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 $\varepsilon$):
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$\lim_{x \to 0} [2x^2 + 5x]$ exists as the sum of the following limits, if the following limits exist: $\lim_{x \to 0} 2x^2$, $\lim_{x \to 0} 5x$.
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$\lim_{x \to 0} 2x^2$ exists as 2 times the following limit, if the following limit exists: $\lim_{x \to 0} x^2$
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$\lim_{x \to 0} 5x$ exists as 5 times the following limit, if the following limit exists: $\lim_{x \to 0} x$
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$\lim_{x \to 0} x^2 = 0$
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$\lim_{x \to 0} x = 0$
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By (5) and (3), $\lim_{x \to 0} 5x$ exists and is equal to $5(0)=0$
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By (4) and (2), $\lim_{x \to 0} 2x^2$ exists and is equal to $2(0)=0$
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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.
Please explain what's going on.
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:
Best Answer
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:
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 $x_2=\frac1{(2n+1)\pi}<x_1<\delta.$ $\cos(1/x_1)=1, \cos(1/x_2)=-1.$ 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.