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Limits, continuity, and differentiability

In Section, we learned how limits can be used to study the trending behavior of a function near a fixed input value.

Introduction

In Section, we learned how limits can be used to study the trending behavior of a function near a fixed input value. In this section, we aim to quantify how the function acts and how its values change near a particular point. If the function has a limit \(L\) at \(x = a\), we will consider how the value of the function \(f(a)\) is related to \(\lim_{x \to a} f(x)\), and whether or not the function has a derivative \(f'(a)\) at \(x = a\).

Exploration

Having a limit at a point

In Section, 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) \(a\). If so, we write \(\lim_{x \to a} f(x) = L\).

Essentially there are two behaviors that a function can exhibit near a point where it fails to have a limit. In Figure, at left we see a function \(f\) whose graph shows a jump at \(a = 1\). 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 \(a = 1\).

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

In the graph of the function \(f\) in Figure, we see that \[\begin{aligned}\end{aligned}\]. 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, 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 \(a = 1\), 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 \(a = 1\), we say that the right-hand limit of \(g\) as \(x \to 1^+\) does not exist, and thus \(\lim_{x \to 1} g(x) \ \text{does not exist}\).

A function \(f\) has limit \(L\) as \(x \to a\) if and only if \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

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 we consider three functions that have a limit at \(a = 1\), 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 \(a = 1\). We will naturally say that \(f\) is not continuous at \(a = 1\). For the function \(g\), we observe that while \(\lim_{x \to 1} g(x) = 3\), the value of \(g(1) = 2\), 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 \(a = 1\). 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, \[\begin{aligned}\end{aligned}\].

With no hole or jump in the graph of \(h\) at \(a = 1\), 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 \(x = a\). If a function is continuous at every point in an interval \([a,b]\), we say the function is continuous on \([a,b]\). 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 \(p(x) = x^2 - 2x + 3\). It can be proved that every polynomial is a continuous function at every real number, and thus if we would like to know \(\lim_{x \to 2} p(x)\), we simply compute \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

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 \((a,f(a))\), since \(f'(a)\) is precisely the slope of this line. In order to even ask if \(f\) has a tangent line at \((a,f(a))\), it is necessary that \(f\) be continuous at \(x = a\): if \(f\) fails to have a limit at \(x = a\), if \(f(a)\) is not defined, or if \(f(a)\) does not equal the value of \(\lim_{x \to a} f(x)\), 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 \(x = a\), then it must be continuous at \(x = a\). So, if \(f\) is not continuous at \(x = a\), then it is automatically the case that \(f\) is not differentiable there. For example, in Figure, both \(f\) and \(g\) fail to be differentiable at \(x = 1\) because neither function is continuous at \(x = 1\). But can a function fail to be differentiable at a point where the function is continuous?

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

But the function \(f\) in Figure 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 \(x = 1\). Alternately, we could use the limit definition of the derivative to attempt to compute \(f'(1)\), and discover that the derivative does not exist. Finally, we can see visually that the function \(f\) in Figure does not have a tangent line. When we zoom in on \((1,1)\) on the graph of \(f\), 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.

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

Condensed — the full section is in Boelkins, Active Calculus.

Summary

A function \(f\) has limit \(L\) as \(x \to a\) if and only if \(f\) has a left-hand limit at \(x = a\), \(f\) has a right-hand limit at \(x = a\), and the left- and right-hand limits are equal. Visually, this means that there can be a hole in the graph at \(x = a\), but the function must approach the same single value from either side of \(x = a\).
A function \(f\) is continuous at \(x = a\) whenever \(f(a)\) is defined, \(f\) has a limit as \(x \to a\), 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 \(x = a\).
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 \(x = a\). 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 \((a,f(a))\).
Of the three conditions discussed in this section (having a limit at \(x = a\), being continuous at \(x = a\), 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 \(x = a\), then \(f\) is also continuous at \(x = a\), and if \(f\) is continuous at \(x = a\), then \(f\) has a limit at \(x = a\).

Practice (11)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Imagine that you are tracking the length of your hair as it grows. Its growth is roughly linear at a rate of 0.4 mm per day. Suppose your hair is 267 mm at the start of this tracking experiment and that after 16 weeks (112 days), you decide to get a haircut. The stylist cuts off 50 mm. Let \(L(t)\) be the function whose output is the length of your hair after \(t\) days.
1. How long was your hair when you entered the salon?
2. How long is your hair when you leave the salon?
3. Sketch a possible graph of \(L(t)\) on the time interval \(0 \leq t \leq 200\).
4. Describe what happens to the graph at the moment the haircut happens.
Explain the situation in the preceding question by using limits.
1. The limit from the left side as \(t\) gets close to \(112\): \[\begin{aligned}\end{aligned}\]
2. The limit from the right side as \(t\) gets close to 112: \[\begin{aligned}\end{aligned}\]
3. What is \(L(112)\)?
4. Describe the graph of \(L(t)\) near \(t=112\).
Come up with two of your own realistic examples of functions that have a jump or a break in their graph.
Consider the graph of the function \(y = p(x)\) that is provided in Figure. Assume that each portion of the graph of \(p\) is a straight line, as pictured.
1. State all values of \(a\) for which \(\lim_{x \to a} p(x)\) does not exist.
2. State all values of \(a\) for which \(p\) is not continuous at \(a\).
3. State all values of \(a\) for which \(p\) is not differentiable at \(x = a\).
4. On the axes provided in Figure, sketch an accurate graph of \(y = p'(x)\).
Avslöja svaret
1. The only value of \(a\) at which \(\lim_{x \to a} p(x)\) does not exist is \(a = 0\). This is because \(\lim_{x \to 0^-} p(x) = 0 \ne -2 = \lim_{x \to 0^+} p(x)\).
2. \(p\) is not continuous at \(a = 0\) because \(\lim_{x \to 0} p(x)\) does not exist. In addition, \(p\) is not continuous at \(a=3\) because \(p(3)\) is not defined.
3. \(p\) is not differentiable at \(a = 0\) and at \(a = 3\) since \(p\) is not continuous at these points. In addition, \(p\) is not differentiable at \(a = -2, 1, 2\) since \(p\) has corner points at these values, making \(p\) not have a tangent line at these values.
4. Note that \(p\) is piecewise linear, so the value of the derivative of \(p\) (where it exists) is the slope of \(p\), which will be constant on each interval where \(p\) is linear. In addition, note that \(p'\) is not defined at \(a = -2, 0, 1, 2, 3\), which is reflected by open circles in the following plot.
For each of the following prompts, give an example of a function that satisfies the stated criteria. A formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why.
1. A function \(f\) that is continuous at \(a = 2\) but not differentiable at \(a = 2\).
2. A function \(g\) that is differentiable at \(a = 3\) but does not have a limit at \(a=3\).
3. A function \(h\) that has a limit at \(a = -2\), is defined at \(a = -2\), but is not continuous at \(a = -2\).
4. A function \(p\) that satisfies all of the following:
  - \(p(-1) = 3\) and \(\lim_{x \to -1} p(x) = 2\)
  - \(p(0) = 1\) and \(p'(0) = 0\)
  - \(\lim_{x \to 1} p(x) = p(1)\) and \(p'(1)\) does not exist
Avslöja svaret
1. The function \(f(x) = |x-2|\) is continuous at \(a = 2\), because \(f(2) = 0 = \lim_{x \to 2} f(x)\), but not differentiable at \(a = 2\) because \(f\) has a corner point at \(a = 2\).
2. This is impossible. Any function that is differentiable at a point must have a limit at the point.
3. Let \(f\) be the function defined to be \(f(x) = 1\) for every value of \(x \ne -2\), and such that \(f(-2) = 4\). This function has a limit of \(1\) as \(x \to -2\), but since \(f(-2) = 4 \ne 1\), the function is not continus at \(a = -2\).
4. A possible function \(p\) that has the desired properties is shown in the following figure.
Let \(h(x)\) be a function whose derivative \(y= h'(x)\) is given by the graph on the right in Figure.
1. Based on the graph of \(y = h'(x)\), what can you say about the behavior of the function \(y = h(x)\)?
2. At which values of \(x\) is \(y = h'(x)\) not defined? What behavior does this lead you to expect to see in the graph of \(y=h(x)\)?
3. Is it possible for \(y = h(x)\) to have points where \(h\) is not continuous? Explain your answer.
4. On the axes provided at left, sketch at least two distinct graphs that are possible functions \(y = h(x)\) that each have a derivative \(y = h'(x)\) that matches the provided graph at right. Explain why there are multiple possibilities for \(y = h(x)\).
Avslöja svaret
1. The function \(h\) must be piecewise linear since its derivative is piecewise constant. That is, wherever \(h'\) is constant, \(h\) must be linear. In particular, \(h\) must be piecewise linear with slope of \(1\) or \(-1\), depending on the interval.
2. \(h'(x)\) is not defined for \(x = -2, 0, 2\), as reflected by the open circles on the graph of \(h'\).
3. It is possible that \(h\) is not continuous at \(x = -2, 0, 2\), since \(h\) is not differentiable at these points; everywhere else, \(h\) must be continuous.
4. There are many different possible graphs for \(h\). Two of them are shown in the following figure. The blue graph is one possible version of \(h\) that is continuous. Note that we can vertically shift that graph and its derivative will still be the given \(h'\). Moreover, we can vertically shift any of the piecewise linear portions of \(h\) to generate a graph such as the one shown in red, which also has derivative \(h'\).

Consider the function \(g(x) = \sqrt{|x|}\).

Use a graph to explain visually why \(g\) is not differentiable at \(x = 0\).
Use the limit definition of the derivative to show that \[\begin{aligned}\end{aligned}\].
Investigate the value of \(g'(0)\) by estimating the limit in (b) using small positive and negative values of \(h\). For instance, you might compute \(\frac{\sqrt{|-0.01|}}{0.01}\). Be sure to use several different values of \(h\) (both positive and negative), including ones closer to 0 than 0.01. What do your results tell you about \(g'(0)\)?
Use your graph in (a) to sketch an approximate graph of \(y = g'(x)\).

Avslöja svaret

As seen at left in the figure below in part (d), \(g\) has a sharp corner point at \(x = 0\), which makes \(g\) not differentiable at this point.
By definition, we know \[\begin{aligned}g'(0) & = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} \\ & = \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ & = \lim_{h \to 0} \frac{\sqrt{|h|} - \sqrt{|0|}}{h} \\ & = \lim_{h \to 0} \frac{\sqrt{|h|}}{h}\end{aligned}\]

In the following table, we record the values of \(\frac{\sqrt{|h|}}{h}\) for small values of \(h\).

\(h\)	\(0.1\)	\(0.01\)	\(0.001\)	\(0.0001\)	\(-0.1\)	\(-0.01\)	\(-0.001\)	\(-0.0001\)
\(\sqrt{\|h\|}/h\)	\(3.162\)	\(10\)	\(31.62\)	\(100\)	\(-3.162\)	\(-10\)	\(-31.62\)	\(-100\)

These results show from a different perspective that \(g'(0)\) does not exist. For small negative \(h\) values, the difference quotient is negative and getting more negative without bound. For small positive values of \(h\), the difference quotient is positive and increasing without bound. This shows that \(\lim_{h \to 0} \frac{\sqrt{|h|}}{h}\) does not exist.

An approximate graph of \(g'\) is shown at right in the figure below.

Consider the graph of the function \(y = p(x)\) that is provided in Figure. Assume that each portion of the graph of \(p\) is a straight line, as pictured.
1. State all values of \(a\) for which \(\lim_{x \to a} p(x)\) does not exist.
2. State all values of \(a\) for which \(p\) is not continuous at \(a\).
3. State all values of \(a\) for which \(p\) is not differentiable at \(x = a\).
4. On the axes provided in Figure, sketch an accurate graph of \(y = p'(x)\).
Avslöja svaret
1. The only value of \(a\) at which \(\lim_{x \to a} p(x)\) does not exist is \(a = 0\). This is because \(\lim_{x \to 0^-} p(x) = 0 \ne -2 = \lim_{x \to 0^+} p(x)\).
2. \(p\) is not continuous at \(a = 0\) because \(\lim_{x \to 0} p(x)\) does not exist. In addition, \(p\) is not continuous at \(a=3\) because \(p(3)\) is not defined.
3. \(p\) is not differentiable at \(a = 0\) and at \(a = 3\) since \(p\) is not continuous at these points. In addition, \(p\) is not differentiable at \(a = -2, 1, 2\) since \(p\) has corner points at these values, making \(p\) not have a tangent line at these values.
4. Note that \(p\) is piecewise linear, so the value of the derivative of \(p\) (where it exists) is the slope of \(p\), which will be constant on each interval where \(p\) is linear. In addition, note that \(p'\) is not defined at \(a = -2, 0, 1, 2, 3\), which is reflected by open circles in the following plot.
For each of the following prompts, give an example of a function that satisfies the stated criteria. A formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why.
1. A function \(f\) that is continuous at \(a = 2\) but not differentiable at \(a = 2\).
2. A function \(g\) that is differentiable at \(a = 3\) but does not have a limit at \(a=3\).
3. A function \(h\) that has a limit at \(a = -2\), is defined at \(a = -2\), but is not continuous at \(a = -2\).
4. A function \(p\) that satisfies all of the following:
  - \(p(-1) = 3\) and \(\lim_{x \to -1} p(x) = 2\)
  - \(p(0) = 1\) and \(p'(0) = 0\)
  - \(\lim_{x \to 1} p(x) = p(1)\) and \(p'(1)\) does not exist
Avslöja svaret
1. The function \(f(x) = |x-2|\) is continuous at \(a = 2\), because \(f(2) = 0 = \lim_{x \to 2} f(x)\), but not differentiable at \(a = 2\) because \(f\) has a corner point at \(a = 2\).
2. This is impossible. Any function that is differentiable at a point must have a limit at the point.
3. Let \(f\) be the function defined to be \(f(x) = 1\) for every value of \(x \ne -2\), and such that \(f(-2) = 4\). This function has a limit of \(1\) as \(x \to -2\), but since \(f(-2) = 4 \ne 1\), the function is not continus at \(a = -2\).
4. A possible function \(p\) that has the desired properties is shown in the following figure.
Let \(h(x)\) be a function whose derivative \(y= h'(x)\) is given by the graph on the right in Figure.
1. Based on the graph of \(y = h'(x)\), what can you say about the behavior of the function \(y = h(x)\)?
2. At which values of \(x\) is \(y = h'(x)\) not defined? What behavior does this lead you to expect to see in the graph of \(y=h(x)\)?
3. Is it possible for \(y = h(x)\) to have points where \(h\) is not continuous? Explain your answer.
4. On the axes provided at left, sketch at least two distinct graphs that are possible functions \(y = h(x)\) that each have a derivative \(y = h'(x)\) that matches the provided graph at right. Explain why there are multiple possibilities for \(y = h(x)\).
Avslöja svaret
1. The function \(h\) must be piecewise linear since its derivative is piecewise constant. That is, wherever \(h'\) is constant, \(h\) must be linear. In particular, \(h\) must be piecewise linear with slope of \(1\) or \(-1\), depending on the interval.
2. \(h'(x)\) is not defined for \(x = -2, 0, 2\), as reflected by the open circles on the graph of \(h'\).
3. It is possible that \(h\) is not continuous at \(x = -2, 0, 2\), since \(h\) is not differentiable at these points; everywhere else, \(h\) must be continuous.
4. There are many different possible graphs for \(h\). Two of them are shown in the following figure. The blue graph is one possible version of \(h\) that is continuous. Note that we can vertically shift that graph and its derivative will still be the given \(h'\). Moreover, we can vertically shift any of the piecewise linear portions of \(h\) to generate a graph such as the one shown in red, which also has derivative \(h'\).

Consider the function \(g(x) = \sqrt{|x|}\).

Use a graph to explain visually why \(g\) is not differentiable at \(x = 0\).
Use the limit definition of the derivative to show that \[\begin{aligned}\end{aligned}\].
Investigate the value of \(g'(0)\) by estimating the limit in (b) using small positive and negative values of \(h\). For instance, you might compute \(\frac{\sqrt{|-0.01|}}{0.01}\). Be sure to use several different values of \(h\) (both positive and negative), including ones closer to 0 than 0.01. What do your results tell you about \(g'(0)\)?
Use your graph in (a) to sketch an approximate graph of \(y = g'(x)\).

Avslöja svaret

As seen at left in the figure below in part (d), \(g\) has a sharp corner point at \(x = 0\), which makes \(g\) not differentiable at this point.
By definition, we know \[\begin{aligned}g'(0) & = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} \\ & = \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ & = \lim_{h \to 0} \frac{\sqrt{|h|} - \sqrt{|0|}}{h} \\ & = \lim_{h \to 0} \frac{\sqrt{|h|}}{h}\end{aligned}\]

In the following table, we record the values of \(\frac{\sqrt{|h|}}{h}\) for small values of \(h\).

\(h\)	\(0.1\)	\(0.01\)	\(0.001\)	\(0.0001\)	\(-0.1\)	\(-0.01\)	\(-0.001\)	\(-0.0001\)
\(\sqrt{\|h\|}/h\)	\(3.162\)	\(10\)	\(31.62\)	\(100\)	\(-3.162\)	\(-10\)	\(-31.62\)	\(-100\)

An approximate graph of \(g'\) is shown at right in the figure below.

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Limits, continuity, and differentiability

What does it mean graphically to say that f has limit L as x \to a? How is this connected to having a left-hand limit at x = a and having a right-hand limit at x = a?
What does it mean to say that a function f is continuous at x = a? What role do limits play in determining whether or not a function is continuous at a point?
What does it mean graphically to say that a function f is differentiable at x = a? How is this connected to the function being locally linear?
How are the characteristics of a function having a limit, being continuous, and being differentiable at a given point related to one another?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Mer information Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests