Sisi Apr 28 '21 at 20:27
  • @Sisi I edited my answer to be more explicit. – Jair Taylor Apr 28 '21 at 23:35
  • Ok thanks and moreover also the step $\lim_{x\to \infty}x\frac{f(x)}{x}=\lim_{x\to \infty}x \lim_{x\to \infty}\frac{f(x)}{x}$ is allowed only if the two limits exist finite, right? – Sisi Apr 29 '21 at 06:32
  • To answer to the question on the title: when I have the limit for $x$ goes to $+\infty$ I never multiply and divide by $x$ for the fact that then I can't split the limit of $x$ from the part divided by $x$, am I understanding well? – Sisi Apr 29 '21 at 06:35
  • Ok thanks! But why BOTH of the limits must be finite? If $f$ is convergent and $g$ diverges I can't write $\lim_{x\to \infty}f-\lim_{x\to \infty}g$? – Sisi Apr 29 '21 at 10:38
  • @sisi It's true you can generalize the rule if you want to include $\infty$ in your number system. You could say things like $\infty + \infty = \infty$ and, in the case you mention, $c - \infty = -\infty$. These should be considered shorthand only. However, the problem in the step I was talking about in my answer you would get $\infty - \infty$ and you cannot say $\infty - \infty = 0$. $\infty - \infty$ is considered indeterminate form. See https://en.wikipedia.org/wiki/Indeterminate_form – Jair Taylor Apr 29 '21 at 15:27
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Note that :

$\lim (f(x) g(x))= ( \lim f(x) ) (\lim g(x))$ is applicable when both the limits that is $\lim f(x)$ and $\lim g(x)$ exist finitely. $\tag 1$

Coming to your question:
Suppose that $\frac{f(x)}{x}\to 1$ as $x\to \infty $.
$\lim f(x)=\lim (x\frac{f(x)}{x})$. Now, you are not allowed to use $(1)$ because $\lim x$ is not finite as $x\to \infty$.

Another example when $x\to 0$, $\lim(\sin x)=\lim(x\frac{\sin x}{x})=0\times 1=0 $ because both the limits that is $\lim x$ and $\lim \frac{\sin x}{x}$ exist finitely hence $(1)$ is applicable.

Koro
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  • Ok thanks so it is not allowed to multiply and divide by $x$ if $x\to\infty$, right? – Sisi Apr 28 '21 at 20:51
  • @Sisi: It is allowed but passing the limit from product (fg) to (f) and (g) that is (1) is not allowed if either of the limits as mentioned in (1) does not exist finitely. For example: $\lim 1=\lim (x\times \frac 1x)=1$ but breaking it into two gives $\lim x\to \infty $ and $\lim \frac 1x \to 0$ as $x \ to \infty$. Also remember that $\lim f(x)$ as $x \to a$ is a statement about values $f(x)$ takes in deleted neighborhood of $a$ and it does not matter what happens at $x=a$. – Koro Apr 28 '21 at 21:00
  • Ok thanks but I can't understand when you say that it is allowed to divide and multiply by $x$ when $x\to \infty$...you mean that I can do this but really it iwould be useless for the fact that I can't split the limit of $x$ from the piece divide by $x$ that remains, right? – Sisi Apr 28 '21 at 21:07
  • @Sisi: Right. The reason why you can’t split in two limits because $(1)$ is not satisfied. – Koro Apr 28 '21 at 21:11
  • Ok very thanks! So to answer to the question on the title: in principle I can multiply and divide by $x$ when $x\to \infty$ but in a limit this is never done for the reason why that then I can't split the limit, right? – Sisi Apr 29 '21 at 06:34
  • @Sisi: I hope you understand what $x\to a$ means and accordingly what $x\to \infty$ means and that you can divide as long as division is defined. With that understanding, you are right. – Koro Apr 29 '21 at 07:00