# Properly arguing for basic limit laws: We have to do it backwards? I mean, we're not sure the limits exist...

Ok dumb questions given all the questions I've asked before on this account, but here goes:

Example/Question 1: When we evaluate things like $$\lim_{x \to 0} [2x^2 + 5x]$$, is it actually improper to say like

$$\lim_{x \to 0} [2x^2 + 5x] = \lim_{x \to 0} 2x^2 + \lim_{x \to 0} 5x$$

$$= 2 \lim_{x \to 0} x^2 + 5\lim_{x \to 0} x$$

$$= 2 (0) + 5 (0) = 0+0 = 0$$

?

Context: This seems to be how it is done in, say, Stewart Calculus.

See the Stewart Calculus limit laws. The limit laws can be used ASSUMING certain limits involved exist in the 1st place.

So for example, I don't see how can we possibly say

$$\lim_{x \to 0} [2x^2 + 5x] = \lim_{x \to 0} 2x^2 + \lim_{x \to 0} 5x$$

when we haven't established that both $$\lim_{x \to 0} 2x^2$$ and $$\lim_{x \to 0} 5x$$ exist

What I think we should do is that the above kind of argument is scratch work and then the proper argument is as follows (similar to the $$\varepsilon-\delta$$ thing where we argue backwards from $$\varepsilon$$ to $$\delta$$ as scratch and then write the formal proof forwards from $$\delta$$ to $$\varepsilon$$):

1. $$\lim_{x \to 0} [2x^2 + 5x]$$ exists as the sum of the following limits, if the following limits exist: $$\lim_{x \to 0} 2x^2$$, $$\lim_{x \to 0} 5x$$.

2. $$\lim_{x \to 0} 2x^2$$ exists as 2 times the following limit, if the following limit exists: $$\lim_{x \to 0} x^2$$

3. $$\lim_{x \to 0} 5x$$ exists as 5 times the following limit, if the following limit exists: $$\lim_{x \to 0} x$$

4. $$\lim_{x \to 0} x^2 = 0$$

5. $$\lim_{x \to 0} x = 0$$

6. By (5) and (3), $$\lim_{x \to 0} 5x$$ exists and is equal to $$5(0)=0$$

7. By (4) and (2), $$\lim_{x \to 0} 2x^2$$ exists and is equal to $$2(0)=0$$

8. By (1), (6) and (7), $$\lim_{x \to 0} [2x^2 + 5x]$$ exists and is equal to $$0+0=0$$.

This seems very weird, unnatural, etc. For some reason ever since elementary calculus this is not what is being done. Yet, I think this should be the case otherwise we may fall into traps like $$\lim_{x \to 0} \frac{x}{1} \frac{1}{x} = \lim_{x \to 0} \frac{x}{1} \lim_{x \to 0} \frac{1}{x} = (1)$$(does not exist) = does not exist. I think I fell for this kind of trap here.

Example/Question 2: (a real multivariable example. I think there's an easy way to do this in single real, but I can't think of an example right now.)

Here, I am trying to argue that $$\lim_{(x,y) \to (0,0)}e^{\frac{y}{x^2+y^2}}\cos(\frac{x}{x^2+y^2})$$ doesn't exist because

$$\lim_{\substack{(x,y) \to (0,0) \\ y=0}}e^{\frac{y}{x^2+y^2}}\cos(\frac{x}{x^2+y^2})$$ doesn't exist because $$\lim_{\substack{(x,y) \to (0,0) \\ y=0}}e^{\frac{y}{x^2+y^2}}\cos(\frac{x}{x^2+y^2}) = \lim_{\substack{x \to 0}}\cos(\frac{1}{x})$$ and then because $$\lim_{\substack{x \to 0}}\cos(\frac{1}{x})$$ doesn't exist.

Actually, how is it even sensible to do this entire long list of limit equalities $$\lim_{\substack{(x,y) \to (0,0) \\ y=0}}e^{\frac{y}{x^2+y^2}}\cos(\frac{x}{x^2+y^2}) = (...) \text{long list of limit equalities} (...) = \lim_{\substack{x \to 0}}\cos(\frac{1}{x})$$

when we're not even sure that the limits exist?

Guess: Perhaps there's some implicit reductio ad absurdum here like 'suppose on the contrary that this limit exists. Then this limit equals (...) that limit. But that limit doesn't exist! Contradiction.'

Example/Question 3: Actually now I want to ask about $$\lim_{\substack{x \to 0}}\cos(\frac{1}{x})$$, but I'm afraid the post will become too broad (if it isn't already)... Update: Asked here.

Maybe related:

• $$\lim_{x\to a}[ f(x)+g(x)]=\lim_{x\to a} f(x)+\lim_{x\to a} g(x)$$ if two of the limits are known to exist. Commented Oct 6, 2021 at 17:10
• Usually you break up a limit of a sum into a sum of limits. If you end up with a sensible answer, then that is the answer to the original limit. If you end up with a senseless answer (e.g., if one of the limits does not exist, or you end up with an indeterminate), then the "break up" was not necessarily valid and you need to go back and start over. One should do scratch work first, and check if $\lim f(x)$ and $\lim g(x)$ each exists, and then say "so $\lim(f+g) = (\lim f)+(\lim g)$..." But in practice that is too onerous and we use the writing short-cut. Commented Oct 6, 2021 at 17:13
• In short: if when you simplify you get something sensible, then the simplification is (almost certainly) okay and you can use it. If when you simplify you get some combination of things that include "does not exist", "infinity", "minus infinity", or "indeterminate", then the simplification is (almost certainly) not okay and you need to go back. Commented Oct 6, 2021 at 17:14
• On the example you have at the end of the first part, it's not correct to conclude, or even write, that $1(\text{does not exist})=\text{does not exist}$. As Arturo says, if you were to use a limit law to express your limit as a sum or product of two other limits, and found that one of the component limit did not exist, then your use of the limit law would be inconclusive. You would not arrive at a false conclusion, you would just have to do something else. Commented Oct 6, 2021 at 17:17
• @KevinLong: I would qualify that you would not necessarily arrive at a false conclusion: what you get at the end may be correct or incorrect: you just have no way of knowing which it is. Commented Oct 6, 2021 at 17:19

Question 1

We only know the limit of the sum is the sum of the limits if we know two of the limits exist.

So if you want to find the limit of $$2x^2+5x,$$ you might try the limits of $$2x^2$$ and the limit of $$5x.$$ But you are correct, you don’t know yet that these two limits exist.

Once you’ve proven the limits of $$2x^2$$ and $$5x$$ exist, you can conclude the limit of $$2x^2+5x$$ exists, and is the sum of the limits.

But you technically can’t write the equality until you’ve proven that both limits exist.

This is a case where people often still write it, informally, before it is proven. It is more an aspiration than an assertion of fact. It is a statement that you intend to prove the two limits exist, or, often, that the writer thinks it is obvious that the two limits exist, and thus is going to skip the proof.

It is certainly a risk to use such informal language, giving the reader a false impression, or even misleading the writer to write a false proof. A careful writer would write:

We want:$$\lim [2x^2+5x]=\lim 2x^2+\lim 5x$$

To give a hint at the aspirational nature of the statement. But this is often not done, even if the rest of the steps dp prove the two limits exists.

Question 2

If $$\lim f(x)g(x)=M$$ exists and $$\lim f(x)=L\neq 0,$$ then you can conclude that $$\lim g(x)$$ exists and is $$M/L.$$

But here, $$\lim_{x\to 0}e^{1/x}$$ actually converges to $$0$$ as $$x\to 0-$$ and diverges to $$+\infty$$ as $$x\to 0+.$$

If $$\cos(1/x)$$ converges to a non-zero value, then $$e^{1/x}\cos(1/x)$$ diverges, by the above comments.

But $$\cos(1/x)$$ could diverge, or $$\cos(1/x)$$ could converge to $$0,$$ and it is still possible for $$e^{1/x}\cos(1/x)$$ to converge.

So you will need another way to solve this problem.

Question 3

We can show that $$\cos(1/x)$$ diverges.

Assume a limit $$L$$ exists. Let $$\epsilon=1.$$ Let $$\delta>0$$ such that $$|\cos(1/x)-L|<1$$ when $$|x|<\delta.$$

Find $$n>0$$ such that $$x_1=\frac1{2n\pi}<\delta.$$ We have $$x_2=\frac1{(2n+1)\pi} $$\cos(1/x_1)=1, \cos(1/x_2)=-1.$$ So: $$2>|\cos(1/x_1)-L|+|\cos(1/x_2)-L|>|1-(-1)|=2.$$ We’ve reached a contradiction.

But that is not enough to show $$\cos(1/x)e^{1/x}$$ diverges.