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I read in Stewart "single variable calculus" page 83 that the limit $$\lim_{x\to 0}{1/x^2}$$ does not exist. How precise is this statement knowing that this limit is $\infty$?. I thought saying the limit does not exist is not true where limits are $\infty$. But it is said when a function does not have a limit at all like $$\lim_{x\to \infty}{\cos x}$$.

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According to some presentations of limits, it is proper to write "$\lim_{x\to 0}\frac{1}{x^2}=\infty$."

This does not commit one to the existence of an object called $\infty$. The sentence is just an abbreviation for "given any real number $M$, there is a real number $\delta$ (which will depend on $M$) such that $\frac{1}{x^2}\gt M$ for all $x$ such that $0\lt |x| \lt \delta$." It turns out that we often wish to write sentences of this type, because they have important geometric content. So having an abbreviation is undeniably useful.

On the other hand, some presentations of limits forbid writing "$\lim_{x\to 0}\frac{1}{x^2}=\infty$." Matter of taste, pedagogical choice. The main reason for choosing to forbid is that careless manipulation of the symbol $\infty$ all too often leads to wrong answers.

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    $\begingroup$ On can write $\lim_{x\to\infty} 1/x^2 = \infty$ and at the same time say "the limit does not exist". And also say $1/x^2$ diverges to $\infty$ as $x$ goes to $\infty$. In the same way, we want to say $\sum_{n=1}^\infty 1/n = \infty$ but also to say $\sum_{n=1}^\infty 1/n$ diverges. $\endgroup$ – GEdgar Apr 3 '12 at 21:53
  • $\begingroup$ @GEdgar do you mean as $x$ goes to 0? $\endgroup$ – Nick T Apr 3 '12 at 23:47
  • $\begingroup$ $x$ goes to $0$, yes. $\endgroup$ – GEdgar Apr 3 '12 at 23:52
  • $\begingroup$ Too bad you can't edit comments :/ $\endgroup$ – john Jan 23 '20 at 4:42
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A limit

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

exists if and only if it is equal to a number. Note that $\infty$ is not a number. For example $\lim_{x\to 0} \frac{1}{x^{2}} = \infty$ so it doesn't exist.

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When a function approaches infinity, the limit technically doesn't exist by the proper definition, that demands it work out to be a number. We merely extend our notation in this particular instance. The point is that the limit may not be a number, but it is somewhat well behaved and asymptotes are usually worth note.

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    $\begingroup$ yes but why in other texts we find sentences like the "limit exists and is finite" $\endgroup$ – palio Apr 3 '12 at 18:11
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    $\begingroup$ @palio This is probably just to emphasize that the limit really isn't infinity or negative infinity. In general the convention is to say that a limit exists only if it is equal to a number (finite numner) $\endgroup$ – Thomas Apr 3 '12 at 18:13
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The term "infinite limit" is actually an oxymoron, like "jumbo shrimp" or "unbiased opinion". True limits are finite.

However, it is okay to write down "lim f(x) = infinity" or "lim g(x) = -infinity", if the given function approaches either plus infinity or minus infinity from BOTH sides of whatever x is approaching, especially to distinguish this from the situation in which it approaches plus infinity on ONE side and minus infinity on the OTHER side, in which case the ONLY correct answer would be "the limit does not exist".

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Note that working in the affinely extended real numbers with the induced order topology this limit exists and equals infinity, unambiguously. We also don't need a "special" definition for infinite limits with this method which is convenient.

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    $\begingroup$ If that's true, then we get the odd result that $f(x) = \frac{1}{(x-2)^2}$ is continuous for all reals if we define $f(2)=\infty$. $\endgroup$ – john Feb 8 '18 at 16:42
  • $\begingroup$ IF we define! but we don't. $\endgroup$ – mike Mar 1 '20 at 19:50
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Generally, we get that

$\lim_{n \rightarrow \infty} x(n) = y \leftrightarrow \forall \epsilon>0: \exists N\in \mathbb{N} : \forall n > N :|x(n) - y|<\epsilon$

but, for example, for $x(n)=n$

it's not true that $\lim_{n \rightarrow \infty} n = \infty $

because it would imply that

$\forall \epsilon>0: \exists N\in \mathbb{N} : \forall n > N :|x(n) - \infty|<\epsilon $

but

$|x(n) - \infty|=\infty$

and it can never be less then $\epsilon$

so $x(n)=n$ does not converge to $\infty$

Real numbers are colloquially defined as limits of sequances, I just shown that $\infty$ is not a real number.

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Ok ... I think there is a difference between saying the limit doesn't exsist and equal to infinity Limit f(x) as x approaches to x' (exsist ) iff lim f(x ) when x approaches to x' from the right = limit f(x) as x reproaches to x' from the left
And in the above limit in your question is exist because limit from the right equal the limit from the left equal to infinity + but if you take f(x)= 1\x in this case the limit doesn't exsist the limit from the right =infinity+ but the limit from the left equal to infinity - Please don't hesitate to discuss this with me.

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