2 Limits Introduction 2.2 The Limit of a Function

2.1 A Preview of Calculus

Coming Concepts

•

Describe the tangent problem and how it led to the idea of a derivative.
•

Explain how the idea of a limit is involved in solving the tangent problem.
•

Recognize a tangent to a curve at a point as the limit of secant lines.
•

Identify instantaneous velocity as the limit of average velocity over a small time interval.
•

Describe the area problem and how it was solved by the integral.
•

Explain how the idea of a limit is involved in solving the area problem.
•

Recognize how the ideas of limit, derivative, and integral led to the studies of infinite series and multivariable calculus.

As we embark on our study of calculus, we shall see how its development arose from common solutions to practical problems in areas such as engineering physics—like the space travel problem posed in the chapter opener. Two key problems led to the initial formulation of calculus: (1) the tangent problem, or how to determine the slope of a line tangent to a curve at a point; and (2) the area problem, or how to determine the area under a curve.

2.1.1 The Tangent Problem and Differential Calculus

Rate of change is one of the most critical concepts in calculus. We begin our investigation of rates of change by looking at the graphs of the three lines $\displaystyle f(x)=-2x-3,g(x)=\frac{1}{2}x+1$ , and $\displaystyle h(x)=2$ , shown in Figure 2.2.

Figure 2.2: The rate of change of a linear function is constant in each of these three graphs, with the constant determined by the slope.

As we move from left to right along the graph of $\displaystyle f(x)=-2x-3$ , we see that the graph decreases at a constant rate. For every 1 unit we move to the right along the $\displaystyle x$ -axis, the $\displaystyle y$ -coordinate decreases by 2 units. This rate of change is determined by the slope ( $\displaystyle-$ 2) of the line. Similarly, the slope of 1/2 in the function $\displaystyle g(x)$ tells us that for every change in $\displaystyle x$ of 1 unit there is a corresponding change in $\displaystyle y$ of 1/2 unit. The function $\displaystyle h(x)=2$ has a slope of zero, indicating that the values of the function remain constant. We see that the slope of each linear function indicates the rate of change of the function.

Compare the graphs of these three functions with the graph of $\displaystyle k(x)=x^{2}$ (Figure 2.3). The graph of $\displaystyle k(x)=x^{2}$ starts from the left by decreasing rapidly, then begins to decrease more slowly and level off, and then finally begins to increase—slowly at first, followed by an increasing rate of increase as it moves toward the right. Unlike a linear function, no single number represents the rate of change for this function. We quite naturally ask: How do we measure the rate of change of a nonlinear function?

Figure 2.3: The function

\displaystyle k(x)=x^{2}

does not have a constant rate of change.

We can approximate the rate of change of a function $\displaystyle f(x)$ at a point $\displaystyle(a,f(a))$ on its graph by taking another point $\displaystyle(x,f(x))$ on the graph of $\displaystyle f(x)$ , drawing a line through the two points, and calculating the slope of the resulting line. Such a line is called a secant line. Figure 2.4 shows a secant line to a function $\displaystyle f(x)$ at a point $\displaystyle(a,f(a))$ .

Figure 2.4: The slope of a secant line through a point

\displaystyle(a,f(a))

estimates the rate of change of the function at the point

\displaystyle(a,f(a))

We formally define a secant line as follows:

Definition.

The secant to the function $\displaystyle f(x)$ through the points $\displaystyle(a,f(a))$ and $\displaystyle(b,f(b))$ is the line passing through these points. Its equation can be given by

	$\displaystyle\displaystyle SL(x)$	$\displaystyle\displaystyle=m_{\sec}(x-a)+f(a)$
	$\displaystyle\displaystyle m_{\sec}$	$\displaystyle\displaystyle=\frac{f(b)-f(a)}{b-a}$

The accuracy of approximating the rate of change of the function with a secant line depends on how close $\displaystyle b$ is to $\displaystyle a$ . As we see in Figure 2.5, if $\displaystyle b$ is closer to $\displaystyle a$ , the slope of the secant line is a better measure of the rate of change of $\displaystyle f(x)$ at $\displaystyle a$ .

Figure 2.5: As

\displaystyle x

gets closer to

\displaystyle a

, the slope of the secant line becomes a better approximation to the rate of change of the function

\displaystyle f(x)

\displaystyle a

The secant lines themselves approach a line that is called the tangent to the function $\displaystyle f(x)$ at $\displaystyle a$ (Figure 2.6). The slope of the tangent line to the graph at $\displaystyle a$ measures the rate of change of the function at $\displaystyle a$ . This value also represents the derivative of the function $\displaystyle f(x)$ at $\displaystyle a$ , or the rate of change of the function at $\displaystyle a$ . This derivative is denoted by $\displaystyle f^{\prime}(a)$ . Differential calculus is the field of calculus concerned with the study of derivatives and their applications.

External Resource.

For an interactive demonstration of the slope of a secant line that you can manipulate yourself, visit this applet (Note: this site requires a Java browser plugin): Math Insight.

Figure 2.6: Solving the Tangent Problem: As

\displaystyle b

approaches

\displaystyle a

, the secant lines approach the tangent line.

Example 2.1.1 illustrates how to find slopes of secant lines. These slopes estimate the slope of the tangent line or, equivalently, the rate of change of the function at the point at which the slopes are calculated.

Example 2.1.1 (Finding Slopes of Secant Lines).

Estimate the slope of the tangent line (rate of change) to $\displaystyle f(x)=x^{2}$ at $\displaystyle x=1$ by finding slopes of secant lines through $\displaystyle(1,1)$ and each of the following points on the graph of $\displaystyle f(x)=x^{2}$ .

(a)

$\displaystyle(2,4)$
(b)

$\displaystyle\left(\frac{3}{2},\frac{9}{4}\right)$

Solution: Use the formula for the slope of a secant line from the definition.

(a)

$\displaystyle m_{\sec}=\frac{4-1}{2-1}=3$
(b)

$\displaystyle m_{\sec}=\frac{\frac{9}{4}-1}{\frac{3}{2}-1}=\frac{5}{2}=2.5$

The point in part b. is closer to the point $\displaystyle(1,1)$ , so the slope of 2.5 is closer to the slope of the tangent line. A good estimate for the slope of the tangent would be in the range of 2 to 2.5 (Figure 2.7).

Figure 2.7: The secant lines to

\displaystyle f(x)=x^{2}

\displaystyle(1,1)

through (a)

\displaystyle(2,4)

and (b)

\displaystyle\left(\frac{3}{2},\frac{9}{4}\right)

provide successively closer approximations to the tangent line to

\displaystyle f(x)=x^{2}

\displaystyle(1,1)

Checkpoint 2.1.1.

Estimate the slope of the tangent line (rate of change) to $\displaystyle f(x)=x^{2}$ at $\displaystyle x=1$ by finding slopes of secant lines through $\displaystyle(1,1)$ and the point $\displaystyle\left(\frac{5}{4},\frac{25}{16}\right)$ on the graph of $\displaystyle f(x)=x^{2}$ .

Hint: Use Example 2.1.1 and Figure 2.7 as a solving guide.

We continue our investigation by exploring a related question. Keeping in mind that velocity may be thought of as the rate of change of position, suppose that we have a function, $\displaystyle s(t)$ , that gives the position of an object along a coordinate axis at any given time $\displaystyle t$ . Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a given time $\displaystyle t=a?$ We start by approximating the instantaneous velocity with an average velocity. First, recall that the speed of an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled. We define the average velocity of an object over a time period to be the change in its position divided by the length of the time period.

Definition.

Let $\displaystyle s(t)$ be the position of an object moving along a coordinate axis at time $\displaystyle t$ . The average velocity of the object over a time interval $\displaystyle[a,t]$ where $\displaystyle a<t$ (or $\displaystyle[t,a]$ if $\displaystyle t<a)$ is

v_{\text{ave}}=\frac{s(t)-s(a)}{t-a}.

(2.1)

As $\displaystyle t$ is chosen closer to $\displaystyle a$ , the average velocity becomes closer to the instantaneous velocity. Note that finding the average velocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function. Furthermore, to find the slope of a tangent line at a point $\displaystyle a$ , we let the $\displaystyle x$ -values approach $\displaystyle a$ in the slope of the secant line. Similarly, to find the instantaneous velocity at time $\displaystyle a$ , we let the $\displaystyle t$ -values approach $\displaystyle a$ in the average velocity. This process of letting $\displaystyle x$ or $\displaystyle t$ approach $\displaystyle a$ in an expression is called taking a limit. Thus, we may define the instantaneous velocity as follows.

Definition.

For a position function $\displaystyle s(t)$ , the instantaneous velocity at a time $\displaystyle t=a$ is the value that the average velocities approach on intervals of the form $\displaystyle[a,t]$ and $\displaystyle[t,a]$ as the values of $\displaystyle t$ become closer to $\displaystyle a$ , provided such a value exists.

Example 2.1.2 illustrates this concept of limits and average velocity.

Example 2.1.2 (Finding Average Velocity).

A rock is dropped from a height of 64 ft. It is determined that its height (in feet) above ground $\displaystyle t$ seconds later (for $\displaystyle 0\leq t\leq 2)$ is given by $\displaystyle s(t)=-16t^{2}+64$ . Find the average velocity of the rock over each of the given time intervals. Use this information to guess the instantaneous velocity of the rock at time $\displaystyle t=0.5$ .

(a)

$\displaystyle[0.49,0.5]$
(b)

$\displaystyle[0.5,0.51]$

Solution: Substitute the data into the formula for the definition of average velocity.

(a)

$\displaystyle v_{\text{ave}}=\frac{s(0.5)-s(0.49)}{0.5-0.49}=-15.84$
(b)

$\displaystyle v_{\text{ave}}=\frac{s(0.51)-s(0.5)}{0.51-0.5}=-16.16$

The instantaneous velocity is somewhere between $\displaystyle-$ 15.84 and $\displaystyle-$ 16.16 ft/sec. A good guess might be $\displaystyle-$ 16 ft/sec.

Checkpoint 2.1.2.

An object moves along a coordinate axis so that its position at time $\displaystyle t$ is given by $\displaystyle s(t)=t^{3}$ . Estimate its instantaneous velocity at time $\displaystyle t=2$ by computing its average velocity over the time interval $\displaystyle[2,2.001]$ .

Hint: Use $\displaystyle v_{\text{ave}}=\frac{s(2.001)-s(2)}{2.001-2}$ .

Key Concepts

•

Differential calculus arose from trying to solve the problem of determining the slope of a line tangent to a curve at a point. The slope of the tangent line indicates the rate of change of the function, also called the derivative. Calculating a derivative requires finding a limit.
•

Integral calculus arose from trying to solve the problem of finding the area of a region between the graph of a function and the $\displaystyle x$ -axis. We can approximate the area by dividing it into thin rectangles and summing the areas of these rectangles. This summation leads to the value of a function called the integral. The integral is also calculated by finding a limit and, in fact, is related to the derivative of a function.
•

Multivariable calculus enables us to solve problems in three-dimensional space, including determining motion in space and finding volumes of solids.

Key Equations

•

Slope of a Secant Line

$\displaystyle m_{\sec}=\frac{f(x)-f(a)}{x-a}$
•

Average Velocity over Interval $\displaystyle[a,t]$

$\displaystyle v_{\text{ave}}=\frac{s(t)-s(a)}{t-a}$

Glossary

average velocity: the change in an object’s position divided by the length of a time period; the average velocity of an object over a time interval $\displaystyle[t,a]$ (if $\displaystyle t<a$ or $\displaystyle[a,t]$ if $\displaystyle t>a)$ , with a position given by $\displaystyle s(t)$ , that is $\displaystyle v_{\text{ave}}=\frac{s(t)-s(a)}{t-a}$
differential calculus: the field of calculus concerned with the study of derivatives and their applications
instantaneous velocity: The instantaneous velocity of an object with a position function that is given by $\displaystyle s(t)$ is the value that the average velocities on intervals of the form $\displaystyle[t,a]$ and $\displaystyle[a,t]$ approach as the values of $\displaystyle t$ move closer to $\displaystyle a$ , provided such a value exists
integral calculus: the study of integrals and their applications
limit: the process of letting $\displaystyle x$ or $\displaystyle t$ approach $\displaystyle a$ in an expression; the limit of a function $\displaystyle f(x)$ as $\displaystyle x$ approaches $\displaystyle a$ is the value that $\displaystyle f(x)$ approaches as $\displaystyle x$ approaches $\displaystyle a$
multivariable calculus: the study of the calculus of functions of two or more variables
secant: A secant line to a function $\displaystyle f(x)$ at $\displaystyle a$ is a line through the point $\displaystyle(a,f(a))$ and another point on the function; the slope of the secant line is given by $\displaystyle m_{\sec}=\frac{f(x)-f(a)}{x-a}$
tangent: A tangent line to the graph of a function at a point $\displaystyle(a,f(a))$ is the line that secant lines through $\displaystyle(a,f(a))$ approach as they are taken through points on the function with $\displaystyle x$ -values that approach $\displaystyle a$ ; the slope of the tangent line to a graph at $\displaystyle a$ measures the rate of change of the function at $\displaystyle a$