# Does a limit at infinity exist?

I use Stewart's (Calculus, 8e) terminology. Infinite limits do not exist. For example we can write $$\lim_{x \rightarrow 0} \frac{1}{x^2} = \infty,$$ but at the same time say that $$\lim_{x \rightarrow 0} \frac{1}{x^2}$$ does not exist. Or at least this is what Stewart (89ff) insists.

My question is this: Do limits at infinity exist? For example we can write $$\lim_{x \rightarrow \infty} \frac{1}{x^2} = 0.$$ But in this case, should we say that $$\lim_{x \rightarrow \infty} \frac{1}{x^2}$$ exists or does not exist? Stewart (126ff) doesn't seem to explicitly address this, which is why I'm asking.

(Related: Why does an infinite limit not exist? --- here the reference is again Stewart, but there seems to be some confusion of the terms "infinite limit" and "limit at infinity".)

• You are asking is it possible to define $\lim_{x\to\infty}f(x)$? Is that correct? Then I recommend you to delete everything before "My question is ... ", and also all references to Stewart and the earlier MSE question. – almagest May 12 '16 at 8:29
• @almagest: As discussed at Why does a limit at infinity not exist?, there is not complete consensus as to whether or not $\lim_{x\rightarrow 0} \frac{1}{x^2}$ exists. So I wanted to provide some context and in particular mention that I'm using Stewart (one of the most popular calculus textbooks) as my reference. And then introduce my question. (Also the other MSE question confusingly has nearly exactly the same title, although in fact my question is subtly different.) – Kenny LJ May 12 '16 at 8:32
• That is not context. It is an entirely different question. I agree that the title of the earlier question is misleading. – almagest May 12 '16 at 8:34
• Just FYI: I've changed the title of the other question to be less misleading. – Daniel Fischer May 12 '16 at 9:36

Any statement (or equation) involving the symbol $\infty$ has a precise meaning not by default (or via knowledge of primary school level math) but via a special definition to interpret such statements. So if you write $$\lim_{x \to 0}\frac{1}{x^{2}} = \infty$$ then it does not mean that the symbol $$\lim_{x \to 0}\frac{1}{x^{2}}$$ is some specific thing and the symbol $\infty$ is another specific thing and both are equal. Rather this equation has a special meaning given by a specific definition which is as follows:

Given any real number $N > 0$, there is a real number $\delta > 0$ such that $$\frac{1}{x^{2}} > N$$ whenever $0 < |x| < \delta$.

Any textbook must define the precise meaning of phrases containing the symbol $\infty$ (and equations containing the symbol $\infty$) before writing such phrases (or equation). If this is not done then the textbook author is guilty of a common crime called "intellectual dishonesty".

On the other hand there are many conventions about the existence of a limit. Some authors prefer to say that a limit exists only when it is finite (I prefer this approach). Some define infinite limits also as a case of existence of the limit. Both the approaches are rigorous and without any fault. It is a matter of taste and individual preference of authors. For a student it is of utmost importance to follow the convention prescribed by his teacher/instructor.

The other limit $$\lim_{x \to \infty}\frac{1}{x^{2}} = 0$$ exists in both the conventions and I don't see any problem here.

• In fact, in some sense it is nicer to work with the affinely-extended reals in analysis, because it is compact, except that it's no longer a field. – user21820 May 12 '16 at 8:43
• @user21820: Well, I said that we need to give meaning to such equations and phrases. They don't have meaning by default. And yes you can define them in such manner these are specific things and are equal. Its a matter of definitions. I am warning against attaching any default meaning in absence of suitable definitions. – Paramanand Singh May 12 '16 at 8:44
• @user21820: I have seen many introductory textbook leave these definitions and hope that the student somehow gets some intuitive meaning out of these equations. This is what I term as intellectual dishonesty. BTW I don't mind your statement. I have interacted with you in past (although I must say it is difficult to remember you by your numeric username) and I am happy to learn about other points of view. – Paramanand Singh May 12 '16 at 8:46
• @KennyLJ: I also strongly doubt that "rigor adds complexity". In my opinion rigor leads to a much better understanding and appreciation of the concepts provided it is given without formalism (i.e. rigor should be given in normal text instead of using too much symbols). – Paramanand Singh May 12 '16 at 9:10
• @KennyLJ: Haha I fully agree with Paramanand that it is practically feasible to teach even weak students with full logically rigour. To me, this doesn't mean that we provide all proofs, since some will be beyond the capability of the student to grasp, but it should mean that the student knows precisely what has been proven and what has not, and knows how to check whether a theorem can be applied, this not requiring knowledge of its proof. I think we both agree that pedagogic soundness includes mathematical correctness even if there are no proofs provided. – user21820 May 13 '16 at 3:18

The only purpose of any textbook insisting that $\lim_{x \to 0} \frac{1}{x^2}$ does not exist is so that the basic limit properties can be stated more simply, such as:

$\lim_{x \to a} ( f(x) + g(x) ) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$ for any function $f$ defined on a punctured neighbourhood of $a$ such that two of these limits exist.

If we choose to say that $\lim_{x \to 0} \frac{1}{x^2}$ exists (for instance as a point on the affinely-extended real line), then the above property about the limit of the sum is no longer true, otherwise we get:

$\color{red}{ \infty = \lim_{x \to 0} \frac{1}{x^2} = \lim_{x \to 0} \frac{2}{x^2} + \lim_{x \to 0} -\frac{1}{x^2} = \infty + (-\infty) = \lim_{x \to 0} \frac{1}{x^2} + \lim_{x \to 0} -\frac{1}{x^2} }$

$\color{red}{ \ = \lim_{x \to 0} 0 = 0 }$. [WRONG!!!]

Now this problem does not arise with limits at $\infty$. The limit properties work as long as the limit values are real numbers.

It may help to list these "different kinds of limits", each with their own formal definition and notation, to get an idea of what is going on and why certain terminilogy about (not) existing is used.

$x$ tends to real, limit is real.

Usually, one begins with defining the situation where, if $x$ tends to $a$ where $a$ is a real number, the function tends to a real number $L$ which we call the limit of $f$ at $a$ and we write this as: $$\lim_{x \to a} f(x) = L$$ When there is no real number $L$ satisfying the formal definition (which I left out), then you say the limit does not exist. This is the "basic case" but it is useful to extend the use of limits to be able to describe other situations.

$x$ tends to real, limit is infinite.

One of the reasons why the limit may not exist, is because the function grows arbitrarily large (or small; negative) when $x$ tends to $a$. This could be useful information which is lost when we would just say that the limit does not exist. To capture this information, we agree to write this as: $$\lim_{x \to a} f(x) = +\infty \quad (\mbox{or}\; -\infty)$$ So in fact we introduce a notation that symbolizes this behavior, but there are reasons to still say that "the limit does not exist". We reserve this for the case where the limit is a real number because with this choice, a lot of properties of limits are more elegant and/or easier to state.

$x$ tends to infinity, limit is real.

Limits are a way to describe the behaviour of function where its value cannot be simply evaluated. It is interesting to be able to describe how a function behaves when $x$ grows arbitrarily large (or small; negative). If a function tends to a real number $L$ when $x$ grows arbitrarily large, we say that the limit of $f$ at (positive) infinity exists and we write: $$\lim_{x \to +\infty} f(x) = L$$ This is a different notation and describes a different situation. The "role of infinity" is not the same: it is not $|f|$ that grows arbitrarily large but rather we choose to let $x$ grow arbitrarily large and introduce a new notation for this. The limit can exist (with the notation above), or not.

$x$ tends to infinity, limit is infinite.

Again, if the limit in the situation above does not exist in the sense that there is no real number $L$ satisfying its (omitted) formal definition, there are different kinds of possible reasons. For example, the sine function does not tend to a fixed real number so the following limit "does not exist": $$\lim_{x \to +\infty} \sin x$$ It is however possible that, like in case #2, the function grows arbitrarily large (or small; negative) when we let $x$ grow arbitrarily large (or small; negative). We capture this behaviour by introducing the notations: $$\lim_{x \to \pm\infty} f(x) = \pm\infty$$ where I used $\pm$ twice to briefly summarize the four possibilities.

Sometimes, when all these cases are in a sense "allowed", text books will explicitly state that in the notation $$\lim_{x \to a} f(x) = L$$ they allow the numbers $a$ and/or $L$ to be $\pm \infty$ as well so they don't have to run down all the different notations/cases. Keep in mind though that these different types of limits all have their own formal definitions.

It does exists

Limit to a value is defined as the value that a function "converges" into as x is approaching the value

Thus limit of your function when x is approaching 0 does not converge into any value, therefore doesn't exists

But however, in case of x is approching infinity, the function is approaching 0, therefore the limit exists and it is 0