Limit Laws

Chapter 2 Review Problems

3.1 Limits

3.1.2 Limit Laws

p

sinx

y

x

1.5 1.5

2 1.5 1 1.5 2

Figure 3.9 The graph of f(x)=^sinπ_x in Example 8.

As the argument of the sine function goes to+∞^or−∞, the function values oscillate between−^{1 and}+1. Therefore, sin^π_x continues to oscillate between−^{1 and}+^{1 as} x →^0.

The behavior exhibited in Example 8 is calleddivergence by oscillation.

Pitfalls The next example is an interesting one that shows other limitations of using a calculator to compute limits.

EXAMPLE 9 Find

lim

x→0

x²+¹⁶−⁴ x²

Solution

The graph of f(x)= √

x²+16−4

x² ,x =0, in Figure 3.10 indicates that the limit exists.

So, on the basis of the graph, we conjecture that the limit is equal to 0.125. If, instead, we use a calculator to produce a table of values of f(x)close to 0, something strange

2 4 0.05

0.15 y

2 4 x

0.05 0.1

(0, 0.125)

x² 16 4 x²

Figure 3.10 The graph of f(x)in Example 9: Asxtends to 0, the function approaches 0.125.

seems to happen:

x 0.01 0.001 0.0001 0.00001 0.000001 0.0000001

f(x) 0.1249998 0.125 0.125 0.125 0.1 0

As we get closer to 0, we first find that f(x)gets closer to 0.125, but when we get very close to 0, f(x)seems to drop to 0. What is going on? First, before you worry too much, note that lim_x→0 f(x) = ⁰.125. In the next section, we will learn how to compute this limit without resorting to the (somewhat dubious) help of the calculator.

The strange behavior of the calculated values happens because, whenxis very small, the difference in the numerator is so close to 0 that the calculator can no longer accurately determine its value. The calculator can compute only a certain number of digits accurately, which is good enough for most cases. Here, however, we need greater accuracy. The same strange thing happens when you try to graph this function on a graphing calculator. When thex range of the viewing window is too small, the graph is no longer accurate. (Try, for instance,−⁰.⁰⁰⁰⁰¹≤x ≤⁰.^{00001 and}−⁰.⁰³≤ y≤⁰.15 as the range for the viewing window.)

At the end of this chapter, we will discuss how limits are formally defined. The formal definition is conceptually similar to the one we used to define limits of the form lim_n→∞an, but we will not use it to compute limits. As in Chapter 2, there are mathematical laws that will allow us to compute limits much more easily.

Limit Laws Suppose thatais a constant and that

x→clim f(x) ^{and lim}

x→cg(x) exist. Then the following rules hold:

1. lim

x→ca f(x)=alim

x→c f(x) 2. lim

x→c[f(x)+g(x)] =^lim

x→c

f(x)+^lim

x→c

g(x) 3. lim

x→c[f(x)·g(x)] =^lim

x→c f(x)·^lim

x→cg(x) 4. lim

x→c

f(x) g(x) =

x→clim f(x)

x→climg(x) provided that lim

x→cg(x)=⁰

You are probably easily convinced that lim

x→c

x =c (3.3)

In Section 3.6, we will use the formal definition of limits to show that this equation is true. For now, we accept (3.3) as a fact. Starting from that equation, we can use the limit laws to compute limits of polynomials and rational functions.

EXAMPLE 10 Find

lim

x→2[x³+⁴x−¹]

Solution

Using Rules 1 and 2, we see that this equation becomes lim

x→2

x³+4 lim

x→2

x−lim

x→21

provided that the individual limits exist. For the first term, we use Rule 3, lim

x→2x³ =

lim

x→2x lim

x→2x lim

x→2x

provided that lim_x→2x exists. From (3.3), it then follows that lim_x→2x = ^{2 and we}

find that

lim

x→2x lim

x→2x lim

x→2x

=(²)(²)(²)=⁸

To compute the second term, we use (3.3) again to obtain lim_x→2x =2. For the last term, we find that lim_x→21 = 1. Now that we have shown that the individual limits exist, we can use Rules 1 and 2 to evaluate

lim

x→2[x³+⁴x−¹] =^lim

x→2x³+^{4 lim}

x→2x−^lim

x→21=⁸+(⁴)(²)−¹=¹⁵ EXAMPLE 11 Find

lim

x→4

x²+1 x−3

Solution

Using Rule 4, we find that lim

x→4

x²+1

x−3 = ^limx→4(x²+1) lim_x→4(x−3)

provided that the limits in the numerator and denominator exist and the limit in the denominator is not equal to 0. Using Rules 2 and 3 in the numerator, we obtain

lim

x→4(x²+1)=

lim

x→4

x²

+

lim

x→41

=(4)(4)+1=17

Breaking up the limit of the sum in the numerator into a sum of limits is justified only after we have shown that the individual limits exist. Using Rules 1 and 2 in the denominator, we get

lim

x→4(x−3)= lim

x→4

x−lim

x→43=4−3=1

Again, using the limit laws is justified only after we have demonstrated that the individual limits exist. Since the limits in both the denominator and the numerator exist and the limit in the denominator is not equal to 0, we obtain

lim

x→4

x²+¹ x−³ = ¹⁷

1 =¹⁷

The computations in Examples 10 and 11 look somewhat awkward, and it appears that what we have done is plug 2 into the expressionx³+⁴x−1 in Example 10 and 4 into the expression^x_x²₋₃⁺¹ in Example 11, even though we emphasized in the informal definition of limits that we are not allowed to simply plugcinto f(x)^when computing lim_x→c f(x). But, in essence, we did the calculation

lim

x→2[x³+4x−1] =2³+(4)(2)−1=15 in Example 10 and the calculation

lim

x→4

x²+¹ x−³ = ¹⁷

1 =¹⁷ in Example 11.

Even though we made a point that we cannot simply substitute the valuecinto f(x)when we take the limit x → c of f(x), the limit laws and (3.3) (which we will prove in Section 3.6) show that we can do just that when we take a limit of a polynomial or a rational function. Let’s summarize this property and then look at two more examples that show how to compute limits of polynomials or rational functions by using these results.

If f(x)is a polynomial, then

x→clim f(x)= f(c) If f(x)is a rational function

f(x)= p(x) q(x)

wherep(x)^andq(x)are polynomials, and ifq(c)=^{0, then}

x→clim f(x)= ^lim

x→c

p(x)

q(x) = p(c)

q(c) = f(c)

EXAMPLE 12 Find

lim

x→3[x²−2x+1]

Solution

Since f(x)=x²−2x+1 is a polynomial, it follows that lim

x→3[x²−²x+¹] =⁹−⁶+¹=⁴ EXAMPLE 13 Find

lim

x→−1

2x³−x+⁵ x²+³x+¹

Solution

Note that

f(x)= ²x³−x+⁵ x²+³x+¹

is a rational function that is defined forx = −1. (The denominator is not equal to 0 when we substitutex = −1.) We find that

lim

x→−1

2x³−x+⁵

x²+³x+¹ = ²(−¹)³−(−¹)+⁵ (−¹)²+³(−¹)+¹ = ⁴

−¹ = −⁴

When you use the limit laws for finding limits of the form

x→clim[f(x)+g(x)] ^{or lim}

x→c[f(x)·g(x)] ^{or lim}

x→c

f(x) g(x)

you need to check first that both lim_x→c f(x)^{and lim}_x→cg(x)exist and, in the case of lim_{x→c g(x)}^f(x), that lim_x→cg(x)=0. The next two examples illustrate the importance of checking the assumptions in the limit laws before applying them.

EXAMPLE 14 Find

lim

x→0 1 x 1 x +1

Solution

We observe that neither lim

x→0

1

x ^nor _x^lim_→0 1

x +1

exist. So we cannot use Rule 4 right away. Multiplying both numerator and denominator byx, however, will help:

lim

x→0 1 x 1

x +1 = ^lim

x→0

1 1+x

Now we have a rational function on the right-hand side, and we can plug in 0 because the denominator, 1+x, will be different from 0. We get

lim

x→0

1

1+x = ¹ 1+⁰ =¹ EXAMPLE 15 Find

lim

x→4

x²−16 x−4

Solution

The function f(x) = ^x_x−4²⁻¹⁶ is a rational function, but since lim_x→4(x−⁴) = ^{0, we} cannot use Rule 4. Instead, we need to simplify f(x)first:

lim

x→4

x²−¹⁶ x−⁴ = ^lim

x→4

(x−⁴)(x+⁴) x−⁴

Becausex=4, we can cancelx−4 in the numerator and denominator, which yields lim

x→4(x+⁴)=⁸

where we used the fact thatx+4 is a polynomial in computing the limit.