Active Calculus

Subsection 1.4.1 Having a limit at a point

In Section 1.1, we learned that \(f\) has limit \(L\) as \(x\) approaches \(a\) provided that we can make the value of \(f(x)\) as close to \(L\) as we like by taking \(x\) sufficiently close (but not equal to) \(a\text{.}\) If so, we write \(\lim_{x \to a} f(x) = L\text{.}\)

Essentially there are two behaviors that a function can exhibit near a point where it fails to have a limit. In Figure 1.4.3, at left we see a function \(f\) whose graph shows a jump at \(a = 1\text{.}\) If we let \(x\) approach 1 from the left side, the value of \(f\) approaches 2, but if we let \(x\) approach \(1\) from the right, the value of \(f\) tends to 3. Because the value of \(f\) does not approach a single number as \(x\) gets arbitrarily close to 1 from both sides, we know that \(f\) does not have a limit at \(a = 1\text{.}\)

For such cases, we introduce the notion of left and right (or one-sided) limits.

In the graph of the function \(f\) in Figure 1.4.3, we see that

Precisely because the left and right limits are not equal, the overall limit of \(f\) as \(x \to 1\) fails to exist.

For the function \(g\) pictured at right in Figure 1.4.3, the function fails to have a limit at \(a = 1\) for a different reason. While the function does not have a jump in its graph at \(a = 1\text{,}\) it is still not the case that \(g\) approaches a single value as \(x\) approaches 1. In particular, due to the infinitely oscillating behavior of \(g\) to the right of \(a = 1\text{,}\) we say that the right-hand limit of \(g\) as \(x \to 1^+\) does not exist, and thus \(\lim_{x \to 1} g(x) \ \text{does not exist} \text{.}\)

To summarize, if either a left- or right-hand limit fails to exist or if the left- and right-hand limits are not equal to each other, the overall limit does not exist.

That is, a function has a limit at \(x = a\) if and only if both the left- and right-hand limits at \(x = a\) exist and have the same value.

In Preview Activity 1.4.1, the function \(f\) given in Figure 1.4.1 fails to have a limit at only two values: at \(a = -2\) (where the left- and right-hand limits are 2 and \(-1\text{,}\) respectively) and at \(x = 2\text{,}\) where \(\lim_{x \to 2^+} f(x)\) does not exist). Note well that even at values such as \(a = -1\) and \(a = 0\) where there are holes in the graph, the limit still exists.

Activity 1.4.2.

Consider a function that is piecewise-defined according to the formula

\begin{equation*} f(x) = \begin{cases}3(x+2)+2 \amp \text{ for \(-3 \lt x \lt -2\) } \\ \frac{2}{3}(x+2)+1 \amp \text{ for \(-2 \le x \lt -1\) } \\ \frac{2}{3}(x+2)+1 \amp \text{ for \(-1 \lt x \lt 1\) } \\ 2 \amp \text{ for \(x = 1\) } \\ 4-x \amp \text{ for \(x \gt 1\) } \end{cases} \end{equation*}

Use the given formula to answer the following questions.

1. For each of the values \(a = -2, -1, 0, 1, 2\text{,}\) compute \(f(a)\text{.}\)

2. For each of the values \(a = -2, -1, 0, 1, 2\text{,}\) determine \(\lim_{x \to a^-} f(x)\) and \(\lim_{x \to a^+} f(x)\text{.}\)

3. For each of the values \(a = -2, -1, 0, 1, 2\text{,}\) determine \(\lim_{x \to a} f(x)\text{.}\) If the limit fails to exist, explain why by discussing the left- and right-hand limits at the relevant \(a\)-value.

4. For which values of \(a\) is the following statement true?

   \begin{equation*} \lim_{x \to a} f(x) \ne f(a) \end{equation*}

5. On the axes provided in Figure 1.4.4, sketch an accurate, labeled graph of \(y = f(x)\text{.}\) Be sure to carefully use open circles (\(\circ\)) and filled circles (\(\bullet\)) to represent key points on the graph, as dictated by the piecewise formula.

Hint

1. Find the interval in which \(a\) lies and evaluate the function there.

2. Remember that for \(\lim_{x \to a^-} f(x)\text{,}\) we only consider values of \(x\) such that \(x \lt a\text{.}\) Find the right formula to use in the piecewise definition for \(f\) to fit the values you are considering.

3. Use your work in (c) and compare left- and right-hand limits.

4. Use your work in (a) and (c).

5. Note that \(f\) is piecewise linear.

Answer

1. \(f(-2) = 1\text{;}\) \(f(-1)\) is not defined; \(f(0) = \frac{7}{3}\text{;}\) \(f(1) = 2\text{;}\) \(f(2) = 2\text{.}\)

2. \begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1 \end{equation*}
   \begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3} \end{equation*}
   \begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3} \end{equation*}
   \begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3 \end{equation*}
   \begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2 \end{equation*}

3. \(\lim_{x \to -2} f(x)\) does not exist. The values of the limits as \(x \to a\) for \(a = -1, 0, 1, 2\) are \(\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}\)

4. \(a = -2\text{,}\) \(a = -1\text{,}\) and \(a = 1\text{.}\)

5. 

Solution

1. \(f(-2) = \frac{2}{3}(-2+2) + 1 = 1\text{;}\) \(f(-1)\) is not defined; \(f(0) = \frac{2}{3}(0+2)+1 = \frac{7}{3}\text{;}\) \(f(1) = 2\) (by the rule); \(f(2) = 4-2 = 2\text{.}\)

2. \begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1 \end{equation*}
   \begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3} \end{equation*}
   \begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3} \end{equation*}
   \begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3 \end{equation*}
   \begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2 \end{equation*}

3. \(\lim_{x \to -2} f(x)\) does not exists because the left-hand limit is \(2\) while the right-hand limit is \(1\text{.}\) All of the other requested limits exist, as left- and right-hand limits exist and are equal in each case. The respective values of the limits as \(x \to a\) for \(a = -1, 0, 1, 2\) are \(\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}\)

4. For \(a = -2\text{,}\) \(a = -1\text{,}\) and \(a = 1\text{,}\) \(\lim_{x \to a} f(x) \ne f(a)\text{.}\) At \(a = -2\text{,}\) the limit fails to exist, but \(f(-2) = 1\text{.}\) At \(a = -1\text{,}\) the limit is \(\frac{5}{3}\text{,}\) but \(f(-1)\) is not defined. At \(a = 1\text{,}\) the limit is 3, but \(f(1) = 2\text{.}\)

5. 

Subsection 1.4.2 Being continuous at a point

Intuitively, a function is continuous if we can draw its graph without ever lifting our pencil from the page. Alternatively, we might say that the graph of a continuous function has no jumps or holes in it. In Figure 1.4.5 we consider three functions that have a limit at \(a = 1\text{,}\) and use them to make the idea of continuity more precise.

First consider the function in the left-most graph. Note that \(f(1)\) is not defined, which leads to the resulting hole in the graph of \(f\) at \(a = 1\text{.}\) We will naturally say that \(f\) is not continuous at \(a = 1\). For the function \(g\text{,}\) we observe that while \(\lim_{x \to 1} g(x) = 3\text{,}\) the value of \(g(1) = 2\text{,}\) and thus the limit does not equal the function value. Here, too, we will say that \(g\) is not continuous, even though the function is defined at \(a = 1\text{.}\) Finally, the function \(h\) appears to be the most well-behaved of all three, since at \(a = 1\) its limit and its function value agree. That is,

With no hole or jump in the graph of \(h\) at \(a = 1\text{,}\) we say that \(h\) is continuous there. More formally, we make the following definition.

Conditions (a) and (b) are technically contained implicitly in (c), but we state them explicitly to emphasize their individual importance. The definition says that a function is continuous at \(x = a\) provided that its limit as \(x \to a\) exists and equals its function value at \(x = a\text{.}\) If a function is continuous at every point in an interval \([a,b]\text{,}\) we say the function is “continuous on \([a,b]\text{.}\)” If a function is continuous at every point in its domain, we simply say the function is “continuous.” Thus, continuous functions are particularly nice: to evaluate the limit of a continuous function at a point, all we need to do is evaluate the function.

For example, consider \(p(x) = x^2 - 2x + 3\text{.}\) It can be proved that every polynomial is a continuous function at every real number, and thus if we would like to know \(\lim_{x \to 2} p(x)\text{,}\) we simply compute

This route of substituting an input value to evaluate a limit works whenever we know that the function being considered is continuous. Besides polynomial functions, all exponential functions and the sine and cosine functions are continuous at every point, as are many other familiar functions and combinations thereof.

Activity 1.4.3.

This activity builds on your work in Preview Activity 1.4.1, using the same function \(f\) as given by the graph that is repeated in Figure 1.4.7.

1. At which values of \(a\) does \(\lim_{x \to a} f(x)\) not exist?

2. At which values of \(a\) is \(f(a)\) not defined?

3. At which values of \(a\) does \(f\) have a limit, but \(\lim_{x \to a} f(x) \ne f(a)\text{?}\)

4. State all values of \(a\) for which \(f\) is not continuous at \(x = a\text{.}\)

5. Which condition is stronger, and hence implies the other: \(f\) has a limit at \(x = a\) or \(f\) is continuous at \(x = a\text{?}\) Explain, and hence complete the following sentence: “If \(f\) at \(x = a\text{,}\) then \(f\) at \(x = a\text{,}\)” where you complete the blanks with has a limit and is continuous, using each phrase once.

Hint

1. Consider the left- and right-hand limits at each value.

2. Carefully examine places on the graph where there's an open circle.

3. Are there locations on the graph where the function has a limit but there's a hole in the graph?

4. Remember that at least one of three conditions must fail: if the function lacks a limit, if the function is undefined, or if the limit exists but does not equal the function value, then \(f\) is not continuous at the point.

5. Note that the definition of being continuous requires the limit to exist.

Answer

1. \(a = -2\text{;}\) \(a = +2\text{.}\)

2. \(a = 3\text{.}\)

3. \(a = -1\text{;}\) \(a = 3\text{.}\)

4. \(a=-2\text{;}\) \(a = 2\text{;}\) \(a = 3\text{;}\) \(a = -1\text{.}\)

5. “If \(f\) is continuous at \(x = a\text{,}\) then \(f\) has a limit at \(x = a\text{.}\)”

Solution

1. \(\lim_{x \to a} f(x)\) does not exist at \(a = -2\) since \(\lim_{x \to -2^-} f(x) = 2 \ne -1 = \lim_{x \to -2^+}\) and \(\lim_{x \to a} f(x)\) does not exist at \(a = +2\) since \(\lim_{x \to 2^+} f(x)\) does not exist due to the infinitely oscillatory behavior of \(f\text{.}\)

2. The only point at which \(f\) is not defined is at \(a = 3\text{.}\)

3. At \(a = -1\text{,}\) note that \(\lim_{x \to -1} f(x)\) exists (and appears to have value approximately \(-3.25\)), but \(f(-1) = 1\text{,}\) and thus \(\lim_{x \to -1} f(x) \ne f(-1)\text{.}\) At \(a = 3\text{,}\) \(\lim_{x \to 3} f(x) = -2.5\text{,}\) but \(f(3)\) is not defined, so the limit exists but does not equal the function value.

4. Based on our work in (a), (b), and (c), \(f\) is not continuous at \(a=-2\) and \(a = 2\) because \(f\) does not have a limit at those points; \(f\) is not continuous at \(a = 3\) since \(f\) is not defined there; and \(f\) is not continuous at \(a = -1\) because at that point its limit does not equal its function value.

5. “If \(f\) is continuous at \(x = a\text{,}\) then \(f\) has a limit at \(x = a\text{,}\)” since one of the defining properties of “being continuous” at \(x = a\) is that the function has a limit at that input value. This shows that being continuous is a stronger condition than having a limit.

Subsection 1.4.3 Being differentiable at a point

We recall that a function \(f\) is said to be differentiable at \(x = a\) if \(f'(a)\) exists. Moreover, for \(f'(a)\) to exist, we know that the function \(y = f(x)\) must have a tangent line at the point \((a,f(a))\text{,}\) since \(f'(a)\) is precisely the slope of this line. In order to even ask if \(f\) has a tangent line at \((a,f(a))\text{,}\) it is necessary that \(f\) be continuous at \(x = a\text{:}\) if \(f\) fails to have a limit at \(x = a\text{,}\) if \(f(a)\) is not defined, or if \(f(a)\) does not equal the value of \(\lim_{x \to a} f(x)\text{,}\) then it doesn't make sense to talk about a tangent line to the curve at this point.

Indeed, it can be proved formally that if a function \(f\) is differentiable at \(x = a\text{,}\) then it must be continuous at \(x = a\text{.}\) So, if \(f\) is not continuous at \(x = a\text{,}\) then it is automatically the case that \(f\) is not differentiable there. For example, in Figure 1.4.5, both \(f\) and \(g\) fail to be differentiable at \(x = 1\) because neither function is continuous at \(x = 1\text{.}\) But can a function fail to be differentiable at a point where the function is continuous?

In Figure 1.4.8, the function has a sharp corner at a point. For the pictured function \(f\text{,}\) we observe that \(f\) is clearly continuous at \(a = 1\text{,}\) since \(\lim_{x \to 1} f(x) = 1 = f(1)\text{.}\)

But the function \(f\) in Figure 1.4.8 is not differentiable at \(a = 1\) because \(f'(1)\) fails to exist. One way to see this is to observe that \(f'(x) = -1\) for every value of \(x\) that is less than 1, while \(f'(x) = +1\) for every value of \(x\) that is greater than 1. That makes it seem that either \(+1\) or \(-1\) would be equally good candidates for the value of the derivative at \(x = 1\text{.}\) Alternately, we could use the limit definition of the derivative to attempt to compute \(f'(1)\text{,}\) and discover that the derivative does not exist. Finally, we can see visually that the function \(f\) in Figure 1.4.8 does not have a tangent line. When we zoom in on \((1,1)\) on the graph of \(f\text{,}\) no matter how closely we examine the function, it will always look like a “V”, and never like a single line, which tells us there is no possibility for a tangent line there.

If a function does have a tangent line at a given point, when we zoom in on the point of tangency, the function and the tangent line should appear essentially indistinguishable¹. Conversely, if we zoom in on a point and the function looks like a single straight line, then the function should have a tangent line there, and thus be differentiable. Hence, a function that is differentiable at \(x = a\) will, up close, look more and more like its tangent line at \((a,f(a))\text{.}\) Therefore, we say that a function that is differentiable at \(x = a\) is locally linear.

To summarize the preceding discussion of differentiability and continuity, we make several important observations.

Activity 1.4.4.

In this activity, we explore two different functions and classify the points at which each is not differentiable. Let \(g\) be the function given by the rule \(g(x) = |x|\text{,}\) and let \(f\) be the function that we have previously explored in Preview Activity 1.4.1, whose graph is given again in Figure 1.4.9.

1. Reasoning visually, explain why \(g\) is differentiable at every point \(x\) such that \(x \ne 0\text{.}\)

2. Use the limit definition of the derivative to show that \(g'(0) = \lim_{h \to 0} \frac{|h|}{h}\text{.}\)

3. Explain why \(g'(0)\) fails to exist by using small positive and negative values of \(h\text{.}\)

4. State all values of \(a\) for which \(f\) is not differentiable at \(x = a\text{.}\) For each, provide a reason for your conclusion.

5. True or false: if a function \(p\) is differentiable at \(x = b\text{,}\) then \(\lim_{x \to b} p(x)\) must exist. Why?

Hint

1. What type of function is \(g\) for all \(x \lt 0\text{?}\) For all \(x > 0\text{?}\)

2. Recall that \(g'(0) = \lim_{h \to 0} \frac{g(0 + h) - g(0)}{h}\text{.}\)

3. What is the value of \(|h|\) when \(h \lt 0\text{?}\)

4. You might start by identifying points where \(f\) is not continuous.

5. What does being differentiable at a point tell you about continuity there?

Answer

1. \(g\) is piecewise linear.

2. Show that

   \begin{equation*} g'(0) = \lim_{h \to 0} \frac{|h|}{h}\text{,} \end{equation*}

   which does not exist.

3. \(a = -3, -2, -1, 1, 2, 3\text{.}\)

4. True.

Solution

1. We know that \(g(x) = |x|\) is given by the formula \(g(x) = -x\) when \(x \lt 0\) and by \(g(x) = x\) when \(x \ge 0\text{.}\) Each of these pieces of \(g\) is a straight line, so at every point other than the point where they meet, the function \(g\) has a well-defined slope, and thus is differentiable.

2. Observe that

   \begin{align*} g'(0) =\mathstrut \amp \lim_{h \to 0} \frac{g(0+h)-g(0)}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|0+h|-|0|}{h}\\ =\mathstrut \amp \lim_{h \to 0} \frac{|h|}{h} \end{align*}

3. Following up on our work in (b), note that whenever \(h > 0\text{,}\) \(|h| = h\text{,}\) and thus

   \begin{equation*} \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\text{,} \end{equation*}

   while whenever \(h \lt 0\text{,}\) \(|h| = -h\text{,}\) and thus

   \begin{equation*} \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \end{equation*}

   Since the right- and left-hand limits are not equal, it follows that

   \begin{equation*} g'(0) = \lim_{h \to 0} \frac{|h|}{h} \end{equation*}

   does not exist.

4. \(f\) is not differentiable at \(a = -2, -1, 2, 3\) because at each of these points \(f\) is not continuous. In addition, \(f\) is not differentiable at \(a = -3\) and \(a = 1\) because the graph of \(f\) has a corner point (or cusp) at each of these values.

5. True: if a function \(p\) is differentiable at \(x = b\text{,}\) then \(\lim_{x \to b} p(x)\) must exist. This is true because we know that if \(p\) is differentiable at a point, then \(p\) is continuous there, and anytime a function is continuous at a point, it must have a limit there.

Subsection 1.4.4 Summary

• A function \(f\) has limit \(L\) as \(x \to a\) if and only if \(f\) has a left-hand limit at \(x = a\text{,}\) \(f\) has a right-hand limit at \(x = a\text{,}\) and the left- and right-hand limits are equal. Visually, this means that there can be a hole in the graph at \(x = a\text{,}\) but the function must approach the same single value from either side of \(x = a\text{.}\)

• A function \(f\) is continuous at \(x = a\) whenever \(f(a)\) is defined, \(f\) has a limit as \(x \to a\text{,}\) and the value of the limit and the value of the function agree. This guarantees that there is not a hole or jump in the graph of \(f\) at \(x = a\text{.}\)

• A function \(f\) is differentiable at \(x = a\) whenever \(f'(a)\) exists, which means that \(f\) has a tangent line at \((a,f(a))\) and thus \(f\) is locally linear at \(x = a\text{.}\) Informally, this means that the function looks like a line when viewed up close at \((a,f(a))\) and that there is not a corner point or cusp at \((a,f(a))\text{.}\)

• Of the three conditions discussed in this section (having a limit at \(x = a\text{,}\) being continuous at \(x = a\text{,}\) and being differentiable at \(x = a\)), the strongest condition is being differentiable, and the next strongest is being continuous. In particular, if \(f\) is differentiable at \(x = a\text{,}\) then \(f\) is also continuous at \(x = a\text{,}\) and if \(f\) is continuous at \(x = a\text{,}\) then \(f\) has a limit at \(x = a\text{.}\)