Chapter 1

Limits and Continuity

Calculus is the mathematical study of continuous change, motion, and the behaviour of functions. This chapter covers the engine that makes all of it possible: the limit.

Lesson 1 · The idea

Getting halfway there, forever

Stand exactly 10 metres from a wall. Every step you take covers half of whatever distance is left. Intuitively the person should eventually get there — but getting halfway there every time will never allow them to arrive.

Figure 1Drag the slider to advance the trajectory, or press Auto-Play. The leaps become infinitesimally small, and the distance remaining never reaches zero.

After $n$ jumps the position is $P(n) = 10\left[1 - \left(\tfrac{1}{2}\right)^n\right]$ . The gap never closes. But watch the number in the readout: it is not wandering. There is exactly one value the position is heading for, and it never gets there.

\lim_{n \to \infty} 10\left[1 - \left(\tfrac{1}{2}\right)^n\right] = 10

Limit: The value a function approaches as its input approaches some target. It is a statement about the trip, not about the destination — the function does not have to arrive, and often does not.

Lesson 1 · Two directions at once

The same behaviour, without the wall

This applies in a lot of cases in math, not just a person running at a wall. Take $f(x) = \frac{1}{x}$ . We cannot evaluate it at $x = 0$ , because dividing by zero is undefined — but we can describe exactly what happens as the input travels infinitesimally close to it.

Figure 2Slide the point left and the output skyrockets. Slide it right and the output shrinks toward nothing. The curve approaches both axes and touches neither.

One function, two limits, pointing in different directions:

As

x

shrinks toward zero

\lim_{x \to 0^+} \frac{1}{x} = +\infty

As

x

grows without bound

\lim_{x \to \infty} \frac{1}{x} = 0

The first is a vertical asymptote, the second a horizontal one. Both are limits. Neither is a value the function ever actually takes.

Lesson 1 · Method 1

Direct Substitution

Infinity does not always have to be involved with the limit. There are relatively normal cases, and most limits you meet are one of them.

Figure 3Drag the slider toward

x = 10

from either side. There are no breaks in the curve, so both approaches land on the same point — the one the function is already sitting on.

You can see that it is approaching a point at $x = 10$ . Because nothing is broken there, we can solve for it by substituting straight into the function.

Evaluating the limit1 / 5

Write the limit $\lim_{x \to 10} \left(-0.04x^3 + 1.5x^2 - 12x + 50\right)$

Direct Substitution: Plug the target in and read off the answer. It works whenever the function is continuous at that point — no hole, no jump, no asymptote — which makes it the first thing to try every single time.

Notation you will meet soon

Writing the whole polynomial inside every limit gets long fast. Once a function has a name, the limit can use the name instead.

f(x) = -0.04x^3 + 1.5x^2 - 12x + 50

Everything above is unchanged; it is the same substitution, written shorter:

\lim_{x \to 10} f(x) = f(10) = 40

You will see limits written this way in every textbook from here on. It is worth getting used to now rather than later.

Lesson 1 · Method 2

When substitution fails: the hole

Great — that was easy. But sometimes the limit is not as friendly. Substitute into this one and the whole thing collapses.

Figure 4The function behaves exactly like the line

y = x + 5

everywhere except at

x = 5

, where it simply stops existing. Both approaches still point at 10.

Evaluating the limit1 / 5

Write the limit $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$

Removable Discontinuity: Sometimes there are just holes in the function, because the function approaches a point that makes $\frac{0}{0}$ , which is undefined. Cancelling the shared factor removes the hole and leaves the function that was underneath it all along.

Lesson 1 · Method 3

When factoring will not get you out

Radicals produce the same $\frac{0}{0}$ , but there is nothing to factor. The move is to multiply by the conjugate — the same expression with the sign flipped — which turns the radical into the thing it is the square root of.

Figure 5Drag the slider to

x = 0

. The denominator is zero there, so a literal hole interrupts an otherwise smooth, slowly decaying curve.

Evaluating the limit1 / 8

Write the limit $\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$

Rationalizing the Numerator: Radicals that produce an indeterminate form require multiplication by the conjugate to simplify the expression and eliminate the discontinuity.

Lesson 2 · One-sided limits

When the rule changes partway through

Sometimes, functions are not always the same function across the entire domain. The function can be different pieces. We call these piecewise functions, and they need a different approach.

Figure 6Drag the slider across

x = 3

. The rule underneath changes from a parabola to a straight line, and the readout says which piece is in force — but the two paths meet at the same coordinate.

Because the mathematical rules change exactly at $x = 3$ , we cannot substitute 3 into just one side and assume it applies to the whole function. We have to check each approach independently.

From the left1 / 3

The rule below 3 is $x^2 - 5$ $\lim_{x \to 3^-} \left(x^2 - 5\right)$

From the right1 / 3

The rule at and above 3 is $2x - 2$ $\lim_{x \to 3^+} (2x - 2)$

Because the left-hand limit and the right-hand limit are identical, the two halves of the function connect. The general limit exists:

\lim_{x \to 3} f(x) = 4

Lesson 2 · Disagreement

The jump discontinuity

Change one of those pieces and the agreement falls apart. Both one-sided limits still exist, and both are perfectly ordinary numbers — they simply point at different places.

Figure 7The open circle at

y = 9

says the parabola never reaches that height. The solid dot at

y = 1

is the value the function actually takes. Between them is the jump.

From the left1 / 3

The rule below 3 is $x^2$ $\lim_{x \to 3^-} x^2$

From the right1 / 3

The rule at and above 3 is $-x + 4$ $\lim_{x \to 3^+} (-x + 4)$

Because $9 \neq 1$ , the approaches never agree on a destination. There is no single value the function is heading for, so there is no limit to report.

\lim_{x \to 3} f(x) = \text{DNE}

DNE: Does Not Exist. Worth being precise about: it does not mean the function is broken or that the two sides are missing. Both one-sided limits here exist and are finite. What does not exist is a single number that both of them agree on.

Lesson 3 · The Squeeze Theorem

Trapping a function that will not settle

Some functions oscillate so violently that they never settle on a value at all, and both substitution and algebra fail. $f(x) = x^2\sin\!\left(\frac{1}{x}\right)$ is one of them: as $x$ gets infinitely small, $\frac{1}{x}$ gets infinitely large, and the sine swings between $-1$ and $1$ endlessly.

Figure 8The blue function is trapped between

g(x) = x^2

above and

h(x) = -x^2

below. Drag toward

x = 0

and watch the two bounds crush the oscillation into a single point.

If we cannot follow the function, we can fence it in. The trick is that we already know something about sine no matter what you feed it.

Squeezing the limit1 / 5

Sine is bounded, whatever its input $-1 \le \sin\left(\frac{1}{x}\right) \le 1$

The Squeeze Theorem: If one function is trapped between two others near a point, and both of those converge to the same value, the trapped function is forced to that value too. It is the one method here that finds a limit without ever evaluating the function.

Lesson 3 · Infinite limits

When the hole cannot be fixed

A limit yielding $\frac{0}{0}$ implies a hidden hole that can usually be fixed. A limit yielding a non-zero number over zero — $\frac{5}{0}$ , say — cannot be. It is a vertical asymptote, and the function does not approach a value there; it runs away.

Figure 9Drag the slider toward the boundary and watch the denominator readout. On the left it is microscopically negative; on the right, microscopically positive. Same numerator, opposite explosions.

Diagnosing the boundary1 / 2

Write the limit $\lim_{x \to 2} \frac{x + 3}{x - 2}$

The $\frac{5}{0}$ confirms a vertical asymptote. It does not say which way. For that, test values microscopically either side and look only at the signs.

From the left, at

x = 1.999

1 / 2

Positive over a microscopic negative $\frac{+4.999}{-0.001}$

From the right, at

x = 2.001

1 / 2

Positive over a microscopic positive $\frac{+5.001}{+0.001}$

One side plummets, the other skyrockets, and there is no consensus:

\lim_{x \to 2} f(x) = \text{DNE}

Lesson 4 · The definition

The three-step continuity checklist

A function is continuous at $x = c$ if there are no holes, jumps, or vertical asymptotes at that exact coordinate. In calculus we do not rely on the informal "can you draw it without lifting your pencil" version. Continuity is tested against three strict conditions, and failing any one of them is enough.

Figure 10The figure cycles through the three ways a function can fail. Click any scenario to take control and hold it there — the checklist beside the graph marks which condition broke.

Testing

f(x)

at a target

c

Does $f(c)$ exist?
You must be able to plug $c$ in and get a real number out. Division by zero or a negative under a square root means the point is not there, and the checklist fails immediately.
FailsFor $f(x) = \frac{x^2 - 9}{x - 3}$ at $c = 3$ , evaluating gives $\frac{0}{0}$ . There is no point to test.
Does $\lim_{x \to c} f(x)$ exist?
The destination has to be agreed on. Approach from the left, approach from the right, and compare. If the two disagree, the function jumps and the general limit does not exist.
FailsA piecewise function whose left side approaches 6 and whose right side approaches 2.
Does $f(c) = \lim_{x \to c} f(x)$ ?
This is the lock. The point that exists has to be the point the curve is aiming at. Both halves can pass on their own and still fail together.
FailsThe limit points at $y = 6$ , but a piecewise rule puts the solid dot at $y = 2$ . The hole stays unfilled.

PassesAll three, together:

\lim_{x \to 3} x^2 = 9

and

f(3) = 9

. The point plugs the curve exactly, so the function is continuous there.

Lesson 4 · Check yourself

Concept check

Two levels. The first covers the definitions; the second is the edge cases, which is where most of the marks are actually lost. Answer, read the explanation whether you were right or not, and try both.

Figure 11Limits and continuity: concept check

Lesson 5 · One last thing

The Intermediate Value Theorem

We are almost done with this chapter. One last thing. A car drives from point A to point B, and a car moves like a continuous function — no teleporting. So if I know it was at A and I know it was at B, I can guarantee it was at every position in between, whether or not anybody was watching.

Figure 12The car starts at

f(0) = 10

and finishes near

f(10) \approx 81.8

. Drag the slider to find the moment it crosses 50 — the theorem promises the moment is there before you go looking for it.

The theorem is an existence guarantee, and that is all it is. It will not tell you where the crossing happens. It uses the unbroken nature of a continuous function to prove that the crossing has to happen somewhere.

To invoke the IVT

$f$ is continuous on the closed interval $[a, b]$
Continuity is the entire engine. Without it there is nothing stopping the function from skipping the value you care about.
The target $K$ lies strictly between $f(a)$ and $f(b)$
The theorem says nothing about values outside that range. It only fills in the gap between the two ends you already know.

PassesThen there is at least one

c

in

(a, b)

with

f(c) = K

. At least one — there may be several, and the theorem does not care which.

Applying it to the car1 / 6

The position function $f(t) = 10 + 8t + 15\sin(t) \quad \text{on } [0, 10]$

The car has no teleportation capabilities. It has to cross the 50 marker to finish the trip.

Chapter 1 · Practice

Sixteen questions, eight topics

Two questions on each method in this chapter, laid out with room to work. Print it, do it by hand, and check yourself against the solutions afterwards — not during.

Practice ExamPDF, formatted for printing SolutionsWorked, step by step