The Limit Definition

Having discussed how you can compute limits, I want to examine the definition of a limit in more detail.

You might wonder why it is necessary to be careful. Suppose you're trying to compute $\displaystyle \lim_{x\to 0} \dfrac{1 - \cos (x^8)}{x^{16}}$ . You might think of drawing a graph; many graphing calculators, for instance, produce a graph like the one below:

$$\hbox{\epsfysize=2in \epsffile{limit-definition-1.eps}}$$

It looks as though the graph is dropping down to 0 near $x = 0$ . From this, you might guess that the limit is 0. In fact,

$$\lim_{x\to 0} \dfrac{1 - \cos (x^8)}{x^{16}} = \dfrac{1}{2}.$$

It's possible to justify this algebraically once you know a little about limits of trig functions.

Pictures can be helpful; so can experimenting with numbers. In many cases, pictures and numerical experiments are inconclusive or even misleading. In these cases, how can you determine whether a proposed answer is correct or not?

Because the limit definition is a bit abstract, I'll start off with an informal definition.

Informal Definition. If $f(x)$ can be made arbitrarily close to L for all x's sufficiently close to c, then

$$\lim_{x \to c} f(x) = L.$$

This statement is like a guarantee. Think of making parts in a factory. Your customers won't buy your parts unless they meet certain specifications. So you might guarantee that your parts will be within 0.01 of the customer's specification.

Likewise, to say that $\displaystyle \lim_{x \to c} f(x) = L$ you must be able to guarantee that you can make $f(x)$ fall within (say) 0.01 of L. But you have to do more: You have to be able to make $f(x)$ fall within any positive tolerance of L --- 0.0001, 0.0000004, and so on, no matter how small.

Another way to think of this is as meeting a challenge; for example:

Challenge: "I challenge you to make $f(x)$ stay within 0.0005 of L."

Your response: "I guarantee that every x within 0.003 of c (except perhaps c itself) will give an $f(x)$ that is within 0.0005 of L."

To prove that $\displaystyle \lim_{x \to c} f(x) = L$ , you must be able to meet the challenge no matter what positive number is used in place of 0.0005.

By the way, notice that $x = c$ is excluded in my guarantee. The reason is that in computing $\displaystyle \lim_{x \to c} f(x)$ , we're concerned with what happens as x approaches c, not what $f(c)$ is.

Before I give some examples, here's an important fact about absolute value:

$$|A - B| = (\hbox{the distance from A to B}).$$

$$\hbox{\epsfxsize=2in \epsffile{limit-definition-2.eps}}$$

We want absolute values, which are always nonnegative, because a distance shouldn't be negative.

Also, notice that

$$|A - B| = |B - A|.$$

That makes sense, because the distance from A to B should be the same as the distance from B to A. For instance,

$$|8 - 2| = |6| = 6 \quad\hbox{and}\quad |2 - 8| = |-6| = 6.$$

Example. By plugging in $x = 4$ , it appears that

$$\lim_{x \to 4} (3 x - 5) = 7.$$

How close should x be to 4 to guarantee that $3 x - 5$ is within 0.01 of 7?

Let's work backwards: I want $3 x - 5$ to be within 0.01 of 7. This means

$$\eqalign{ |(3 x - 5) - 7| & < 0.01 \cr |3 x - 12| & < 0.01 \cr 3 |x - 4| & < 0.01 \cr |x - 4| & < \dfrac{0.01}{3} \cr}$$

The last inequality says that the distance from x to 4 should be less than $\dfrac{0.01}{3}$ . So if x lies within $\dfrac{0.01}{3}$ of 4, I can guarantee that $3 x - 5$ will be within 0.01 of 7.

A formal proof would just reverse the steps above:

$$\eqalign{ |x - 4| & < \dfrac{0.01}{3} \cr 3 |x - 4| & < 0.01 \cr |3 x - 12| & < 0.01 \cr |(3 x - 5) - 7| & < 0.01 \cr}$$

Can you see that if I'm challenged to make $3 x - 5$ lie within 0.00001 of 7, I should make x lie within $\dfrac{0.00001}{3}$ of 4? Just replace 0.01 with 0.00001 in the discussion above.

And similarly, I can make $3 x - 5$ lie within any tolerance FOO of 7 by making x lie within $\dfrac{\hbox{FOO}}{3}$ of 4.

This shows that I can meet any challenge, since I can just take the challenge tolerance and plug it in for FOO. This proves that

$$\lim_{x \to 4} (3 x - 5) = 7.\quad\halmos$$

Example. The graph of a function $y = f(x)$ is shown below.

$$\hbox{\epsfysize=2in \epsffile{limit-definition-3a.eps}}$$

It appears that $\displaystyle \lim_{x \to 4} f(x) = 3$ .

A (grey) horizontal strip of width 0.5 is drawn around $y = 3$ . Draw a picture to show a range of x-values around 4 for which the corresponding $f(x)$ -values lie in the horizontal strip.

Use it to estimate the width of a symmetric vertical strip around 4 representing x-values whose corresponding $f(x)$ -values lie in the horizontal strip.

Suppose I'm challenged to make $f(x)$ fall within 0.5 of 3. That is, I want my y-values to fall within the grey strip in the picture.

On the right side of 4, the graph stays within the grey strip as far as 4.25; on the left side of 4, the graph stays within the grey strip as far as 3.

$$\hbox{\epsfysize=2in \epsffile{limit-definition-3b.eps}}$$

If I want a strip that's symmetric about 4, I use the closer of the two values, which is 4.25. Now 4.25 is 0.25 units from 4, so my answer is: If x is within 0.25 of 4, then $f(x)$ will be within 0.5 of 3.

If I can meet such a challenge with any positive number in place of 0.5, then I will have proved that $\displaystyle \lim_{x \to 4} f(x) = 3$ .

Example. (Disproving a limit) Consider the function $y = f(x)$ whose graph is show below.

$$\hbox{\epsfysize=2in \epsffile{limit-definition-4a.eps}}$$

Suppose that Calvin Butterball thinks that $\displaystyle \lim_{x \to 3} f(x) = 4$ . Use the limit definition to disprove it.

To disprove Calvin's claim, I'll make a challenge that Calvin can't meet.

I challenge Calvin to make $f(x)$ fall within 0.5 of 4. This means that he must find a range of x's around 3 so that the corresponding part of the graph lies within the grey strip shown below:

$$\hbox{\epsfysize=2in \epsffile{limit-definition-4b.eps}}$$

You can see that there's no way to do this. (Note: He's not allowed to use $x = 3$ alone. Remember that what the function does at $x = 3$ has no bearing on the value of the limit.)

Since this challenge can't be met, $\displaystyle \lim_{x \to 3} f(x) \ne 4$ . In fact, $\displaystyle \lim_{x \to 3} f(x)$ is undefined.

Example. Suppose

$$f(x) = \cases{ 5 - 2 x & if $x < 1$ \cr 4 x - 1 & if $x \ge 1$ \cr}.$$

It appears that $\displaystyle \lim_{x \to 1} f(x) = 3$ . How close should x be to 1 in order to guarantee that $f(x)$ will be within 0.0008 of 3?

As in an earlier example, I'll work backwards.

From the left side, I'd need

$$\eqalign{ |(5 - 2 x) - 3| & < 0.0008 \cr |2 - 2 x| & < 0.0008 \cr |2 x - 2| & < 0.0008 \cr |x - 1| & < 0.0004 \cr}$$

The last inequality says that x should be within 0.0004 of 1.

From the right side, I'd need

$$\eqalign{ |(4 x - 1) - 3| & < 0.0008 \cr |4 x - 4| & < 0.0008 \cr |x - 1| & < 0.0002 \cr}$$

This means that x should be within 0.0002 of 1.

To satisfy the two requirements at the same time, I'll use the smaller of the two numbers. So I'll require that x should be within 0.0002 of 1, which means

$$|x - 1| < 0.0002.$$

Here is the "real" proof, which I get by writing the scratch work in the reverse order.

Suppose $|x - 1| < 0.0002$ . If $x \ge 1$ , I have

$$\eqalign{ |x - 1| & < 0.0002 \cr |4 x - 4| & < 0.0008 \cr |(4 x - 1) - 3| & < 0.0008 \cr |f(x) - 3| & < 0.0008 \cr}$$

Now

$$|x - 1| < 0.0002 < 0.0004.$$

So if $x < 1$ , I have

$$\eqalign{ |x - 1| & < 0.0004 \cr |2 x - 2| & < 0.0008 \cr |2 - 2 x| & < 0.0008 \cr |(5 - 2 x) - 3| & < 0.0008 \cr |f(x) - 3| & < 0.0008 \cr}$$

(From the second to the third line, I used the fact that $|A - B| = |B - A|$ .)

Thus, if x is within 0.0002 of 1, then $f(x)$ will be within 0.0008 of 3.

I'm almost ready to give the formal definition of a limit, but I need to mention something first as a matter of honesty. It's a technical issue, and it won't arise in the majority of problems and examples (so you can ignore it without much harm if you wish).

A technical point. In discussing $\displaystyle \lim_{x \to c} f(x)$ , I'll usually assume that f is defined on an open interval containing c. That is, there are numbers a and b such that $a < c < b$ and f is defined (at least) on $a < x < b$ .

For one-sided limit (which I'll discuss later), $f(x)$ should be defined on an open interval with c as an endpoint.

To understand why you want to do this, consider the function

$$f(x) = \cases{ \ln x & if $x > 0$ \cr 42 & if $x = -10$ \cr}.$$

(So, for instance, f is simply not defined at $x = -1$ , or at $x = -57$ .)

In the definition of $\displaystyle \lim_{x \to -10} f(x)$ , the "if" part of the definition would hold vacuously (for small open intervals around -10), because there would be no values of x near -10 for which f was defined. Thus, the limit L could be anything!

The condition on the domain of f is made to avoid silly cases like this one.

In order to avoid cluttering the statements of the definition or of proofs of limit properties, I usually won't state this assumption about the domains of functions in limits explicitly.

Now I'll give the formal definition of a limit, and show how to use it to do $\epsilon-\delta$ proofs.

Definition. $\displaystyle \lim_{x \to c} f(x) = L$ means:

For every $\epsilon > 0$ , there is a $\delta$ , such that for all x in the domain of f, if $\delta > |x - c| > 0$ , then $\epsilon > |f(x) - L|$ .

"$\epsilon$ " is the Greek letter epsilon. It is the "challenge number", the tolerance or maximum error you have to meet. $\delta$ is the Greek letter delta. It is the "response number", the setting on x which meets the challenge. The Greek letters are used in this definition for traditional reaons; there is nothing otherwise special about them.

Let's see how proofs of limits work using the definition.

Example. Prove that $\displaystyle \lim_{x \to 4} (7 x - 3) = 25$ .

In this problem, 4 corresponds to c, $7 x - 3$ corresponds to $f(x)$ , and 25 corresponds to L in the limit definition.

I have to show that, given any $\epsilon > 0$ , there is a $\delta$ , such that

$$\hbox{if}\quad \delta > |x - 4| > 0, \quad\hbox{then}\quad \epsilon > |(7 x - 3) - 25|.$$

Notice that I'm given $\epsilon$ , but I'm not told its value (which was the case in earlier examples). All I can assume is that it's some positive number. I have to come up with a $\delta$ that meets the condition above. To do this, I work backwards as I did in earlier examples. This is "scratchwork", and doesn't count as the "real" proof, which will come afterward.

Scratchwork. I want $\epsilon > |(7 x - 3) - 25|$ . I'll work backwards from this and try to get something that looks like "$(\hbox{whatever}) > |x - 4|$ ". Then I'll set $\delta = (\hbox{whatever})$ and try to do the real proof.

$$\eqalign{ \epsilon & > |(7 x - 3) - 25| \cr \epsilon & > |7 x - 28| \cr \epsilon & > 7 |x - 4| \cr \noalign{\vskip2 pt} \dfrac{\epsilon}{7} & > |x - 4| \cr}$$

Okay --- I'll try $\delta = \dfrac{\epsilon}{7}$ .

The real proof. Let $\delta = \dfrac{\epsilon}{7}$ . I must show that

$$\hbox{if}\quad \delta > |x - 4| > 0, \quad\hbox{then}\quad \epsilon > |(7 x - 3) - 25|.$$

When you are proving an "if-then" statement, you get to assume the "if" part, and you prove the "then" part. So assume

$$\dfrac{\epsilon}{7} = \delta > |x - 4|.$$

The rest of the proof is easy: I just reverse the steps I did on scratchwork:

$$\eqalign{ \dfrac{\epsilon}{7} & > |x - 4| \cr \noalign{\vskip2 pt} \epsilon & > 7 |x - 4| \cr \epsilon & > |7 x - 28| \cr \epsilon & > |(7 x - 3) - 25| \cr}$$

Therefore, by the limit definition,

$$\lim_{x \to 4} (7 x - 3) = 25.\quad\halmos$$

A similar approach works for limits of the form $\displaystyle \lim_{x \to c} (a x + b)$ . Here is a harder example.

Example. Prove that $\displaystyle \lim_{x \to 3} (x^2 + 5 x) = 24$ .

In this case, 3 corresponds to c, $x^2 + 5 x$ corresponds to $f(x)$ , and 24 corresponds to L.

Scratchwork. I want $\epsilon > |(x^2 + 5 x) - 24|$ . I'll work backwards from this and try to get something that looks like "$(\hbox{whatever}) > |x - 3|$ ". Then I'll set $\delta = (\hbox{whatever})$ and try to do the real proof.

$$\eqalign{ \epsilon & > |(x^2 + 5 x) - 24| \cr \epsilon & > |x^2 + 5 x - 24| \cr \epsilon & > |(x + 8)(x - 3)| \cr \epsilon & > |x + 8| |x - 3| \cr}$$

I can't just divide both sides by $|x + 8|$ (like I divided by 7 in the last example:

$$\dfrac{\epsilon}{|x + 8|} > |x - 3|.$$

The problem is that I can't set $\delta = \dfrac{\epsilon}{|x + 8|}$ , because I would need to know x in order to know $\delta$ --- but $\delta$ is supposed to determine the range of x's.

Instead, I need to make a "preliminary" setting of $\delta$ . I'll provisionally set $\delta = 1$ . Then

$$\eqalign{ 1 = \delta & > |x - 3| \cr 2 < &\ x < 4 \cr}$$

$$\hbox{\epsfxsize=2 in \epsffile{limit-definition-5.eps}}$$

Remember that you have complete control over $\delta$ . Setting $\delta$ to 1 is like adjusting a setting on an instrument, where you make an initial rough setting, then fine-tune it. We'll see how this works out when we write the "real proof".

Adding 8 to each term, I get

$$\eqalign{ 2 < &\ x < 4 \cr 10 < &\ x + 8 < 12 \cr |x + 8| & < 12 \cr}$$

Remember that I want the inequality $\epsilon > |x + 8| |x - 3|$ .

If I could get $\epsilon > 12 |x - 3|$ , then I'd have

$$\eqalign{ 12 & > |x + 8| \cr 12 |x - 3| & > |x + 8| |x - 3| \cr \epsilon > 12 |x - 3| & > |x + 8| |x - 3| \cr}$$

But

$$\eqalign{ \epsilon & > 12 |x - 3| \cr \noalign{\vskip2 pt} \dfrac{\epsilon}{12} & > |x - 3| \cr}$$

It looks like I should try $\delta = \dfrac{\epsilon}{12}$ ... but then, I remember I needed to set $\delta = 1$ earlier. How can I get both of these things to happen? The idea is to make $\delta$ the smaller of the two numbers 1 and $\dfrac{\epsilon}{12}$ --- in symbols,

$$\delta = \min \left(1, \dfrac{\epsilon}{12}\right).$$

("min" stands for "minimum".) This means that

$$1 \ge \delta \quad\hbox{and}\quad \dfrac{\epsilon}{12} \ge \delta.$$

The real proof. Let $\delta = \min \left(1, \dfrac{\epsilon}{12}\right)$ . I must show that:

$$\hbox{if}\quad \delta > |x - 3| > 0, \quad\hbox{then}\quad \epsilon > |(x^2 + 5 x) - 24|.$$

So I may assume $\delta > |x - 3| > 0$ , and I have to prove $\epsilon > |(x^2 + 5 x) - 24|$ .

As I noted in the scratchwork, I know that

$$1 \ge \delta \quad\hbox{and}\quad \dfrac{\epsilon}{12} \ge \delta.$$

Take $1 \ge \delta$ first. Then

$$\eqalign{ 1 \ge \delta & > |x - 3| \cr 4 > & x > 2 \cr 12 > & x + 8 > 10 \cr 12 & > |x + 8| \cr}$$

Next, I'll use $\dfrac{\epsilon}{12} \ge \delta$ . Multiply this inequality and the inequality $12 > |x + 8|$ to get

$$\eqalign{ \epsilon = 12 \cdot \dfrac{\epsilon}{12} & > |x + 8| |x - 3| \cr \epsilon & > |x^2 + 5 x - 24| \cr \epsilon & > |(x^2 + 5 x) - 24| \cr}$$

Therefore, I've proved that $\displaystyle \lim_{x \to 3} (x^2 + 5 x) = 24$ .

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