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


Given that $\lim_{x \to 0} \frac{x}{1} \frac{1}{x}$ exists even though $\lim_{x \to 0} \frac{1}{x}$ doesn't exist, I guess we can't just say (for extended real number $a$) that $\lim_{x \to a} f(x)g(x)$ does not exist just because $\lim_{x \to a} f(x)$ or $\lim_{x \to a} g(x)$ doesn't exist.

  1. I could be remembering wrong, but I could swear I've done a million times in calculus class and even in higher classes like say in measure theory or probability theory a lot of things like

$$\lim_{x \to a} f(x)g(x)$$ does not exist because

$$\lim_{x \to a} f(x)g(x)$$

$$ = \lim_{x \to a} f(x) \lim_{x \to a}g(x)$$

But $\lim_{x \to a}g(x)$ does not exist. Therefore, $\lim_{x \to a} f(x)g(x)$ does not exist.

This really doesn't make sense because the law about limit of products is product of limits $$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a}g(x)$$ assumes that each of $\lim_{x \to a} f(x)$ and $\lim_{x \to a}g(x)$ exist.

  • (I don't think you can even do the argument as above when $\lim_{x \to a} f(x)$ and $\lim_{x \to a}g(x)$ exist...I think it's not very rigorous. I think one should 1st prove the existence of $\lim_{x \to a} f(x)$ and $\lim_{x \to a}g(x)$ before using the limit law. Plan to ask about this in another question. Update: here.)

Question 1: Are there some conditions where we can somehow say that the limit of the product doesn't exist because limit of 1 of the factors doesn't exist?


  1. An example in real multivariable: Here, I initially tried arguing $$\lim_{(x,y) \to (0,0)}e^{\frac{y}{x^2+y^2}}\cos(\frac{x}{x^2+y^2})$$ doesn't exist because $\lim_{(x,y) \to (0,0)} e^{\frac{y}{x^2+y^2}}$ doesn't exist...

Question 2: But this is incorrect reasoning...right?

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  • $\begingroup$ There are specific indeterminate forms for limits of products; e.g. $0\cdot\infty$ is a 'classic' one. (For instance the $\frac1x\cdot\frac x1$ example). If one of $f()$ or $g()$ tends to a non-zero limit but the other doesn't, then you can show that their product doesn't either. (See if you can figure out how — a proof by contradiction works nicely here.) $\endgroup$ Oct 6, 2021 at 16:39

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Question 1: Are there some conditions where we can somehow say that the limit of the product doesn't exist because limit of 1 of the factors doesn't exist?

Yes, indeed you can. Frequently, if the limit of exactly one of the two terms exists, then their product (or sum) will also fail to exist, but there are caveats. Formally, if $\lim_{x\to a} f(x)$ exists and is non-zero, and $\lim_{x \to a} g(x)$ does not exist, then $\lim_{x \to a} f(x)g(x)$ does not exist too.

Proving this is a straightforward proof by contradiction. We can write $$g(x) = \frac{f(x)g(x)}{f(x)}.$$ If $\lim_{x \to a} f(x)g(x)$ exists, then by the algebra of limits, given once again the non-zero limit of the denominator, then the limit of $g(x)$ would exist and we would have $$\lim_{x \to a} g(x) = \frac{\lim_{x \to a} f(x)g(x)}{\lim_{x \to a} f(x)},$$ which contradicts the limit of $g$ not existing. So, our assumption that $\lim_{x \to a} f(x) g(x)$ exists is wrong.

A similar result holds for $f(x) + g(x)$, except we need not assume $\lim_{x \to a} f(x) \neq 0$.

In your example, $\frac{x}{1} \cdot \frac{1}{x}$, the problem is, of course, that your $f(x) = \frac{x}{1}$ tends to the one and only forbidden limit: $0$. This provides a suitable counterexample to show that the $\lim_{x \to a} f(x) \neq 0$ condition is necessary.

I initially tried arguing$$\lim_{(x,y) \to (0,0)}e^{\frac{y}{x^2+y^2}}\cos\left(\frac{x}{x^2+y^2}\right)$$doesn't exist because $\lim_{(x,y) \to (0,0)} e^{\frac{y}{x^2+y^2}}$ doesn't exist...

Question 2: But this is incorrect reasoning...right?

This is indeed incorrect, as the limit of neither term exists. Instead, try considering the limit along the two axes.

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  • $\begingroup$ re question 1: ok done. thanks. unfortunately your good answer brings me more questions than answers. unfortunately further i'm not quite sure what questions to ask. $\endgroup$
    – BCLC
    Oct 6, 2021 at 17:25
  • $\begingroup$ re question 2: right so my later argument is to do $x=0$ or $y=0$. wait what do you mean sensible limit? You mean the limit actually exists (as an extended real number)?!?!?! (now double checking on wolfram alpha. i'm so stupid to not do this on the 1st place) wolfram says: does not exist...? $\endgroup$
    – BCLC
    Oct 6, 2021 at 17:29
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    $\begingroup$ @JohnSmithKyon Sorry re: question 2. That was from a previous draft in which I made a stupid error. It absolutely does not have a sensible limit. $\endgroup$ Oct 6, 2021 at 17:32
  • $\begingroup$ Theo Bendit haha don't have to be sorry. thanks a lot! $\endgroup$
    – BCLC
    Oct 6, 2021 at 17:32
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    $\begingroup$ @JohnSmithKyon Well, if $\lim_{(x, y) \to (0, 0)} f(x, y)$ exists, then $\lim_{x \to 0} f(x, 0)$ and $\lim_{y \to 0} f(0, y)$ exist too, and are equal to the same limit. So, if you find one or the other does not exist (or perhaps they do exist, but are equal to different limits), then the original limit does not exist either. We can see $\lim_{x \to 0} e^{1/y}$ doesn't exist because $e^{1/y}$ tends to $\infty$ as $y \to 0^+$ and $0$ as $y^\to 0^-$ (once again, in order for the two-sided limit to exist, both one-sided limits must exist and be equal). $\endgroup$ Oct 6, 2021 at 17:43

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