2.1 A Preview of Calculus

Big idea

Algebra and trigonometry handle quantities that hold still. Calculus handles quantities that change, and it does so with one move repeated everywhere: replace the thing you cannot compute with a sequence of things you can, then watch where the answers go. This section is a tour of that move before the machinery arrives.

Two old problems motivate everything that follows. The first is the tangent problem. You know how to find the slope of a line through two points on a curve, but the slope of the curve at a single point has no two points to work with. The fix is to take a second point nearby, compute the slope of the line through the pair, and then slide the second point toward the first. Each slope is an honest computation. The number those slopes approach, if there is one, is what you agree to call the slope at the point.

The second is the area problem. You know the area of a rectangle, and a region bounded by a curve is not one. The fix is the same in spirit: fill the region with rectangles whose areas you can add, then use more and thinner rectangles. Each total is an honest computation, and the number those totals approach is what you agree to call the area. The tangent problem gives you the derivative and the area problem gives you the integral, and the surprise of the course is that these two questions turn out to be inverse to each other.

Both fixes have the same shape: a quantity you want, a family of approximations that get better under some parameter, and a single number they close in on. That number is a limit. Nothing in this section computes a limit rigorously, because the definition comes later. What matters now is recognizing the pattern, and getting comfortable with the idea that a value can be perfectly well determined by what happens near a point even when nothing happens at it.

Decoder

A limit is the single number that a family of approximations closes in on as the approximation is refined, whether or not any one approximation ever equals it.

The phrase carrying the content is “closes in on.” Not “reaches” and not “gets somewhat near.” The approximations may never hit the limit; every secant slope you compute uses two distinct points, so none of them is the tangent slope. What the limit records is the destination, not an arrival.

The other loaded word is “single.” A family of approximations that wanders, or that heads to one number from the left and a different number from the right, has no limit. The whole theory rests on the destination being unambiguous, which is why the next sections spend so long on one-sided behavior and on functions that fail the test.

The classic mistake here is to read a limit as a value of the function. In the tangent calculation the difference quotient is undefined at the very point you are approaching: the two points collide and you divide by zero. That is not a flaw in the method, it is the reason the method exists. A limit describes the neighborhood of a point, and it deliberately ignores the point itself.

Definitions and results

Secant line. A secant line through a curve $y = f(x)$ is the straight line through two of its points, $(a, f(a))$ and $(x, f(x))$. Its slope is the difference quotient

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

Average rate of change. The same quotient, read as a rate, is the average rate of change of $f$ over the interval from $a$ to $x$: total change in output divided by total change in input. Its units are output units per input unit.

The increment form. Writing $x = a + h$ turns the quotient into $\frac{f(a+h) - f(a)}{h}$, and sliding $x$ toward $a$ becomes sending $h$ toward $0$. The two forms are the same computation; the increment form is usually easier to simplify.

Tangent line. The tangent line to $y = f(x)$ at $x = a$ is the line through $(a, f(a))$ whose slope is the limiting value of the secant slopes as the second point approaches the first. When that limiting value exists, the tangent is the line that best matches the curve near the point.

Instantaneous velocity. If $s(t)$ gives position at time $t$, the average velocity over an interval is the same difference quotient applied to $s$. The instantaneous velocity at a single instant is the value the average velocities approach as the interval shrinks to that instant.

Approximating an area. To estimate the area under a nonnegative curve from $x = a$ to $x = b$, cut the interval into $n$ equal pieces of width $\Delta x = (b-a)/n$, build a rectangle on each piece using a height taken from the curve, and add the areas. Using left endpoints and using right endpoints give two different estimates that bracket the truth when the curve is monotonic.

The area under a curve. The area is the number the rectangle totals approach as $n$ grows without bound and $\Delta x$ shrinks to zero. That number is the definite integral, developed in a later chapter.

Reading a limit from a table or graph. Evaluate the function at inputs marching toward the target from both sides, or trace the graph in from both sides. If both marches settle on the same number, that number is the limit. Evidence of this kind is persuasive, not conclusive; the algebraic and formal tools come next.

Worked examples

Secant slopes closing in on a tangent

Take $f(x) = x^2$ and the point $(2, 4)$. The secant slope through $(2,4)$ and a nearby point $(x, x^2)$ is

$$ \frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \qquad (x \ne 2) $$

At $x = 3$ the slope is $5$, at $x = 2.1$ it is $4.1$, at $x = 1.99$ it is $3.99$. From either side the slopes head for $4$, so the tangent line at $(2,4)$ has slope $4$ and equation $y = 4x - 4$.

Notice what the cancellation did. The original quotient is undefined at $x = 2$, since both parts are zero there. After cancelling, the expression $x + 2$ agrees with the quotient everywhere except at $x = 2$, and it is that agreement nearby, not at the point, that pins the value down.

Average velocity shrinking to an instant

A ball dropped from rest falls $s(t) = 5t^2$ meters in $t$ seconds. Its average velocity from $t = 2$ to $t = 2 + h$ is

$$ \frac{5(2+h)^2 - 5(2)^2}{h} = \frac{20 + 20h + 5h^2 - 20}{h} = 20 + 5h $$

Over the next full second the average is $25$ m/s; over the next tenth it is $20.5$ m/s; over the next thousandth it is $20.005$ m/s. Taking $h$ negative gives averages just under $20$. The instantaneous velocity at $t = 2$ is $20$ m/s.

Check the units and the sign. The quotient is meters divided by seconds, so $20$ is a speed, and it is positive because the ball is moving in the direction of increasing $s$. It is also larger than the average velocity over the whole first two seconds, which is $20/2 = 10$ m/s, exactly as it should be for something speeding up.

Rectangles under a parabola

Estimate the area under $y = x^2$ from $x = 0$ to $x = 1$ with four rectangles of width $1/4$. Using right endpoints, the heights are $(1/4)^2, (1/2)^2, (3/4)^2, 1^2$, so the total is

$$ \frac{1}{4}\left(\frac{1}{16} + \frac{4}{16} + \frac{9}{16} + \frac{16}{16}\right) = \frac{1}{4}\cdot\frac{30}{16} = \frac{15}{32} \approx 0.469 $$

Left endpoints use heights $0, (1/4)^2, (1/2)^2, (3/4)^2$ and give $\frac{1}{4}\cdot\frac{14}{16} = \frac{7}{32} \approx 0.219$. Since $x^2$ increases on this interval, the true area sits between the two.

With $n$ rectangles and right endpoints, the sum can be written in closed form as $\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}$. At $n = 4$ that is $0.469$, matching the hand computation, and as $n$ grows the two correction terms vanish. The area is $1/3$.

A limit read off a table

Let $g(x) = \dfrac{x^3 - 8}{x - 2}$. At $x = 2$ the formula is $0/0$, so the function has no value there, but nearby it does. Factoring gives $g(x) = x^2 + 2x + 4$ for $x \ne 2$, which is what the table shows:

At $x = 1.9$, $g = 11.41$; at $x = 1.99$, $g = 11.9401$; at $x = 2.01$, $g = 12.0601$; at $x = 2.1$, $g = 12.61$.

Both columns converge on $12$, and the factored form confirms it, since $2^2 + 2(2) + 4 = 12$. The gap in the domain at $x = 2$ is invisible to the limit.

Practice

Start with the difference quotient itself. These drills ask for the average rate of change of a function over an interval, and for the slope of the secant line that the rate describes.

Then read limits off pictures and tables. You are given a graph or a list of values and asked what the outputs approach as the input closes in from each side.

Quiz

Five items on average rates of change, secant and tangent slopes, and limits estimated from graphs and tables.