# What exactly is going on when we're finding a limit?

I understand that there are point discontinuities in certain functions, and that there are sometimes values that these functions approach from both sides of the discontinuity. Then the limit exists, and it's a useful thing to know.

Assuming that's right, I still have a problem with this process of finding limits. Is the point to muck about with the function algebraically until there isn't a discontinuity - and that's it? So far in my classes it just seems like we play around until there are no zero divisors and then poof it's all working. It just seems a little fishy.

What should I read about to gain a deeper appreciation for what's going on?

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First of all, if two functions agree everywhere except for at one point $c$ (in some open interval), then their limits are the same. The limit is about where the function is headed, so the actual value of the function at $c$ does not affect the limit. So, if we want the limit as $x$ approaches $11$, it doesn't matter what the function is at $11$. It only matters how the function behaves right around $11$.

Now, we need an example. Let's look at a very simple example. Let

$$f(x) = \frac{7(x - 11)}{x - 11}$$

If we are thinking of the function itself, and not a limit, we can not just cancel the $x - 11$ on top and bottom. That would change the function. $f(x)$, as is, is not defined at $x = 11$. If we cancelled the $x - 11$, we end up with just $7$, a function that is defined everywhere. However, the function $g(x) = 7$ and the function $f(x)$ only differ at $x = 11$. In reality,

$$f(x) = \begin{cases} 7 & x \neq 11 \\ \text{undefined} & x = 11 \end{cases}$$

because we can cancel the $x - 11$ as long as $x$ is not 11. Therefore, the function $f(x)$ and the function $g(x) = 7$ are exactly the same for all $x$ but $11$. So, if we take the limit as $x \to 11$, the limit will be the same with both functions, because the limit does not depend on the value of the function at $x = 11$, only $x$ values around $11$.

Essentially what all this means is that it is perfectly legitimate to cancel out factors in a fraction when taking limits. If you're just looking at a function, with no limit, it is not necessarily legitimate because it could change where the function is defined. But, if we're talking limits, it is. So, if we wanted to calculate the limit of $f(x)$, we simply cancel the $x-11$

$$\lim_{x \to 11} \frac{7(x - 11)}{x - 11} = \lim_{x \to 11} 7 = 7$$

Or, if we wanted to do a more complicated limit, we can still make things simpler by cancelling common factors:

$$\lim_{x \to 3} \frac{x^2 - 7x + 12}{x^2 - 8x + 15} = \lim_{x \to 3} \frac{(x - 3)(x - 4)}{(x - 3)(x - 5)} = \lim_{x \to 3} \frac{x-4}{x-5} = \frac{3 - 4}{3 - 5} = \frac{1}{2}$$

where we can simply plug in the 3 at the end because we no longer have a division by 0.

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I like this explanation, but it's still missing a piece that I'm looking for. Perhaps my question should be, how does one go about proving that your new g(x) is actually identical in every way to f(x) except for at x=11? In this case is it just by inspection or is there something else that can be said? – Henry Sep 15 '12 at 6:50
@Henry I think you're just making it too complicated. It's as simple as the fact that $\frac{x - 11}{x - 11}$ is equal to 1 for every single $x$ value you plug in, other than 11. I think if you plug in a few different $x$ values to it, you'll get the point. Something divided by itself, as long as it it's not 0 (or $\pm \infty$), is always 1. At $x = 11$, we have division by 0, so there's a problem. But, every where else, that function is $1$. So, it looks like a complicated function but it's just 1 whenever $x \neq 11$ and it's undefined at $x = 11$. – Graphth Sep 15 '12 at 13:09
of course I realize that, it's certainly not a very complicated example. I'm sure you've sometimes wanted to understand some more fundamental reasoning behind simple facts. What if you said 'you're making it too complicated, rocks just fall on the ground', you see? – Henry Sep 15 '12 at 13:34
@Henry That's not the same. If we hadn't learned about gravity, it would be very mysterious. I'm telling you, here, there is nothing more to this than the fact that $\frac{a}{a} = 1$ as long as $a \neq 0$. You already know this fact, don't you? If you do, then apply it and you're done. There's nothing mysterious. I'm not trying to be condescending or anything like that, so I apologize if that's how it sounded. The theorem in Hurkyl's answer is the one that I mentioned in my answer, though I didn't state it in the same fashion. – Graphth Sep 15 '12 at 14:05
@Henry Can you tell me at what step you don't understand? Do you know that $\frac{a}{a} = 1$ as long as $a \neq 1$? If so, then the function $\frac{7(x-11)}{x-11}$ must be $7$ as long as $x \neq 11$, right? $\frac{7(x - 11)}{x - 11}$ isn't anything special, it's almost exactly like $\frac{a}{a}$. Instead of plugging in any $a \neq 0$, you plug in any $x \neq 11$. So, if $\frac{7(x - 11)}{x - 11}$ is $7$ as long as $x \neq 11$, and since a limit does not depend on the value of the function at the point, only those around it, the limit is the same as $\lim_{x \to 11} 7 = 7$. – Graphth Sep 15 '12 at 18:10

Theorem: Suppose that $f(x) = g(x)$ for all $x \neq a$. Then, if either limit in the equation $$\lim_{x \to a} f(x) = \lim_{x \to a} g(x)$$ exists, then they both exist, and the equation holds.

This is the theorem you are implicitly using when you write an equation like

$$\lim_{x \to 0} \frac{x}{x} = \lim_{x \to 0} 1$$

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thank you, this is definitely pointing me in a better direction. I'm struggling with 'it just works' and this helped me understand why. The point seems to be proving that this concept of f(x) = g(x) for all x != a as x->a is applicable to the changes you make to the original limit function. Thanks again. – Henry Sep 15 '12 at 6:48

I think a good example is the following function which Spivak presents in his chapter devoted to Limits.

Define $f:(0,1)\to\mathbb R$ as $$f(x)=\begin{cases} 0\text{ ; if }x \text{ is irrational}\\\frac 1 p \text{ ; if } x=\frac q p \text{ an irreducible fraction}\end{cases}$$

This function is very "strange". You can try plotting it for some values on $[0,1]$. Try taking $q=1,2,3,\dots$ to make it easier. The functions has a nice pattern.

What we want to prove is that, for every $a\in(0,1)$, we have $$\lim_{x\to a}f(x)=0$$

from where we'll see it is only continuous at the irrational points. So, let's pick any $a\in(0,1)$, and let us be given $\epsilon >0$. Choose $n$ so that $1/n\leq \epsilon$.

First, we note that the only points where it might be false that $|f(x)-0|<\epsilon$ are $$\frac{1}{2};\frac{1}{3},\frac{2}{3};\frac{1}{4},\frac{3}{4};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}; \cdots ;\frac{1}{n}, \cdots ,\frac{{n - 1}}{n}$$ If $a$ is rational, then $a$ could be one of those numbers. But, as many as there can be, the amount of these numbers are finite. Thus, for some $p/q$ in that list, the number $$|a-p/q|$$ is least. If $a$ is one of these numbers, consider $p/q\neq a$. Then, take $\delta$ as this distance. It will then be the case that if $0<|x-a|<\delta$, then $x$ will be none of the numbers

$$\frac{1}{2};\frac{1}{3},\frac{2}{3};\frac{1}{4},\frac{3}{4};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}; \cdots ;\frac{1}{n}, \cdots ,\frac{{n - 1}}{n}$$

thus it will be true that $|f(x)-0|<\epsilon$.

Another example, closely related to this one, is:

For each $n$, let $A_n$ be a finite subset of $(0,1)$. That is, each $A_n$ is a set consisting of numbers with $0<x<1$, and let them be such that if $n\neq m$, $A_n$ and $A_m$ have no elements in common. We define then $f$ as

$$f(x)=\begin{cases} 0\text{ ; if }x \text{ is not in any }A_n \\\frac 1 n \text{ ; if } x\in A_n\end{cases}$$

Now, we want to prove again that, for every $a\in(0,1)$, we have $$\lim_{x\to a}f(x)=0$$

Now, if $\lim\limits_{x\to a}f(x)=0$, this means that for every $\epsilon >0$ there is a $\delta >0$ such that $0<|x-a|<\delta$ implies $|f(x)|<\epsilon$. Since $A_1$ is finite, there must exist a $\delta_1>0$ such that no $x\in A_1$ is in $(a-\delta_1,a+\delta_1)-\{a\}$. Thus, we must have $|f(x)|\leq 1/2$ for any $x$ with $0<|x-a|<\delta_1$. Similarily there exists a $\delta_2>0$ such that no $x\in A_2$ is in $(a-\delta_2,a+\delta_2)-\{a\}$ from which it is $|f(x)|\leq 1/3$ in $0<|x-a|<\delta_2$. Analogously, there exists a $\delta_n >0$ such that no $x\in A_n$ is in $(a-\delta_n,a+\delta_n)-\{a\}$ from where $|f(x)|\leq 1/n$ for $0<|x-a|<\delta_n$. We're ready to finish it off. Let $\epsilon >0$ be given and choose $n_0$ such that $1/n_0\leq \epsilon$. Let $n\geq n_0$ and $\delta =\min\{\delta_1,\dots,\delta_n\}$. Then, whenever $0<|x-a|<\delta$, we must have $|f(x)|<\epsilon$, from where it must be $$\lim_{x\to a}f(x)=0$$ for every $a\in(0,1)$.

Note how in this two examples, the finiteness of the sets in the second case and of the "bad" numbers in the first was key. Note also that we really have no explicity formula for $f$ that allows any algebraic manipulation whatsoever. (We can actually put $$f\left( x \right) = \sum\limits_{n \in {\Bbb N}} {\frac{1}{n}{\chi _{{A_n}}}\left( x \right)}$$ in the second case, but not much is revealed).

Let alone is the recommendation that you read Spivaks exposition, if you can.

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You'll need to know some basic topology.

A limit is a point that an infinite sequence converges to. In other words, if we have an infinite sequence $\{p_n\}$, and a limit point $p$, then we can draw a neighborhood (a ball) around $p$ such that every $p_n$ in the sequence beyond some point $p_m$ is inside that ball; in other words, only finitely many points are outside that ball.

So that means that in some functions, even arbitrary "boring" points are limits. The limit of $f(x)= x$ going to $x=4$ is just 4. Nothing interesting about it.

Finding the limit using the tools taught in basic calculus is essentially a less-detailed way to show that you can construct an infinite sequence $\{f(x_n)\}$ such that as $n \to \infty$, $f(x_n) \to y$. These tools that you learn in basic calculus insulate you from having to actually construct and analyze that infinite sequence. The details of how they work are just hidden from you.

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You probably want to correct that second paragraph. It’s not enough that each open ball contain infinitely many terms of the sequence: consider $\langle 0,1,0,1,\dots\rangle$. – Brian M. Scott Sep 15 '12 at 2:43
@BrianM.Scott Thanks. I updated it; hopefully, I kept the explanation informal enough. – Arkamis Sep 15 '12 at 2:52
The original version is still a little misleading; why not say that all but finitely many of the $p_n$ are inside that nbhd, or simply that all of them from some point on are inside it? – Brian M. Scott Sep 15 '12 at 2:55

The examples you see when first encountering limits are chosen to yield to these simple algebraic techniques. Later on you may encounter limits that are much more challenging.

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