Chap 1

Introduction to Limits Chapter 1 August 1999

1

Limits

The study of calculus is essentially the study of the use of limits. Over the entire calculus sequence, you will see the notion of limit developed to study rates of change (the derivative, which we will study in this course), the area problem (the integral) and approximation (sequences and series). The triumph of Newton and Liebniz is largely the recognition of the intimate connections between these seemingly different ideas. Let’s take a quick look at examples that illustrate what a limit is. If we consider the sequence of numbers 1, 1/2, 1/3, 1/4, ..., 1/n, ... it is clear that the farther out in this sequence we progress, the smaller the numbers become. In fact, given any positive number c, by choosing a point far enough out in the sequence, we can guarantee that all terms in the remainder of the sequence will be smaller than c. We say that the sequence has a limit of zero, even though no term of the sequence will ever be 0. If f is a function, we say that the limit of f as x approaches a is L, written lim f (x) = L

x→a

if, given any c > 0, we can find a d > 0 so that, for all x satisfying |x − a| < d, |f (x) − L| < c. So, in the example above, we write 1 lim = 0. n→∞ n

2

The geometric series

While you will not begin a formal study of series this semester, the notion of a geometric series is fairly straightforward and serves as another good example of the notion of limit. Suppose that a ball dropped from a certain height h rebounds to half its original height. Starting from the first time the ball hits the floor, how far will the ball travel (both up and down) before it comes to a rest? Since the first time the ball bounces, it rebounds to a height h/2, and then falls, it will travel h units between the first and second bounces. Between the second and third bounce, it will travel h/2units (h/4 up, h/4 down). Between the third and fourth, h/4, and so on. So, the total distance traveled will be   h h h 1 1 1 h + + + ... + n + ... = h ∗ 1 + + + ... + n + ... 2 4 2 2 4 2 But how can we add up the terms in parentheses? Let’s add up a few terms and see what’s happening.   1 1 2047 h ∗ 1 + + ... + 10 = ≈ 1.9990 2 2 1024   1 1 2097151 h ∗ 1 + + ... + 20 = ≈ 1.9999990 2 2 1048576 1

This is very close to (but not quite) 2. Will there ever be enough terms to exceed 2? To see that the answer is no, we will use the following formula from algebra. 1 − xn = (1 − x)(1 + x + x2 + x3 + ... + x(n−1) ) which implies that (1 + x + x2 + x3 + ... + x(n−1) ) =

(1 − xn ) 1−x

In our case, we are interested in x = 1/2. So, 1+

1 1 1 + + ... + n−1 2 4 2

= =

1 − 1/2n 1 − 1/2   1 2 1− n 2

But, as we add more and more terms, the sum gets closer and closer to 2, since 1/2n ; gets closer and closer to 0 as n gets large. We therefore say that the limit as n goes to infinity of the sum 1+

1 1 1 + + ... + n 2 4 2

is 2, or more compactly,   1 1 1 lim 1 + + + ... + n = 2. n→∞ 2 4 2

3

Instantaneous rate of change: speed

We all have an intuitive notion of what is meant by speed: If I travel at a speed of s miles per hour for t hours, then I will go s · t miles. Or, what is more common, if I drive for t hours and cover m miles, I might say that I averaged m/t miles per hour. That is, our notion of speed is fundamentally based on a measure of change of position over a period of time. We want to extend that to a notion of instantaneous rate of change. Galileo (1564-1642), through his famous experiments at Pisa, discovered that an object in free fall near the earth’s surface, ignoring air resistance, would fall a distance of 16t2 feet in t seconds. So, during the first second, an object will fall 16 feet. During the second second, 64 − 16 = 48 feet. During the third, 144 − 64 = 80 feet. But how fast is the ball traveling at the end of each second? It is clear that, during the first second, the average speed is 16 feet per second. And, since it started with a speed of 0, we can be certain that the speed at the end of that second is greater than 16. To examine this more carefully, let’s look at the average speed over smaller intervals. t (in seconds) 3/4 7/8 15/16 31/32 63/64

Distance traveled from t to 1
9 16 − 16 16 =7
15 16 − 16 49 64 = 4
31 16 − 16 225 256 = 16
63 961 16 − 16 1024 =
64 3969 127 16 − 16 4096 = 256

Average speed (t to 1) 7 1−3/4 = 28 15/4 1−7/8 = 30 31/16 1−15/16 = 31 63/64 63 1−31/32 = 2 = 31.5 127/256 127 1−63/64 = 4 = 31.75

It appears that, as we take smaller and smaller time intervals, the average speed approaches 32. We are therefore led to conclude that the speed at the end of 1 second is 32 feet/second.

2

4

More on speed

Here is a function which will compute the average speed of an object in free fall between two times s and t. V (s, t) =

16s2 − 16t2 s−t

Use this to determine the speed at the end of 2 and 3 seconds. What do you think the general case is? Using algebra, justify your answer. Note that 16s2 − 16t2 s−t

= = =

s2 − t2 s−t (s − t) (s + t) 16 s−t 16 (s + t) , for s 6= t. 16

So, what happens as t gets closer and closer to s?

5

The tangent problem

One of the fundamental problems of calculus is the problem of computing the slope of the tangent line to the graph of a function at some point. Here is a picture which shows graph of the function f (x) = sin x with a tangent line shown at the point (π/6, 1/2).

Figure 1: Graph of sin x with tangent line at (π/6, 1/2). The tangent line coincides with the graph of y = sin x at the point (π/6, 1/2), and at that point, has the same slope. But, how does one compute the slope of the tangent line at a point? You can probably guess that this will have something to do with the notion of limit. Let’s start by computing slopes of lines passing through distinct points on the graph of the function. (These lines are called secant lines.) Recall from your algebra class that the slope of a line passing through two points (x0 , y0 ) and (x1 , y1 ) is given by m=

y1 − y0 . x1 − x0

Once we have computed the slope, the equation of the line is y − y0 = m(x − x0 ). 3

In our example, we will want to use the point (π/6, 1/2) as one of the points, and another point, which we will vary, will have the form (x, sin x). So, we can compute the slope of the desired secant line as m = m(x) =

sin x − 1/2 x − π/6

Now we want to ask, what happens as x gets close to√π/6? Use the above formula to compute the slope for several values of x close to π/6, and compare your results to 3/2 ≈ .8660, the true slope of the tangent line. Formally, we are attempting to compute the limit as x approaches π/6 of m(x), that is, lim m(x) x→π/6

We want to be able to determine the exact value, however, not just approximations as above. In order to handle this sort of problem efficiently, we need to develop some theory and rules for determining limits. That is the next order of business.

6

Formal calculation of limits and continuity of functions

Let’s start with an example. Suppose that we wish to determine the limit lim x2 .

x→1

From a plot of y = x2

Figure 2: Graph of y = x2 on the interval [0, 2]. it seems fairly obvious that the closer x gets to 1, the closer x2 gets to 1, which is of course simply the value of x at x = 1. We want to emphasize that, though in many cases a limit may not be so easy to determine, we insist that our formal procedures comfirm our intuition in this example. (By the way, we will say that the function f (x) = x2 is continuous.) Def: We say that the function f is continuous at x = a if 2

lim f (x) = f (a).

x→a

Let’s look at another example. Suppose that we want to consider what happens to the expression x2 − 9 x−3 4

as x gets close to 3. Note that, at x = 3, the expression makes no sense (it has the form 0/0, which is undefined). However, for values of x other than 3, (x − 3) (x + 3) x2 − 9 = = (x + 3) x−3 x−3 and it is clear that, as x gets close to 3, x + 3 gets close to 6. We therefore conclude that x2 − 9 x→3 x − 3 lim

= =

lim (x + 3)

x→3

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We can formalize the above: If f (x) = g(x), except possibly at x = a, and lim g(x) = L

x→a

then lim f (x) = L.

x→a

6.1

On your own

Compute the following limits: 16x2 − 16 x→1 x−1 lim

x3 − 27 x→3 x − 3 lim

Limits are not always finite. For example, the function f (x) = 1/x2 is undefined at x = 0, but it’s behaviour near 0 is still predictable: as x gets close to 0, f (x) gets bigger and bigger. We formalize this by saying that lim

x→0

1 = ∞. x2

On the other hand, the function g(x) = 1/x also has predictable behaviour near 0, but that behaviour depends on whether x > 0 or x < 0. This points to a need for one sided limits. We will say that the limit of 1/x as x approaches 0 from the right is ∞, written 1 lim = ∞, x→0+ x and the limit of 1/x as x approaches 0 from the left is −∞, written lim

x→0−

1 = −∞ x

Compute the following limits. 1 x+1  x2 , if x > 1 lim+ f (x), where f (x) = x − 2, if x ≤ 1 x→1 lim

x→−1+

Other limits may be done by comparison to known functions. Here is an example: lim x sin

x→0

5

1 x

The behaviour of sin(1/x) is erratic near x = 0. However, the sine function is bounded above by 1, and bounded below by −1. So, we will always have − |x| ≤ x sin (1/x) ≤ |x| Since lim |x| = lim (− |x|) = 0,

x→0

x→0

we will also have lim x sin (1/x) = 0.

x→0

This is the pinching or squeeze theorem. Formally, we have: If f (x) < g(x) < h(x), lim f (x)

= L, and

lim h(x)

= L,

x→a

x→a

then we must also have lim g(x) = L.

x→a

We will use this result later to develop important results regarding limits of trigonometric functions.

7

Further remarks on limits and continuity

The text provides a list of properties regarding limits (p. 61) and continuity of functions (p. 82-86). You should note that many of the properties are similar. For example, ”the limit of a sum is the sum of the limits” and ”the sum of continuous functions is continuous.” This should not be surprising, since continuity is formally defined in terms of the existence of limits. Of particular importance are the facts regarding polynomials and rational functions: polynomials are continuous everywhere, while rational functions are continuous at all points in their domain, i.e. at all points for which they are defined. So, for example, the polynomial p(x) = x3 − 3x2 + 5x + 1 is continuous everywhere, and furthermore we will always have lim p(x) = p(a).

x→a

The rational function

x2 + x − 3 x+1 will be continuous at all values of x except x = −1, the only value not in the domain of r. Hence, except for a = −1, r(x) =

lim r(x) = r(a).

x→a

Note that there are different categories of discontinuities. For example, while the function f (x) =

x2 − 9 x−3

is not continuous at x = 3 (it is not even defined there), lim f (x)

x→3

x2 − 9 x→3 x − 3 (x − 3) (x + 3) = lim x→3 x−3 = lim (x + 3) =

lim

x→3

=

6 6

And, a plot of f should look like the graph of the line y = x + 3, with a hole at x = 3.

Figure 3: Graph of f (x) =

x2 −9 x−3

on the interval [0, 6].

This kind of discontinuity is called a removable discontinuity, because we can redefine the function at a single point and produce a continuous function. On the other hand, the function g(x) = 1/x has a vertical asymptote at x = 0. This type of discontinuity is called an infinite discontinuity, or pole. Finally, we have functions such as h(x) = x / |x|. Note that, if x < 0, h(x) = −1, while h(x) = 1 for x > 0. (Plot it.) This type of discontinuity is called a jump discontinuity.

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