The Limit Definition

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

Example. Why is it 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:

$$\hbox{\epsfysize=2in \epsffile{limdef1.eps}}$$

(I produced the graph using a program called Mathematica.)

It looks as though the graph is dropping down to 0 near $x = 0$ . 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; at best, they suggest rather than {\it prove}. Sometimes understanding requires more precision.


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

means that $f(x)$ can be made arbitrarily close to L for all x's sufficiently close to c.

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:

Challenge: "I challenge you to make $f(x)$ fall 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. Reason: As I noted earlier, in computing $\displaystyle \lim_{x\to c} f(x)$ , you only consider what happens as x approaches c, not what $f(c)$ is.

Example. You can see by plugging in that

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

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

Remember that

$$|\hbox{BLIP} - \hbox{BLAP}| = (\hbox{the distance from BLIP to BLAP}).$$

I want $3x - 5$ to be within 0.01 of 7. This means

$$|(3x - 5) - 7| < 0.01, \quad\hbox{or}\quad |3x - 12| < 0.01.$$


$$3|x - 4| < 0.01, \quad |x - 4| < \dfrac{0.01}{3}.$$

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 $3x - 5$ will be within 0.01 of 7.

Can you see that if I'm challenged to make $3x - 5$ 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 $3x - 5$ 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

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

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

$$\hbox{\epsfysize=2in \epsffile{limdef2a.eps}}$$

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

$$\hbox{\epsfysize=2in \epsffile{limdef2b.eps}}$$

I'll 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.

$$\hbox{\epsfysize=2in \epsffile{limdef3a.eps}}$$

Suppose that Calvin Butterball thinks that $\displaystyle \lim_{x\to 3} f(x) = 4$ . I can use the limit definition to disprove it; to do so, 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:

$$\hbox{\epsfysize=2in \epsffile{limdef3b.eps}}$$

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

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

It is true 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?

From the left side, I'd need

$$|(5 - 2x) - 3| < 0.0008, \quad\hbox{or}\quad |2 - 2x| < 0.0008.$$

This is the same as $|2x - 2| < 0.0008$ , or $|x - 1| < 0.0004$ . The last inequality says that x should be within 0.0004 of 1.

From the right side, I'd need

$$|(4x - 1) - 3| < 0.0008, \quad\hbox{or}\quad |4x - 4| < 0.0008.$$

This gives $|x - 1| < 0.0002$ , which 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. Thus, if x is within 0.0002 of 1, then $f(x)$ will be within 0.0008 of 3.

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