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The notion of limit

In we used a function, s(t), to model the location of a moving object at a given time.

Introduction

In we used a function, \(s(t)\), to model the location of a moving object at a given time. Functions can model other interesting phenomena, such as the rate at which an automobile consumes gasoline at a given velocity, or the reaction of a patient to a given dosage of a drug. We can use calculus to study how a function value changes in response to changes in the input variable.

Think about the falling ball whose position function is given by \(s(t) = 64 - 16t^2\). Its average velocity on the interval \([1,x]\) is given by \[\begin{aligned}\end{aligned}\].

Note that the average velocity is a function of \(x\). That is, the function \(g(x) = \frac{16 - 16x^2}{x-1}\) tells us the average velocity of the ball on the interval from \(t = 1\) to \(t = x\). To find the instantaneous velocity of the ball when \(t = 1\), we need to know what happens to \(g(x)\) as \(x\) gets closer and closer to \(1\). But also notice that \(g(1)\) is not defined, because it leads to the quotient \(0/0\).

This is where the notion of a limit comes in. By using a limit, we can investigate the behavior of \(g(x)\) as \(x\) gets arbitrarily close, but not equal, to \(1\). We first use the graph of a function to explore points where interesting behavior occurs.

Exploration

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 \(a\). To begin, think about a recent example.

In Preview Activity, 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 \(1\), not \(4\). 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, What follows here is not what mathematicians consider the formal definition of a limit. To be completely precise, it is necessary to quantify both what it means to say as close to \(L\) as we like and sufficiently close to \(a\). That can be accomplished through what is traditionally called the epsilon-delta definition of limits. The definition presented here is sufficient for the purposes of this text. we say the following.

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

\(a\)
\(a\)

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.

An important lesson to take from Example 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.

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

Instantaneous Velocity

Suppose that we have a moving object whose position at time \(t\) is given by a function \(s\). We know that the average velocity of the object on the time interval \([a,b]\) is \(AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}\). We define the instantaneous velocity at \(a\) to be the limit of average velocity as \(b\) approaches \(a\). Note particularly that as \(b \to a\), 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 \(t = a\), and thus \[\begin{aligned}\end{aligned}\].

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

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.

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 \(\lim_{x \to a} f(x) = L\), as the limit of \(f\) as \(x\) approaches \(a\) is \(L\), 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 \(a\).
To find \(\lim_{x \to a} f(x)\) for a given value of \(a\) and a known function \(f\), we can estimate this value from the graph of \(f\), or we can make a table of function values for \(x\)-values that are closer and closer to \(a\). 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 \(x \to a\).
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.

Practice (11)

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

Focusing on the ant, the \(x\)-value that the ant is directly above is the input value to the function, and the \(y\)-value at the ant's height (directly across from the ant) is the corresponding output value of the function. In the figure here, the current ant's position corresponds approximately to the point \((1.6, f(1.6)) = (1.6, 2.6).\)
If the ant is at the point on the graph directly above \(x=3\) what is the ant's height? In other words, what is the corresponding \(y\)-value, \(y = f(3)\)? State your answer in the form \(f(3) =\).
If the ant walks along the curve so that it gets closer and closer to the location where \(x = 2\), what happens to the \(y\)-value that corresponds to the ant's height? Does it matter if the ant approaches the location where \(x = 2\) from either the left side or the right side of \(x = 2\)? Discuss what you see on the graph near the point where \(x = 2\).
Finally, what happens if the ant walks in such a way that the corresponding \(x\)-value gets closer and closer to \(x = -1\)? What do you notice about the corresponding output values of the function near the point where \(x = -1\)?

Consider the function whose formula is \(f(x) = \frac{16-x^4}{x^2-4}\).

\(f\)
\(x\)\(a = 2\)\(\lim_{x \to 2} f(x)\)
\(\frac{16-x^4}{x^2-4}\)\(\lim_{x \to 2} f(x)\)
\(f(2) = -8\)
\(\frac{16-x^4}{x^2-4} = -4-x^2\)\(f\)
\(y = f(x)\)\([1,3]\)\(\lim_{x \to 2} \frac{16-x^4}{x^2-4}\)

Показати відповідь

\(f\) is defined for every value of \(x\) except those that make \(x^2 - 4 = 0\). Hence, the domain of \(f\) is the set of all real numbers except \(x = \pm 2\).

Using a spreadsheet, we generate the following table.

\(x\)	\(f(x)\)
\(2.1\)	\(-8.41\)
\(2.01\)	\(-8.0401\)
\(2.001\)	\(-8.004001\)
\(1.999\)	\(-7.996001\)
\(1.99\)	\(-7.9601\)
\(1.9\)	\(-7.61\)

The numerical trend suggests that \(\lim_{x \to 2} f(x) = -8\).

By factoring the numerator and denominator of \(f(x)\), we observe that \[\begin{aligned}\end{aligned}\]. Taking the limit as \(x \to 2\), we take values closer and closer, but not equal to 2, and therefore \[\begin{aligned}\end{aligned}\], since \(\frac{2-x}{x-2} = -1\) for all \(x \ne 2\). Finally, evaluating the last limit as \(x \to 2\), we see that \[\begin{aligned}\end{aligned}\].
False. In part (a), we saw that \(f(2)\) is not defined. We now know that \(f\) has a limit of \(-8\) as \(x\) approaches \(2\), but it is still the case that \(f(2)\) is undefined.
False. The noted equality only holds for \(x\)-values other than \(2\)or \(-2\). We can use this equality when evaluating the limit, since in so doing we never let \(x\) actually equal \(2\), but only approach \(2\).
Since \(f(2)\) is not defined but \(f\) has a limit of \(-8\) as \(x \to 2\), we see that the graph has a hole in it at the point \((2,-8)\), as seen in the following figure.

Let \(g(x) = -\frac{|x+3|}{x+3}\).

\(g\)
\(a = -3\)\(\lim_{x \to -3} g(x)\)
\(\frac{|x+3|}{x+3}\)\(\lim_{x \to -3} g(x)\)\(|a| = a\)\(a \ge 0\)\(|a| = -a\)\(a \lt 0\)
\(g(-3) = -1\)
\(-\frac{|x+3|}{x+3} = -1\)\(g\)
\(y = g(x)\)\([-4,-2]\)\(\lim_{x \to -3} g(x)\)

Показати відповідь

The domain of \(g\) is all real numbers except \(x = -3\).

Using a spreadsheet, we generate the following table.

\(x\)	\(g(x)\)
\(-2.9\)	\(-1\)
\(-2.99\)	\(-1\)
\(-2.999\)	\(-1\)
\(-3.001\)	\(1\)
\(-3.01\)	\(-1\)
\(-3.1\)	\(-1\)

As we approach \(-3\) from the right, the function values approach (stay constant at) \(1\); but as we approach \(-3\) from the left, the function values approach (stay constant at) \(-1\). As such, the function values do not get arbitrarily close to a single value, and hence the limit does not exist.

We observe that \(|x+3| = (x+3)\) whenever \(x \ge -3\), and that \(|x+3| = -(x+3)\) whenever \(x \lt -3\). So, if \(x \gt -3\), we see that \[\begin{aligned}\end{aligned}\] whereas if \(x \lt -3\), it follows that \[\begin{aligned}\end{aligned}\]. Because the function value is \(-1\) to the right of \(-3\) and \(+1\) to the left of \(-3\), it follows that the limit does not exist.
False, since \(-3\) does not belong to the domain of \(g\), \(g(-3)\) is undefined.
False. This equality is only true when \(x \gt -3\), as seen in our work in (c).
Since \(g\) approaches different output values as \(x\) approaches \(-3\) from different directions, \(g\) does not have a limit as \(x \to -3\), as seen in the following figure.

For each of the following prompts, sketch a graph on the provided axes of a function that has the stated properties.
1. \(y = f(x)\) such that
  - \(f(-2) = 2\)\(\lim_{x \to -2} f(x) = 1\)
  - \(f(-1) = 3\)\(\lim_{x \to -1} f(x) = 3\)
  - \(f(1)\)\(\lim_{x \to 1} f(x) = 0\)
  - \(f(2) = 1\)\(\lim_{x \to 2} f(x)\)
2. \(y = g(x)\) such that
  - \(g(-2) = 3\)\(g(-1) = -1\)\(g(1) = -2\)\(g(2) = 3\)
  - \(x = -2, -1, 1\)\(2\)\(g\)
  - \(g(0)\)\(\lim_{x \to 0} g(x)\)
Показати відповідь
1. One possible graph follows.
2. One possible graph follows.

A bungee jumper dives from a tower at time \(t=0\). Her height \(s\) in feet at time \(t\) in seconds is given by \(s(t) = 100\cos(0.75t) \cdot e^{-0.2t}+100\).

\([1,1+h]\)
\(h \to 0\)
What is the meaning of the value of the limit in (b)? What are its units?

Показати відповідь

Using the standard formula for average velocity, we see that \[\begin{aligned}AV_{[1,1+h]} &= \frac{s(1+h)-s(1)}{h} \\ &= \frac{(100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)}+100) - (100\cos(0.75) \cdot e^{-0.2}+100)}{h} \\ &= \frac{100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)}+100 - 100\cos(0.75) \cdot e^{-0.2}-100)}{h} \\ &= \frac{100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)} - 100\cos(0.75) \cdot e^{-0.2}}{h}\end{aligned}\]

Using a spreadsheet and small values of \(h\) in the expression \(AV_{[1,1+h]}\), we see that

\(h\)	\(AV_{[1,1+h]}\)
\(0.1\)	\(-54.50145823\)
\(0.01\)	\(-53.90917396\)
\(0.001\)	\(-53.84429251\)
\(0.0001\)	\(-53.83774757\)
\(-0.1\)	\(-53.83629174\)
\(-0.01\)	\(-53.82973418\)
\(-0.001\)	\(-53.76359015\)
\(-0.0001\)	\(-53.04508149\)

and hence \[\begin{aligned}\end{aligned}\].

Since \(\lim_{h \to 0} AV_{[1, 1+h]} \approx -53.837\) this tells us that the instantaneous velocity of the bungee jumper at the moment \(t = 1\) is approximately \(-53.837\) ft/sec.

Consider the function whose formula is \(f(x) = \frac{16-x^4}{x^2-4}\).

\(f\)
\(x\)\(a = 2\)\(\lim_{x \to 2} f(x)\)
\(\frac{16-x^4}{x^2-4}\)\(\lim_{x \to 2} f(x)\)
\(f(2) = -8\)
\(\frac{16-x^4}{x^2-4} = -4-x^2\)\(f\)
\(y = f(x)\)\([1,3]\)\(\lim_{x \to 2} \frac{16-x^4}{x^2-4}\)

Показати відповідь

\(f\) is defined for every value of \(x\) except those that make \(x^2 - 4 = 0\). Hence, the domain of \(f\) is the set of all real numbers except \(x = \pm 2\).

Using a spreadsheet, we generate the following table.

\(x\)	\(f(x)\)
\(2.1\)	\(-8.41\)
\(2.01\)	\(-8.0401\)
\(2.001\)	\(-8.004001\)
\(1.999\)	\(-7.996001\)
\(1.99\)	\(-7.9601\)
\(1.9\)	\(-7.61\)

The numerical trend suggests that \(\lim_{x \to 2} f(x) = -8\).

By factoring the numerator and denominator of \(f(x)\), we observe that \[\begin{aligned}\end{aligned}\]. Taking the limit as \(x \to 2\), we take values closer and closer, but not equal to 2, and therefore \[\begin{aligned}\end{aligned}\], since \(\frac{2-x}{x-2} = -1\) for all \(x \ne 2\). Finally, evaluating the last limit as \(x \to 2\), we see that \[\begin{aligned}\end{aligned}\].
False. In part (a), we saw that \(f(2)\) is not defined. We now know that \(f\) has a limit of \(-8\) as \(x\) approaches \(2\), but it is still the case that \(f(2)\) is undefined.
False. The noted equality only holds for \(x\)-values other than \(2\)or \(-2\). We can use this equality when evaluating the limit, since in so doing we never let \(x\) actually equal \(2\), but only approach \(2\).
Since \(f(2)\) is not defined but \(f\) has a limit of \(-8\) as \(x \to 2\), we see that the graph has a hole in it at the point \((2,-8)\), as seen in the following figure.

Let \(g(x) = -\frac{|x+3|}{x+3}\).

\(g\)
\(a = -3\)\(\lim_{x \to -3} g(x)\)
\(\frac{|x+3|}{x+3}\)\(\lim_{x \to -3} g(x)\)\(|a| = a\)\(a \ge 0\)\(|a| = -a\)\(a \lt 0\)
\(g(-3) = -1\)
\(-\frac{|x+3|}{x+3} = -1\)\(g\)
\(y = g(x)\)\([-4,-2]\)\(\lim_{x \to -3} g(x)\)

Показати відповідь

The domain of \(g\) is all real numbers except \(x = -3\).

Using a spreadsheet, we generate the following table.

\(x\)	\(g(x)\)
\(-2.9\)	\(-1\)
\(-2.99\)	\(-1\)
\(-2.999\)	\(-1\)
\(-3.001\)	\(1\)
\(-3.01\)	\(-1\)
\(-3.1\)	\(-1\)

We observe that \(|x+3| = (x+3)\) whenever \(x \ge -3\), and that \(|x+3| = -(x+3)\) whenever \(x \lt -3\). So, if \(x \gt -3\), we see that \[\begin{aligned}\end{aligned}\] whereas if \(x \lt -3\), it follows that \[\begin{aligned}\end{aligned}\]. Because the function value is \(-1\) to the right of \(-3\) and \(+1\) to the left of \(-3\), it follows that the limit does not exist.
False, since \(-3\) does not belong to the domain of \(g\), \(g(-3)\) is undefined.
False. This equality is only true when \(x \gt -3\), as seen in our work in (c).
Since \(g\) approaches different output values as \(x\) approaches \(-3\) from different directions, \(g\) does not have a limit as \(x \to -3\), as seen in the following figure.

For each of the following prompts, sketch a graph on the provided axes of a function that has the stated properties.
1. \(y = f(x)\) such that
  - \(f(-2) = 2\)\(\lim_{x \to -2} f(x) = 1\)
  - \(f(-1) = 3\)\(\lim_{x \to -1} f(x) = 3\)
  - \(f(1)\)\(\lim_{x \to 1} f(x) = 0\)
  - \(f(2) = 1\)\(\lim_{x \to 2} f(x)\)
2. \(y = g(x)\) such that
  - \(g(-2) = 3\)\(g(-1) = -1\)\(g(1) = -2\)\(g(2) = 3\)
  - \(x = -2, -1, 1\)\(2\)\(g\)
  - \(g(0)\)\(\lim_{x \to 0} g(x)\)
Показати відповідь
1. One possible graph follows.
2. One possible graph follows.

A bungee jumper dives from a tower at time \(t=0\). Her height \(s\) in feet at time \(t\) in seconds is given by \(s(t) = 100\cos(0.75t) \cdot e^{-0.2t}+100\).

\([1,1+h]\)
\(h \to 0\)
What is the meaning of the value of the limit in (b)? What are its units?

Показати відповідь

Using the standard formula for average velocity, we see that \[\begin{aligned}AV_{[1,1+h]} &= \frac{s(1+h)-s(1)}{h} \\ &= \frac{(100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)}+100) - (100\cos(0.75) \cdot e^{-0.2}+100)}{h} \\ &= \frac{100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)}+100 - 100\cos(0.75) \cdot e^{-0.2}-100)}{h} \\ &= \frac{100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)} - 100\cos(0.75) \cdot e^{-0.2}}{h}\end{aligned}\]

Using a spreadsheet and small values of \(h\) in the expression \(AV_{[1,1+h]}\), we see that

\(h\)	\(AV_{[1,1+h]}\)
\(0.1\)	\(-54.50145823\)
\(0.01\)	\(-53.90917396\)
\(0.001\)	\(-53.84429251\)
\(0.0001\)	\(-53.83774757\)
\(-0.1\)	\(-53.83629174\)
\(-0.01\)	\(-53.82973418\)
\(-0.001\)	\(-53.76359015\)
\(-0.0001\)	\(-53.04508149\)

and hence \[\begin{aligned}\end{aligned}\].

Since \(\lim_{h \to 0} AV_{[1, 1+h]} \approx -53.837\) this tells us that the instantaneous velocity of the bungee jumper at the moment \(t = 1\) is approximately \(-53.837\) ft/sec.

Symbols used here

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

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

\sin,\ \cos,\ \tan

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

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

\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…

\infty

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

\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: The notion of limit

What is the mathematical notion of limit and what role do limits play in the study of functions?
What is the meaning of the notation \lim_{x \to a} f(x) = L?
How do we go about determining the value of the limit of a function at a point?
How do we manipulate average velocity to compute instantaneous velocity?

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.

Спробуйте власну

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

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