Active Calculus

Subsection 1.1.1 The Notion of Limit

Limits give us a way to identify a trend in the values of a function as its input variable approaches a particular value of interest. We need a precise understanding of what it means to say “a function \(f\) has limit \(L\) as \(x\) approaches \(a\text{.}\)” To begin, think about a recent example.

In Preview Activity 1.1.1, we saw that as \(x\) gets closer and closer (but not equal) to 0, \(g(x)\) gets as close as we want to the value 4. At first, this may feel counterintuitive, because the value of \(g(0)\) is \(1\text{,}\) not \(4\text{.}\) But limits describe the behavior of a function arbitrarily close to a fixed input, and the value of the function at the fixed input does not matter. More formally,¹ we say the following.

For any function \(f\text{,}\) there are typically three ways to answer the question “does \(f\) have a limit at \(x = a\text{,}\) and if so, what is the limit?” The first is to reason graphically as we have just done with the example from Preview Activity 1.1.1. If we have a formula for \(f(x)\text{,}\) there are two additional possibilities:

1. Evaluate the function at a sequence of inputs that approach \(a\) on either side (typically using some sort of computing technology), and ask if the sequence of outputs seems to approach a single value.
2. Use the algebraic form of the function to understand the trend in its output values as the input values approach \(a\text{.}\)

The first approach produces only an approximation of the value of the limit, while the latter can often be used to determine the limit exactly.

For each of the following functions, we'd like to know whether or not the function has a limit at the stated \(a\)-values. Use both numerical and algebraic approaches to investigate and, if possible, estimate or determine the value of the limit. Compare the results with a careful graph of the function on an interval containing the points of interest.

1. \(f(x) = \frac{4-x^2}{x+2}\text{;}\) \(a = -1\text{,}\) \(a = -2\)
2. \(g(x) = \sin\left(\frac{\pi}{x}\right)\text{;}\) \(a = 3\text{,}\) \(a = 0\)

Solution

a. We first construct a graph of \(f\) along with tables of values near \(a = -1\) and \(a = -2\text{.}\)

\(x\)	\(f(x)\)
\(-0.9\)	\(2.9\)
\(-0.99\)	\(2.99\)
\(-0.999\)	\(2.999\)
\(-0.9999\)	\(2.9999\)
\(-1.1\)	\(3.1\)
\(-1.01\)	\(3.01\)
\(-1.001\)	\(3.001\)
\(-1.0001\)	\(3.0001\)

\(x\)	\(f(x)\)
\(-1.9\)	\(3.9\)
\(-1.99\)	\(3.99\)
\(-1.999\)	\(3.999\)
\(-1.9999\)	\(3.9999\)
\(-2.1\)	\(4.1\)
\(-2.01\)	\(4.01\)
\(-2.001\)	\(4.001\)
\(-2.0001\)	\(4.0001\)

From Table 1.1.5, it appears that we can make \(f\) as close as we want to 3 by taking \(x\) sufficiently close to \(-1\text{,}\) which suggests that \(\lim_{x \to -1} f(x) = 3\text{.}\) This is also consistent with the graph of \(f\text{.}\) To see this a bit more rigorously and from an algebraic point of view, consider the formula for \(f\text{:}\) \(f(x) = \frac{4-x^2}{x+2}\text{.}\) As \(x \to -1\text{,}\) \((4-x^2) \to (4 - (-1)^2) = 3\text{,}\) and \((x+2) \to (-1 + 2) = 1\text{,}\) so as \(x \to -1\text{,}\) the numerator of \(f\) tends to 3 and the denominator tends to 1, hence \(\lim_{x \to -1} f(x) = \frac{3}{1} = 3\text{.}\)

The situation is more complicated when \(x \to -2\text{,}\) because \(f(-2)\) is not defined. If we try to use a similar algebraic argument regarding the numerator and denominator, we observe that as \(x \to -2\text{,}\) \((4-x^2) \to (4 - (-2)^2) = 0\text{,}\) and \((x+2) \to (-2 + 2) = 0\text{,}\) so as \(x \to -2\text{,}\) the numerator and denominator of \(f\) both tend to 0. We call \(0/0\) an indeterminate form. This tells us that there is somehow more work to do. From Table 1.1.6 and Figure 1.1.7, it appears that \(f\) should have a limit of \(4\) at \(x = -2\text{.}\)

To see algebraically why this is the case, observe that

\begin{align*} \lim_{x \to -2} f(x) = \amp \lim_{x \to -2} \frac{4-x^2}{x+2}\\ = \amp \lim_{x \to -2} \frac{(2-x)(2+x)}{x+2}\text{.} \end{align*}

It is important to observe that, since we are taking the limit as \(x \to -2\text{,}\) we are considering \(x\) values that are close, but not equal, to \(-2\text{.}\) Because we never actually allow \(x\) to equal \(-2\text{,}\) the quotient \(\frac{2+x}{x+2}\) has value 1 for every possible value of \(x\text{.}\) Thus, we can simplify the most recent expression above, and find that

\begin{equation*} \lim_{x \to -2} f(x) = \lim_{x \to -2} 2-x\text{.} \end{equation*}

This limit is now easy to determine, and its value clearly is \(4\text{.}\) Thus, from several points of view we've seen that \(\lim_{x \to -2} f(x) = 4\text{.}\)

b. Next we turn to the function \(g\text{,}\) and construct two tables and a graph.

\(x\)	\(g(x)\)
\(2.9\)	\(0.84864\)
\(2.99\)	\(0.86428\)
\(2.999\)	\(0.86585\)
\(2.9999\)	\(0.86601\)
\(3.1\)	\(0.88351\)
\(3.01\)	\(0.86777\)
\(3.001\)	\(0.86620\)
\(3.0001\)	\(0.86604\)

\(x\)	\(g(x)\)
\(-0.1\)	\(0\)
\(-0.01\)	\(0\)
\(-0.001\)	\(0\)
\(-0.0001\)	\(0\)
\(0.1\)	\(0\)
\(0.01\)	\(0\)
\(0.001\)	\(0\)
\(0.0001\)	\(0\)

First, as \(x \to 3\text{,}\) it appears from the table values that the function is approaching a number between \(0.86601\) and \(0.86604\text{.}\) From the graph it appears that \(g(x) \to g(3)\) as \(x \to 3\text{.}\) The exact value of \(g(3) = \sin(\frac{\pi}{3})\) is \(\frac{\sqrt{3}}{2}\text{,}\) which is approximately 0.8660254038. This is convincing evidence that

\begin{equation*} \lim_{x \to 3} g(x) = \frac{\sqrt{3}}{2}\text{.} \end{equation*}

As \(x \to 0\text{,}\) we observe that \(\frac{\pi}{x}\) does not behave in an elementary way. When \(x\) is positive and approaching zero, we are dividing by smaller and smaller positive values, and \(\frac{\pi}{x}\) increases without bound. When \(x\) is negative and approaching zero, \(\frac{\pi}{x}\) decreases without bound. In this sense, as we get close to \(x = 0\text{,}\) the inputs to the sine function are growing rapidly, and this leads to increasingly rapid oscillations in the graph of \(g\) betweem \(1\) and \(-1\text{.}\) If we plot the function \(g(x) = \sin\left(\frac{\pi}{x}\right)\) with a graphing utility and then zoom in on \(x = 0\text{,}\) we see that the function never settles down to a single value near the origin, which suggests that \(g\) does not have a limit at \(x = 0\text{.}\)

How do we reconcile the graph with the righthand table above, which seems to suggest that the limit of \(g\) as \(x\) approaches \(0\) may in fact be \(0\text{?}\) The data misleads us because of the special nature of the sequence of input values \(\{0.1, 0.01, 0.001, \ldots\}\text{.}\) When we evaluate \(g(10^{-k})\text{,}\) we get \(g(10^{-k}) = \sin\left(\frac{\pi}{10^{-k}}\right) = \sin(10^k \pi) = 0\) for each positive integer value of \(k\text{.}\) But if we take a different sequence of values approaching zero, say \(\{0.3, 0.03, 0.003, \ldots\}\text{,}\) then we find that

\begin{equation*} g(3 \cdot 10^{-k}) = \sin\left(\frac{\pi}{3 \cdot 10^{-k}}\right) = \sin\left(\frac{10^k \pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866025\text{.} \end{equation*}

That sequence of function values suggests that the value of the limit is \(\frac{\sqrt{3}}{2}\text{.}\) Clearly the function cannot have two different values for the limit, so \(g\) has no limit as \(x \to 0\text{.}\)

An important lesson to take from Example 1.1.4 is that tables can be misleading when determining the value of a limit. While a table of values is useful for investigating the possible value of a limit, we should also use other tools to confirm the value.

Recall that our primary motivation for considering limits of functions comes from our interest in studying the rate of change of a function. To that end, we close this section by revisiting our previous work with average and instantaneous velocity and highlighting the role that limits play.

Subsection 1.1.2 Instantaneous Velocity

Suppose that we have a moving object whose position at time \(t\) is given by a function \(s\text{.}\) We know that the average velocity of the object on the time interval \([a,b]\) is \(AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}\text{.}\) We define the instantaneous velocity at \(a\) to be the limit of average velocity as \(b\) approaches \(a\text{.}\) Note particularly that as \(b \to a\text{,}\) the length of the time interval gets shorter and shorter (while always including \(a\)). We will write \(IV_{t=a}\) for the instantaneous velocity at \(t = a\text{,}\) and thus

Equivalently, if we think of the changing value \(b\) as being of the form \(b = a + h\text{,}\) where \(h\) is some small number, then we may instead write

Again, the most important idea here is that to compute instantaneous velocity, we take a limit of average velocities as the time interval shrinks.

The closing activity of this section asks you to make some connections among average velocity, instantaneous velocity, and slopes of certain lines.

Activity 1.1.4.

For the moving object whose position \(s\) at time \(t\) is given by the graph in Figure 1.1.11, answer each of the following questions. Assume that \(s\) is measured in feet and \(t\) is measured in seconds.

1. Use the graph to estimate the average velocity of the object on each of the following intervals: \([0.5,1]\text{,}\) \([1.5,2.5]\text{,}\) \([0,5]\text{.}\) Draw each line whose slope represents the average velocity you seek.
2. How could you use average velocities or slopes of lines to estimate the instantaneous velocity of the object at a fixed time?
3. Use the graph to estimate the instantaneous velocity of the object when \(t = 2\text{.}\) Should this instantaneous velocity at \(t = 2\) be greater or less than the average velocity on \([1.5,2.5]\) that you computed in (a)? Why?

Hint

1. Remember that average velocity on an interval computes the quotient of “change in \(s\) over change in \(t\text{.}\)” This is the slope of the line between the corresponding two points on the graph of \(s\text{.}\)
2. Think about shorter and shorter time intervals and drawing the lines whose slopes represent average velocity.
3. Think about zooming in on the graph at \(t = 2\) and drawing a line that, up close, looks just like the curve \(s(t)\text{.}\) What is the approximate slope of that line?

Answer

1. \(AV_{[0.5,1]} = \frac{1-1}{1-0.5} = 0\text{,}\) \(AV_{[1.5,2.5]} = \frac{3-1}{2.5-1.5} = 2\text{,}\) and \(AV_{[0,5]} = \frac{5-0}{5-0} = 1\text{.}\)
2. Take shorter and shorter time intervals and draw the lines whose slopes represent average velocity. If those lines' slopes are approaching a single number, that number represents the instantaneous velocity.
3. The instantaneous velocity at \(t = 2\) is greater than the average velocity on \([1.5,2.5]\text{.}\)

Solution

1. The average velocity on \([0.5,1]\) is the slope of the line joining the points \((0.5,s(0.5))\) and \((1,s(1))\text{,}\) which is \(AV_{[0.5,1]} = \frac{1-1}{1-0.5} = 0\text{.}\) On \([1.5,2.5]\text{,}\) we similarly find \(AV_{[1.5,2.5]} = \frac{3-1}{2.5-1.5} = 2\text{,}\) and on \([0,5]\text{,}\) we have \(AV_{[0,5]} = \frac{5-0}{5-0} = 1\text{.}\)
2. Take shorter and shorter time intervals and draw the lines whose slopes represent average velocity. If those lines' slopes are approaching a single number, that number represents the instantaneous velocity. For example, to estimate the instantaneous velocity at \(t = 2\text{,}\) we might consider average velocities on \([2,3]\text{,}\) \([2,2.5]\text{,}\) and \([2,2.25]\text{.}\)
3. If we draw the line through \((2,2)\) and \((2.1,s(2.1))\text{,}\) it looks like the line's slope is approximately 2.5: if we go over one grid-width, we appear to go up about 2.5. The slope of this line is clearly greater than the slope of the line through \((1.5, s(1.5))\) and \((2.5, s(2.5))\text{,}\) which is 2. Hence the instantaneous velocity at \(t = 2\) is greater than the average velocity on \([1.5,2.5]\text{.}\)

Subsection 1.1.3 Summary

• Limits enable us to examine trends in function behavior near a specific point. In particular, taking a limit at a given point asks if the function values nearby tend to approach a particular fixed value.

• We read \(\lim_{x \to a} f(x) = L\text{,}\) as “the limit of \(f\) as \(x\) approaches \(a\) is \(L\text{,}\)” which means that we can make the value of \(f(x)\) as close to \(L\) as we want by taking \(x\) sufficiently close (but not equal) to \(a\text{.}\)

• To find \(\lim_{x \to a} f(x)\) for a given value of \(a\) and a known function \(f\text{,}\) we can estimate this value from the graph of \(f\text{,}\) or we can make a table of function values for \(x\)-values that are closer and closer to \(a\text{.}\) If we want the exact value of the limit, we can work with the function algebraically to understand how different parts of the formula for \(f\) change as \(x \to a\text{.}\)

• We find the instantaneous velocity of a moving object at a fixed time by taking the limit of average velocities of the object over shorter and shorter time intervals containing the time of interest.