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

(Context is this.)

  1. Am I correct to say it's wrong to argue $\lim_{\substack{x \to 0}}\cos(\frac{1}{x})$ doesn't exist because $\lim_{\substack{x \to 0}}\cos(\frac{1}{x})$ $= \cos(\lim_{\substack{x \to 0}}\frac{1}{x})$ $ = \cos (\text{does not exist})$ $= \text{does not exist}$ (where we justify the 1st equality like '$\cos$ is continuous on $\mathbb R$')?

Sure we can define $g$ to have range all of $\mathbb R$ like $g(x)=\frac 1 x, g: \mathbb R \ \setminus \ 0 \to \mathbb R$ and thus $g$'s range includes $0$. So we might use the fact that $g$'s range includes $0$ to combine with the fact that $\cos: \mathbb R \to \mathbb R$ is (defined and) continuous at $x=0$. But technically $g$'s image doesn't include $0$. So I think it's nonsensical to say that we can put the limit inside because the outer function $\cos$ is continuous at a value that is in the range of the inner function because the inner function can have anything in its range.

  1. If yes to (1) (i.e. I'm correct that it's wrong), then how again do we argue $\lim_{\substack{x \to 0}}\cos(\frac{1}{x})$ doesn't exist? Really from scratch/definition or something?

  2. If yes to (1) (i.e. I'm correct that it's wrong), then what about in general when we have

$\lim_{x \to a}f(g(x))$ where $\lim_{x \to a}g(x)$ doesn't exist (but I guess $f$ is say...defined and continuous on all of $\mathbb R$)?

Wait I just realised:

I don't think we can say like

$\lim_{x \to a}f(g(x))$ doesn't exist just because $\lim_{x \to a}g(x)$ doesn't exist.

I think of $a=0, f(x) = \frac 1 x = g(x), f,g: \mathbb R \ \setminus 0 \to \mathbb R \ \setminus 0$.

BCLC
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    You are correct that you cannot say that $\lim_{x\to a}f(g(x))$ does not exist just because $\lim_{x\to a}g(x)$ does not exist. Take $g(x) = |x|/x$, $f(u)=u^2$, and $a=0$. The $\lim_{x\to 0}g(x)$ does not exist, but $\lim_{x\to 0}f(g(x)) = \lim_{x\to 0}|x|^2/x^2 = 1$. – Arturo Magidin Oct 06 '21 at 17:33
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    To argue your initial limit does not exist, it suffices to show that for every $\delta\gt 0$, there are values of $x$ within $\delta$ of $0$ where $\cos(1/x)=1$, and values where $\cos(1/x) = -1$. That will show that no matter how close to $0$ you are, the values of $\cos(1/x)$ do not approach any particular, single number. And to achieve that, just take a sufficiently large $N$ and look at $x=1/2N\pi$, and $x=/(2N+1)\pi$. – Arturo Magidin Oct 06 '21 at 17:36
  • thanks @ArturoMagidin 1 - is my example at the end of the post about $\frac 1 x \circ \frac 1 x$ correct? 2 - so when you argue why $\lim_{x \to 0} \cos(\frac 1 x)$ doesn't exist, the argument is not necessarily directly related to why $\lim_{x \to 0} \frac 1 x$ exists and so we should do $\lim_{x \to 0} \cos(\frac 1 x)$ from scratch? aaaand same for $\lim_{y \to 0} e^{\frac 1 y}$? – BCLC Oct 06 '21 at 17:38
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    2- The limit law you are trying to apply in both cases is related to continuity. It is a conditional equality (see comment in your other question). It says: "IF $\lim_{x\to a}g(x) = L$ and $f$ is continuous at $L$, then $\lim_{x\to a}f(g(x)) = f(\lim_{x\to a}g(x)) = f(L)$." Or alternatively, "IF $f(x)$ is continuous everywhere and $\lim_{x\to a}g(x)$ exists, then $\lim_{x\to a}f(g(x)) = f(\lim_{x\to a}g(x))$." These laws are predicated on the assumption that $\lim_{x\to a}g(x)$ exists. If it doesn't exist, then you cannot apply it.(cont) – Arturo Magidin Oct 06 '21 at 17:40
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    (cont) Not sure what you mean when you say "from scratch". It's certainly true that you cannot go that way; but there may be other results that you can bring into play. For $\lim_{y\to 0}e^{1/y}$, note that $\lim_{y\to 0^+} 1/y=\infty$, so $\lim_{y\to 0^+}e^{1/y} = \lim_{x\to\infty}e^x=\infty$ does not exist, so the two-sided limit does not exist. (By contrast, $\lim_{y\to 0^-}e^{1/y} = 0$, even though $\lim_{y\to 0^-}1/y$ does not exist). The fact that $\lim_{y\to 0^+}e^{1/y} = \lim_{x\to\infty}e^x$ (unconditionally, valid even if they don't exist) is related to the continuity of $e^x$. – Arturo Magidin Oct 06 '21 at 17:42
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  • Yes: $\lim_{x\to 0}1/(1/x) = \lim_{x\to 0}x = 0$, because the function $1/(1/x)$ and the function $x$ take the same values everywhere except at $0$, so the first equality is valid unconditionally.
    • – Arturo Magidin Oct 06 '21 at 17:42
  • @ArturoMagidin ok thanks so 3 - basically $\lim f(g(x))$ may or may not exist when $\lim g(x)$ doesn't exist and 4 - if $\lim f(g(x))$ doesn't exist, then it may or may not be related to why $\lim g(x)$ doesn't exist? – BCLC Oct 06 '21 at 17:52
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    Well... yes; because you are asking "if $X$ happens, then either (yes or no). And if $Y$ happens it may either be related or unrelated to $Z$". Both statements are tautologies. – Arturo Magidin Oct 06 '21 at 18:19
  • @ArturoMagidin right ok ummm...rephrase 3 - 3.1. - it is wrong to say $\lim f \circ g$ necessarily doesn't exist? 3.2 - it is wrong to say $\lim f \circ g$ necessarily exists? 4 - 4.1 - if $\lim f \circ g$ doesn't exist then it is wrong to say that the non-existence is necessarily related to the non-existence of $\lim g$ 4.2 - if $\lim f \circ g$ doesn't exist then it is wrong to say that non-existence is necessarily not related to the non-existence of $\lim g$? – BCLC Oct 06 '21 at 18:22
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    3: If $\lim g(x)$ does not exist, this, in and of itself, does not tell you anything about whether $\lim f(g(x))$ exists or not. The limit may exist, or may fail to exist; the information that $\lim g(x)$ does not exist is insufficient to resolve the problem. 4. Yes it is incorrect; you know $\lim f(g(x))$ exists if two things happen: $\lim g(x)$ exists, and $f$ is continuous at $\lim g(x)$. So if the limit does not exist, you know those two things are not both true... but you don't know which one is false (or whether both are). 4.2: too many negatives; I can't parse it. – Arturo Magidin Oct 06 '21 at 18:55
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    Example: let $f(x) = |x|/x$ and let $g(x)=\sin(x)$. Then $\lim_{x\to 0} g(x)$ certainly exists, but $\lim_{x\to 0}f(g(x))$ does not; you can't blame $g$ for the non existence of the latter limit. On the other hand, the nonexistence of $\lim_{x\to 0}e^{1/x^2}$ is related to the nonexistence of $\lim_{x\to 0}1/x^2$. – Arturo Magidin Oct 06 '21 at 18:58
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    Thinks of it this way: you know that if your alarm goes off and your car starts in the morning, you will get to the office on time. If your alarm doesn't go off, do you know for a fact that you got to the office late? No; you may or may not; maybe you got up on time anyway on your own, or speeded when you were late. If you got late to the office, we know that either your alarm didn't go off or your car didn't start (or both); but we don't know which, and just from knowing you came in late we cannot deduce which went wrong, just that at least one of them did. – Arturo Magidin Oct 06 '21 at 19:39
  • @ArturoMagidin THANK YOU – BCLC Oct 07 '21 at 12:27