## The Precise Definition of a Limit

The intuitive definition of a limit is inadequate for some
*purposes because such phrases as “x is close to 2” and *

*“f(x) gets closer and closer to L” are vague.*

In order to be able to prove conclusively that or

we must make the definition of a limit precise.

## The Precise Definition of a Limit

To motivate the precise definition of a limit, let’s consider the function

*Intuitively, it is clear that when x is close to 3 but x* ≠ 3, then
*f(x) is close to 5, and so lim*_{x}_{→ 3 }*f(x) = 5.*

*To obtain more detailed information about how f(x) varies *

*The distance from x to 3 is | x – 3 | and the distance from *
*f(x) to 5 is |f(x) – 5 |, so our problem is to find a number *δ
such that

*| f(x) – 5 | < 0.1 if |x – 3 | <* δ *but x* ≠ 3

*If |x – 3 | > 0, then x* ≠ 3, so an equivalent formulation of our
problem is to find a number δ such that

δ

## The Precise Definition of a Limit

## The Precise Definition of a Limit

*Notice that if 0 < | x – 3 | < (0.1)/2 = 0.05, then*

*| f(x) – 5 | = | (2x – 1) – 5 | = |2x – 6 | *

*= 2| x – 3 | < 2(0.05) = 0.1*
that is,

*| f(x) – 5 | < 0.1 if 0 < | x – 3 | < 0.05*

## The Precise Definition of a Limit

If we change the number 0.1 in our problem to the smaller
number 0.01, then by using the same method we find that
*f(x) will differ from 5 by less than 0.01 provided that x*

differs from 3 by less than (0.01)/2 = 0.005:

*| f(x) – 5 | < 0.01 if 0 < | x – 3 | < 0.005 *
Similarly,

*| f(x) – 5 | < 0.001 if 0 < | x – 3 | < 0.0005 *

## The Precise Definition of a Limit

*For 5 to be the precise limit of f(x) as x approaches 3, we *
*must not only be able to bring the difference between f(x) *
and 5 below each of these three numbers; we must be able
*to bring it below any positive number. *

And, by the same reasoning, we can! If we write *ε (the *

Greek letter epsilon) for an arbitrary positive number, then we find as before that

## The Precise Definition of a Limit

*This is a precise way of saying that f(x) is close to 5 when x*
is close to 3 because (1) says that we can make the values
*of f(x) within an arbitrary distance ε from 5 by restricting the *
*values of x to be within a distance ε/2 from 3 (but x ≠ 3).*

Note that (1) can be rewritten as follows: if

3 – δ *< x < 3 +* δ *(x* ≠ 3)
then

## The Precise Definition of a Limit

*By taking the values of x (≠ 3) to lie in the interval *

(3 – δ, 3 + δ*) we can make the values of f(x) lie in the *
interval (5 – *ε, 5 + ε).*

Using (1) as a model, we give a precise definition of a limit.

## The Precise Definition of a Limit

*Since | x – a | is the distance from x to a and | f(x) – L | is the *
*distance from f(x) to L, and since ε can be arbitrarily small, *
the definition of a limit can be expressed in words as

follows:

lim_{x}_{→ a}*f(x) = L*

*means that the distance between f(x) and L can be made*
*arbitrarily small by requiring that the distance from x to a be *
sufficiently small (but not 0).

## The Precise Definition of a Limit

Alternatively,

lim_{x}_{→ a}*f(x) = L*

*means that the values of f(x) can be made as close as we *
*please to L by requiring x to be close enough to a (but not *
*equal to a).*

## The Precise Definition of a Limit

We can also reformulate Definition 2 in terms of intervals
*by observing that the inequality |x – a | < *δ is equivalent
to –δ *< x – a < *δ, which in turn can be written as

*a –* δ *< x < a + *δ.

*Also 0 < | x – a | is true if and only if x – a* *≠ 0, that is, x ≠ a. *

## The Precise Definition of a Limit

*Similarly, the inequality | f(x) – L | < ε is equivalent to the *
*pair of inequalities L –* *ε < f(x) < L + ε. Therefore, in terms *
of intervals, Definition 2 can be stated as follows:

lim_{x}_{→ a}*f(x) = L*

means that for every *ε > 0 (no matter how small ε is) we *
can find δ *> 0 such that if x lies in the open interval *

*(a –* δ*, a + *δ*) and x* *≠ a, then f(x) lies in the open interval *
*(L –* *ε, L + ε).*

## The Precise Definition of a Limit

We interpret this statement geometrically by representing a
*function by an arrow diagram as in Figure 2, where f maps *
a subset of onto another subset of .

**Figure 2**

## The Precise Definition of a Limit

The definition of limit says that if any small interval

*(L –* *ε, L + ε) is given around L, then we can find an interval *
*(a –* δ*, a + *δ*) around a such that f maps all the points in *

*(a –* δ*, a + *δ*) (except possibly a) into the interval *
*(L –* *ε, L + ε). (See Figure 3.)*

## The Precise Definition of a Limit

Another geometric interpretation of limits can be given in
terms of the graph of a function. If *ε > 0 is given, then we *
*draw the horizontal lines y = L + ε and y = L – ε and the *
*graph of f. (See Figure 4.)*

## The Precise Definition of a Limit

If lim_{x}_{→ a}*f(x) = L, then we can find a number *δ > 0 such that
*if we restrict x to lie in the interval (a –* δ*, a + *δ) and take

*x* *≠ a, then the curve y = f(x) lies between the lines *

*y = L –* *ε and y = L + ε (See Figure 5.) You can see that if *
such a δ has been found, then any smaller δ will also work.

## The Precise Definition of a Limit

It is important to realize that the process illustrated in

*Figures 4 and 5 must work for every positive number ε, no *
matter how small it is chosen. Figure 6 shows that if a

smaller *ε is chosen, then a smaller* δ may be required.

## Example 1

*Since f(x) = x*^{3} *– 5x + 6 is a polynomial, we know from the *
Direct Substitution Property that

lim_{x→1}*f(x) = f(1) = 1*^{3} – 5(1) + 6 = 2.

Use a graph to find a number δ *such that if x is within *δ of
*1, then y is within 0.2 of 2, that is,*

*if | x – 1 | < *δ *then |(x*^{3} *– 5x + 6) – 2 | < 0.2*

In other words, find a number δ that corresponds to *ε = 0.2 *

*Example 1 – Solution*

*A graph of f is shown in Figure 7; we are interested in the *
region near the point (1, 2).

Notice that we can rewrite the inequality

**Figure 7**

*Example 1 – Solution*

*So we need to determine the values of x for which the *
*curve y = x*^{3} *– 5x + 6 lies between the horizontal lines *
*y = 1.8 and y = 2.2.*

*Therefore we graph the curves y = x*^{3} *– 5x + 6, y = 1.8, and *
*y = 2.2 near the point (1, 2) in Figure 8.*

cont’d

*Example 1 – Solution*

*Then we use the cursor to estimate that the x-coordinate of *
*the point of intersection of the line y = 2.2 and the curve *
*y = x*^{3} *– 5x + 6 is about 0.911. *

*Similarly, y = x*^{3} *– 5x + 6 intersects the line y = 1.8 when *
*x* ≈ 1.124. So, rounding toward 1 to be safe, we can say
that

*if 0.92 < x < 1.12 then 1.8 < x*^{3} *– 5x + 6 < 2.2*

cont’d

*Example 1 – Solution*

We can choose δ to be the smaller of these numbers, that is, δ = 0.08.

Then we can rewrite our inequalities in terms of distances as follows:

*if | x – 1 | < 0.08 then |(x*^{3} *– 5x + 6) – 2 | < 0.2 *
*This just says that by keeping x within 0.08 of 1, we are *
*able to keep f(x) within 0.2 of 2.*

cont’d

## Example 2

Prove that

Solution:

**1. Preliminary analysis of the problem (guessing a value ***for *δ).

Let *ε be a given positive number. We want to find *
a number δ such that

*if 0 < | x – 3 | < *δ *then |(4x – 5) – 7 | < ε*

*Example 2 – Solution*

Therefore we want δ such that

*if 0 < | x – 3 | < *δ *then 4| x – 3 | < ε*
*that is, if 0 < | x – 3 | < *δ *then |x – 3 | < *

This suggests that we should choose δ = *ε/4.*

cont’d

*Example 2 – Solution*

* 2. Proof (showing that this *δ

*works). Given ε > 0, choose*δ =

*ε/4. If 0 < |x – 3| <*δ, then

*| (4x – 5) – 7 | = | 4x – 12 | = 4| x – 3 | < 4*δ = = *ε*
Thus

*if 0 < | x – 3 | < *δ *then | (4x – 5) – 7 | < ε*

cont’d

*Example 2 – Solution*

Therefore, by the definition of a limit,

This example is illustrated by Figure 9.

cont’d

## The Precise Definition of a Limit

The intuitive definitions of one-sided limits can be precisely reformulated as follows.

## The Precise Definition of a Limit

## Example 3

Use Definition 4 to prove that

*Example 3 – Solution*

* 1. Guessing a value for *δ. Let

*ε be a given positive number.*

*Here a = 0 and L = 0, so we want to find a number *δ
such that

*if 0 < x < *δ then | – 0 | < *ε*
that is,

*if 0 < x < *δ then < *ε*

or, squaring both sides of the inequality < *ε, we get*

*Example 3 – Solution*

* 2. Showing that this *δ

*works. Given ε > 0, let*δ =

*ε*

^{2}. If

*0 < x <*δ, then

so | – 0 | < *ε*

According to Definition 4, this shows that

cont’d

## The Precise Definition of a Limit

If lim_{x}_{→ a}*f(x) = L and lim*_{x}_{→ a}*g(x) = M both exist, then *

## Infinite Limits

## Infinite Limits

Infinite limits can also be defined in a precise way.

## Infinite Limits

*This says that the values of f(x) can be made arbitrarily large *
*(larger than any given number M) by requiring x to be close *
*enough to a (within a distance *δ, where δ *depends on M, but *
*with x* *≠ a). A geometric illustration is shown in Figure 10.*

## Infinite Limits

*Given any horizontal line y = M, we can find a number *
δ *> 0 such that if we restrict x to lie in the interval*

*(a –* δ*, a + *δ*) but x* *≠ a, then the curve y = f(x) lies above *
*the line y = M.*

*You can see that if a larger M is chosen, then a smaller *δ
may be required.

## Example 5

Use Definition 6 to prove that

Solution:

Let M be a given positive number. We want to find a number δ such that

*if 0 < | x – 0 | < *δ *then 1/x*^{2} *> M*
But