A Formal Definition of Limits (Optional)

Section 3.5 Problems

3.6 A Formal Definition of Limits (Optional)

The ancient Greeks used limiting procedures to compute areas, such as the area of a circle, by the “method of exhaustion.” In this method, a region was covered (or

“exhausted”) as closely as possible by triangles. Adding the areas of the triangles then yielded an approximation of the area of the region of interest. Newton and Leibniz, the inventors of calculus, were aware of the importance of taking limits in their development of the subject; however, they did not give a rigorous definition of the procedure. The French mathematician Augustin-Louis Cauchy (1789–1857) was the first to develop a rigorous definition of limits; the definition we will use goes back to the German mathematician Karl Weierstrass (1815–1897).

Before we write the formal definition, let’s return to the informal one. In that definition, we stated that lim_x→c f(x) = L means that the value of f(x)^{can be} made arbitrarily close toLwheneverxis sufficiently close toc. But just how close is sufficient? Take Example 1 from Section 3.1: Suppose we wish to show that

lim

x→2

x² =4

without using the continuity of y = x², which itself was based on lim_x→cx = c [Equation (3.3)]. What would we have to do? We would need to show thatx²can be made arbitrarily close to 4 for all values ofxsufficiently close, but not equal, to 2. (In what follows, we will always excludex =2 from the discussion, since the value ofx² atx = 2 is irrelevant in finding the limit.) Suppose we wish to makex²within 0.01 of 4; that is, we want|x²−⁴| <⁰.01. Does this inequality hold for allxsufficiently close, but not equal, to 2? We begin with

|x²−⁴|<⁰.⁰¹

which is equivalent to

−⁰.⁰¹<x²−⁴<⁰.⁰¹ 3.⁹⁹<x²<⁴.⁰¹ 3.⁹⁹<|x|<

4.⁰¹ Now,

3.99=1.997498. . . and

4.01=2.002498. . .. We therefore find that values ofx =2 in the interval(¹.⁹⁹⁸,².⁰⁰²)^satisfy|x²−⁴|<⁰.01. (We chose a somewhat smaller interval than indicated, to get an interval that is symmetric about 2.) That is, for all values ofx within 0.002 of 2 but not equal to 2 (i.e., 0<|x−²|<⁰.^002),x² is within the prescribed precision—that is, within 0.01 of 4.

You might think about this example in the following way: Suppose that you wish to stake out a square of area 4 m². Each side of your square is 2 m long. You bring along a stick, which you cut to a length of 2 m. We can then ask: How accurately do we need to cut the stick so that the area will be within a prescribed precision? Our prescribed precision was 0.01, and we found that if we cut the stick within 0.002 of 2 m, we would be able to obtain the prescribed precision.

There is nothing special about 0.01; we could have chosen any other degree of precision and would have found a corresponding interval ofx-values. We translate this procedure into a formal definition of limits. (See Figure 3.24.)

f(x)

d d

x y

c L

Figure 3.24 The –δdefinition of limits.

Definition The statement

x→clim f(x)= L

means that, for every >0, there exists a numberδ >0 such that

|f(x)−L|< ^{whenever 0}<|x−c|< δ

Note that, as in the informal definition of limits, we exclude the valuex=cfrom the statement. (This is done in the inequality 0<|x−c|.) To apply the formal definition, we first need to guess the limiting valueL. We then choose an >0, the prescribed precision, and try to find aδ >0 such that f(x)^{is within of}Lwheneverxis within δ^ofcbut not equal toc. [In our example, f(x)= x²,c =^2, L = ^4, = ⁰.^{01, and} δ=⁰.^002.]

EXAMPLE 1 Show that

lim

x→1

(2x−3)= −1

Solution

We let f(x)=²x−3. Our guess for the limiting value isL = −^{1. Then}

|f(x)−L| = |²x−³−(−¹)|

= |²x−²|

=2|x−1|

We now choose >^{0. (} is arbitrary, and we do not specify it because our statement needs to hold for all > 0.) Our goal is to find a δ > 0 such that 2|x −¹| <

wheneverxis withinδof 1 but not equal to 1; that is, 0<|x−¹|< δ. The value ofδ will typically depend on our choice of . Since|x−1|< δimplies that 2|x−1|<2δ, we should try 2δ= . If we chooseδ= /2, then, indeed,

|f(x)−L| =²|x−¹|<²δ=² 2 =

This means that, for every >0, we can find a numberδ >^{0 (namely,}δ= /^{2) such} that

|f(x)−(−¹)|< ^{whenever 0}<|x−¹|< δ But this is exactly the definition of

lim

x→1(2x−3)= −1 EXAMPLE 2 We promised in Section 3.2 that we would show that

x→climx =c

Solution

Let f(x)= x. We need to show that, for every > 0, there corresponds a number δ >0 such that

|x−c|< ^{whenever 0}<|x−c|< δ ^(3.9) This immediately suggests that we should chooseδ = , and, indeed, ifδ = ^{, then} (3.9) holds.

Let’s look at an example in which f(x)is not linear.

EXAMPLE 3 Use the formal definition of limits to show that lim

x→0

x³ =0

Solution

We need to show that, for every >0, there corresponds a numberδ >0 such that

|x³|< whenever 0<|x|< δ (3.10) Now,|x³|< is equivalent to

− <x³<

− ^1/3<x < ^1/3

This pair of inequalities suggests that we setδ= ^1/3. Accordingly, if 0<|x|< ^1/3, then

− ^1/3<x < ^1/3 or

− <x³<

which is the same as|x³|< .

We can also use the formal definition to show that a limit does not exist.

EXAMPLE 4 Show that

lim

x→0

|x| x does not exist.

4 2 2 4

2 1

2 ^x

x

y

x

Figure 3.25 The graph of f(x)= ^|x|_x in Example 4: The limit of^|x|_x asxtends to 0 does not exist.

Solution

Showing that this limit does not exist is tricky. (See Figure 3.25.) The approach is as follows: First, we set f(x)= |x|/x,x =0. Then we assume that the limit exists and try to find a contradiction.¹Suppose, then, that there exists anLsuch that

lim

x→0

|x| x = L

If we look at Figure 3.25, we see that if, for instance, we chooseL=1, then we cannot get close toLwhenx is less than 0. Similarly, we see that, for any value ofL, either the distance to+1 exceeds 1 or the distance to−1 exceeds 1. That is, regardless of the value ofL, if <1, we will not be able to find a value ofδsuch that if 0<|x| < δ^, then|f(x)−L|< ^{, since} f(x)takes on both the values+^{1 and}−^{1 for 0}<|x|< δ^. Therefore, lim_x→0^|x|_x does not exist.

In the previous section, we considered an example in which lim_x→c f(x)= ∞^. This statement can be made precise as well.

Definition The statement

x→clim f(x)= ∞

means that, for everyM >0, there exists aδ >0 such that f(x) > M whenever 0<|x−c|< δ

Similar definitions hold for the case when f(x)decreases without bound asx → cand for one-sided limits. We will not give definitions for all possible cases; rather, we illustrate how we would use such a definition.

(1)This approach is called “indirect proof” or “reductio ad absurdum.” We assume the opposite of what we wish to prove, and then we show that assuming the opposite leads to a contradiction. Therefore, what we originally sought to prove must be true.

EXAMPLE 5 Show that

lim

x→0

1 x² = ∞

1.5

2 1 0.5 0 0.5 1 1.5 2

1 x²

80 100

60 y

x M

d d

Figure 3.26 The function f(x)= _x¹2in Example 5: The limit of _x¹2asxtends to 0 does not exist.

Solution

The graph of f(x)=¹/x²,x =0, is shown in Figure 3.26. We fixM >^{0. (Again,} M is arbitrary, because our solution must hold for allM >0.) We need to find aδ > ⁰ such that f(x) > M whenever 0 < |x| < δ. (Note thatc = 0.) We start with the inequality f(x) > Mand try to determine how to chooseδ^{. We have}

1

x² >M is the same as x²< ¹ M Taking square roots on both sides, we find that

|x|< √¹ M This suggests that we should chooseδ = ¹/√

M. Let’s try that value: Given M >0, we chooseδ=1/√

M. If 0<|x|< δ, then x²< δ², ^{or 1}

x² > ¹ δ² =M That is, 1/x²> Mwhenever 0<|x|< δ=¹/√

M.

There is also a formal definition whenx→ ∞(and a similar one forx → −∞).

This definition is analogous to that in Chapter 2.

Definition The statement

lim

x→∞ f(x)= L

means that, for every >0, there exists anx₀>0 such that

|f(x)−L|< ^wheneverx >x₀

Note that, in the definition,x₀is a real number.

EXAMPLE 6 Show that

x→∞lim x 0.⁵+x =1

Solution

This limit is illustrated in Figure 3.27. You can see that f(x)= x/(⁰.⁵+x)^,x ≥^0, is in the strip of width 2 and centered at the limiting valueL = 1 for all values of x greater thanx₀. (We assume that 0< <1, since, when ≥1, the choicex₀= ¹ works.) We now determinex₀when <1. To do this, we try to solve

x

0.⁵+x −¹ <

for >0. This inequality is equivalent to the pair of inequalities

− < x

0.5+x −¹<

or, after adding 1 to all three parts, 1− < x

0.5+x <¹+

Since_0.5+^x _x <^{1 for}x>0, the right-hand inequality always holds. We therefore need only consider

1− < x 0.⁵+x

Because we are interested in the behavior of f(x)^as x → ∞, we need only look at large values ofx. Multiplying by 0.⁵+x (and noticing that we can assume that 0.⁵+x >0, because we letx → ∞), we obtain

(¹− )(⁰.⁵+x) <x Solving forxyields

(¹− )(⁰.⁵) <x−x(¹− ) (¹− )(⁰.⁵) < x

1− 2 <x For instance, if =⁰.1 (as in Figure 3.27), then

x > ⁰.9 0.² =4.5

That is, we would setx₀=4.5 and conclude that, forx >4.5,|f(x)−1|<0.1.

0 2 4 6 8 10

0 0.2 0.4 0.6 0.8 1 1.2 y

x x0

x 0.5 x

Figure 3.27 The function f(x)=_0.5^x_+x ^{in Example} 6: The limit of f(x)asxtends to infinity is 1.

More generally, we find that, for every 0< <1, there exists an x₀ = ¹−

2 such that

|f(x)−¹|< ^whenever x >x₀