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Suppose that $L$ is a real number and $f$ is a real-valued function defined on some interval $(b, \infty)$. We say that $\displaystyle{\lim_{x \to \infty} f(x) =L}$ if for every positive real number $\epsilon$, there is a real number $M$ such that if $x>M$ then $|f(x) -L| < \epsilon$.

Is this statement correct, or should it be amended to imply that a limit can exist at L (i.e. it is possible for a limit to exist at L), but does not have to be the limit of the function? For example, we can prove from this definition that $\displaystyle{\lim_{x \to \infty} \frac{4}{x^2}=0}$, but can't one also prove that $\displaystyle \lim_{x \to \infty} \frac{4}{x^2}=-0.001$, $\displaystyle \lim_{x \to \infty} \frac{4}{x^2}=-0.0001$ and other false claims by application of this definition?

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What do you mean by a limit can exist at $L$? Also, you need to write out how you go about "proving" $\displaystyle \lim_{x \rightarrow \infty} \frac4{x^2} = -0.0001$ for us to show where you are making the mistake. – user17762 Mar 11 '12 at 4:32
Remember, the statement requires that for every $\epsilon>0$ you can find such an $M$. – Alex Becker Mar 11 '12 at 4:32
up vote 6 down vote accepted

The statement is correct.

(Note also that we usually talk about a limit existing at $a$ to refer to the point that the variable $x$ is approaching, rather than what the values of the function are approaching; to refer to what the function is approaching, we talk about the limit being $L$, or equaling $L$).

In your example, you cannot prove that $\lim\limits_{x\to\infty}\frac{4}{x^2} = -.0001$: given any $L\gt 0$, let $\epsilon = \frac{L}{2}$. Then for any $N\gt 0$, pick $x\gt\max\{N, \sqrt{\frac{8}{L}}\}$. Then $$\frac{8}{L}\lt x^2,\text{ therefore }\frac{4}{x^2}\lt\frac{L}{2}.$$ And therefore, we have that $$\left|L-\frac{4}{x^2}\right| = L - \frac{4}{x^2} \gt L-\frac{L}{2} = \epsilon.$$ We have therefore proven that if $L\gt 0$, then:

For every $N\gt 0$ there exists $x\gt N$ such that $|L-f(x)|\gt\frac{L}{2}$.

This proves that the limit definition cannot be satisfied, since the condition fails for at least one $\epsilon$.

If $L\lt 0$, pick $\epsilon=\frac{-L}{2}$ and a similar computation shows that you can always find $x$ greater than any given $N$ that will show the property is not satisfied.

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Think of it this way. A function gets "trapped" in a neighborhood $(L+\epsilon, L-\epsilon)$ of $L$ if eventually $f(x)$ is always in that neighborhood past some marker $M$ on the real line. To say the limit is $L$ is to say it gets trapped in arbitrarily small neighborhoods of $L$; if this were possible for two different points then we could just select neighborhoods of each small enough that they are disjoint - and then where will the points be!? Contradiction. Graphically,

$\hskip 1in$ trap

Once the green and blue bubbles around green $L_1$ and $L_2$ are small enough to be disjoint, you can't have points inside both of them simultaneously. This is why the limit must be unique.

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VERY good explaination!!! – Mathemagician1234 Sep 25 '13 at 4:19

The limit is unique. Choose $\epsilon=0.000001$ and then try to find $M$ satysfying the required property for $L=-0.001$.

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Is it possible to prove that a limit claim is true algebraically, using the definition and without trial and error? – amy Mar 11 '12 at 4:34
@Jaydon: Yes; you can invoke theorems that establish the desired result. Or you can do things without guessing because you understand what is going on... – Arturo Magidin Mar 11 '12 at 4:40

If you prove that the limit is L for some number L, based in the definition, is not possible that the limit is another number H.Because you can take as epsilon for example the third part of the distance between L and H , and if for a neighborhood of L the values of the function are there for all X>M, then for all N>0 there will exist values x>N that are not in the neighborhood of N with radius epsilon.

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As a matter of fact, you cannot actually prove your other two results. The key is that my $\epsilon$ can be any positive real number, no matter how small, and my limit must satisfy the inequality for all such $\epsilon$. With sufficient playing around and reasoning, you will be able to pick an $\epsilon$ so that the assertions $\displaystyle \lim_{x \to \infty} \frac{4}{x^2}=-0.001$ or $\displaystyle \lim_{x \to \infty} \frac{4}{x^2}=-0.0001$ fail, as azarel describes in his answer.

More generally, one can show that if our pointwise limit exists for a real-valued function (such as our $\frac{4}{x^2}$), it is unique. We even have a much stronger statement: for any sequence in a Hausdorff space, there is at most one limit.

If the concept of a Hausdorff space (or a general topological space) is unrelated to your current knowledge and interest, the takeaway can be that when I have a sufficiently nice space (of which $\mathbb{R}$ is one), we will not be able to prove false claims such as the ones you describe. Of course, you don't have to take this unwillingly on faith from me! Try taking a case where a limit exists at a point, claiming it equals two different real numbers, and arriving at a contradiction from your definition of a limit.

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