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 limx → 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:
limx → 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,
limx → 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:
limx → 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 limx → af(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) = x3 – 5x + 6 is a polynomial, we know from the Direct Substitution Property that
limx→1 f(x) = f(1) = 13 – 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 |(x3 – 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 = x3 – 5x + 6 lies between the horizontal lines y = 1.8 and y = 2.2.
Therefore we graph the curves y = x3 – 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 = x3 – 5x + 6 is about 0.911.
Similarly, y = x3 – 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 < x3 – 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 |(x3 – 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 | < 4δ = = 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 = ε2. If 0 < x < δ, then
so | – 0 | < ε
According to Definition 4, this shows that
cont’d
The Precise Definition of a Limit
If limx → a f(x) = L and limx → 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/x2 > M But
