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I know a question With the exact same title has an answer but it hasn’t areally answered mine so please hear me out.

I know what it means for the limit At a point to “ not exist”; the function doesn’t approach a certain value as x approaches some value. But i have seen this phrase used in 2 different contexts. One where the one-sided limits are different and

the second where function approaches infinity or negative infinity from both sides.

The notation for the second, “infinite limit, is always written out in the regular limit notation by the author, who says that it is “describing the way in which limit does not exist”. What are the differences in these 2 types of non-existences in context of the precise definition of a limit not existing?

Secondly, the author explains how The limit of a quotient cannot be computed by the quotient of the limits when the limit of the denominator is equal to zero and that of the numerator is Positive and proceeds to say that the “limit does not exist” without mentioning the limit notation used for the “infinite limits”, so I assume that he’s implying that this is the first kind of non existence. Shoudn’t this be the second type, since this is no longer an intermediate form and should clearly approach a very large positive number?

Sorry for not formatting, I post on a tablet and dont know how to format.

José Carlos Santos
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Vulgar Mechanick
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  • Usually, we sloppy say the limit is $\infty$ or $-\infty$, so I guess it is meant that the limits do not coincide, which in this case means that the signs from left and right are distinct. – Peter Jun 11 '20 at 09:23
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    To say that e.g. a certain sequence of numbers converges to a limit $L$ means that there is a specific behavior of the sequence (a behavior specified by the definition of limit). The same for the limit of a function $f(x)$ at a certain point $x_0$. To say that the limit does not exist means that the condition specified by the definition is not satisfied. This can happen in different cases. – Mauro ALLEGRANZA Jun 11 '20 at 09:31
  • See Does-a-limit-at-infinity-exist ? and Why-does-an-infinite-limit-not-exist ? – Mauro ALLEGRANZA Jun 11 '20 at 09:33
  • What about the second part of the question? Which case is that? – Vulgar Mechanick Jun 11 '20 at 09:39
  • What part precisely ? – Mauro ALLEGRANZA Jun 11 '20 at 09:41
  • Starts with “secondly” – Vulgar Mechanick Jun 11 '20 at 09:41
  • Consider sequences (maybe is more simple): can you "see" why $1,2,1,2,1,2,\ldots$ does not converge ? Can you "see" why $1,2,3,4,...$ does not converge ? If so, can you check why they do not satisfy the defintion of convergence to a limit $L$ ? – Mauro ALLEGRANZA Jun 11 '20 at 09:42
  • That is not my question. I understand that if the numerator is positive but denominator is 0, the limit does not exist. But this isnt an intermediate form, like a 1/0, is it? Isnt it clear that it will be positive infinity, not that infinity is a number – Vulgar Mechanick Jun 11 '20 at 09:46
  • Basically, if we have $\lim_{x \to 0} \dfrac A x$ when $A$ is a positive quantity, there is no way to find a number $L$ (i.e. a finite magnitude) such that whenever we choose $x$ we have that $ \dfrac A x$ is "sufficiently close" to $L$. This case corresponds (for functions) exactly to the case $1,2,3,\ldots$ for sequences. – Mauro ALLEGRANZA Jun 11 '20 at 09:50
  • As far the limit of a function is concerned there are 5 mutually exhaustive and exclusive possibilities. Either it tends to a limit, or diverges to $\infty $, or diverges to $-\infty $, oscillates finitely, or oscillates infinitely. – Paramanand Singh Jun 11 '20 at 10:41

2 Answers2

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We say the the limit $\lim_{x\to a}f(x)$ is a real number $l$ when$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in D_f):|x-a|<\delta\implies|f(x)-l|<\varepsilon,$$we say that $\lim_{x\to a}f(x)=\infty$ when$$(\forall M\in\Bbb R)(\exists\delta>0)(\forall x\in D_f):|x-a|<\delta\implies f(x)>M,$$and we say that $\lim_{x\to a}f(x)=-\infty$ when$$(\forall M\in\Bbb R)(\exists\delta>0)(\forall x\in D_f):|x-a|<\delta\implies f(x)<M.$$The limits $\lim_{x\to a^+}f(x)$ and $\lim_{x\to a^-}f(x)$ are similar, but then $|x-a|<\delta$ becomes $0<x-a<\delta$ and $-\delta<x-a<0$ respectively. In any case, $\lim_{x\to a}f(x)$ exists if and only if both limits $\lim_{x\to a^+}f(x)$ and $\lim_{x\to a^-}f(x)$ exist.

When someone says that the limit $\lim_{x\to a}f(x)$ exists, that person should let it clear whether he or she is talking about existence in $\Bbb R$ or about existence in $\Bbb R\cup\{\pm\infty\}$. Assuming that the context here is just the existence in $\Bbb R$, how can a limit fail to exist? These are the possibilities:

  • One of the limits $\lim_{x\to a^\pm}f(x)$ does not exist in $\Bbb R$.
  • Both of them exist, but they are distinct.
José Carlos Santos
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We might summarize by saying that

  • the limit is said to not exist when there is no real number which is the limit (either because the function does not converge to a single real, or because it is unbounded); e.g. $\sin\frac1x$ or $\frac1x$ or $\frac1{x^2}$ at $0$.

  • the infinite limit is said to not exist if the function does not converge to infinity nor minus infinity; e.g. $\frac1x\sin\frac1x$ or $\frac1x$ at $0$.

  • How can you use the definite article THE for something that does not exist? – Jochen Jun 11 '20 at 10:06
  • @Jochen: because it is the limit in question. Like when you say that the square root of $-1$ does not exist. –  Jun 11 '20 at 10:12
  • I do not say the square root of $-1$ does not exist. Abuse of the definite article causes a lot of confusion. – Jochen Jun 11 '20 at 10:19
  • @Jochen: would you say "the above limit does not exist" or "an above limit does not exist" ? –  Jun 11 '20 at 10:22
  • I would say $\sin(1/x)$ does not have a limit at $0$. – Jochen Jun 11 '20 at 10:31
  • @Jochen: you don't answer, it is assumed that the expression of the limit was written above. –  Jun 11 '20 at 10:48