Mathematics guide, Unit 1, step 10 of 33Previous step Next step

Step 10: Continuity, Asymptotes and the Intermediate Value Theorem

By the EconTutorials team. 10-minute read with 6 practice questions.

SyllabusAP 1.10–1.16 · IB HL 5.12

In 30 seconds

$ f $ is continuous at $ x = a $ when three things hold: $ f(a) $ is defined, $ \lim_{x\to a} f(x) $ exists, and the two are equal.
Continuity can fail in three standard ways: a removable discontinuity (a hole), a jump, or an infinite discontinuity (a vertical asymptote).
Infinite limits give vertical asymptotes; limits as $ x \to \pm\infty $ give horizontal asymptotes. For a rational function, compare the degrees of the top and bottom.
Exam point: the Intermediate Value Theorem needs continuity on a closed interval $ [a, b] $. State it before you use the theorem, or you lose the justification mark.

Video lesson coming soonA short animated lesson for this step will appear here.

What continuity means

This step completes AP Calculus Unit 1, Limits and Continuity, covering topics 1.10 to 1.16; Step 9 covered 1.1 to 1.9. IB students meet continuity formally at HL (5.12), and the asymptote work links to IB SL topic 2. Cambridge 9709 does not test continuity as a topic, but the sign-change method for locating roots in Papers 2 and 3 rests on the theorem at the end of this page.

Informally, a function is continuous if you can draw its graph without lifting your pen. Limits make that precise, one point at a time.

Key termContinuous at $ x = a $: (1) $ f(a) $ is defined, (2) $ \lim_{x\to a} f(x) $ exists, and (3) $ \lim_{x\to a} f(x) = f(a) $. All three must hold.

Each way of breaking continuity breaks a different condition.

Type	What the graph shows	Which condition fails
Removable	a hole, or a single point out of place; the limit exists	(1) or (3)
Jump	the one-sided limits exist but differ	(2)
Infinite	the function is unbounded near $ a $; a vertical asymptote	(2)

A function is continuous on an open interval $ (a, b) $ if it is continuous at every point inside it. On a closed interval $ [a, b] $ we also need the one-sided limits at the ends to match the end values: $ \lim_{x\to a^+} f(x) = f(a) $ and $ \lim_{x\to b^-} f(x) = f(b) $.

You rarely check this point by point. Polynomials, rational functions, roots, exponentials, logarithms, sine and cosine are continuous at every point of their domains, and so are their sums, products, composites and quotients (where the denominator is not zero). So the only places to worry about are where a denominator is zero, where a root or logarithm stops being defined, and where a piecewise function changes formula.

Removing a discontinuity and joining pieces

A removable discontinuity is fixed by defining, or redefining, one value: set $ f(a) $ equal to the limit. A piecewise function with a parameter is made continuous by choosing the parameter so the pieces meet.

Worked example

(a) $ f(x) = \dfrac{x^2-2x-8}{x-4} $ for $ x \neq 4 $, and $ f(4) = k $. Find $ k $ so that $ f $ is continuous at $ x = 4 $. (b) $ g(x) = kx^2-1 $ for $ x < 2 $ and $ g(x) = 3x + k $ for $ x \ge 2 $. Find $ k $ so that $ g $ is continuous for all $ x $.

Step 1(a) Factorise: $ \dfrac{(x-4)(x + 2)}{x-4} = x + 2 $ for $ x \neq 4 $, so $ \lim_{x\to4} f(x) = 6 $.

Step 2Condition (3) needs $ f(4) = 6 $, so $ k = 6 $.

Step 3(b) Each piece is a polynomial, so the only doubtful point is $ x = 2 $. Left limit: $ 4k-1 $. Right limit and $ g(2) $: $ 6 + k $.

Step 4Set them equal: $ 4k-1 = 6 + k $, so $ 3k = 7 $ and $ k = \tfrac73 $. Check: both pieces give $ \tfrac{25}{3} $ at $ x = 2 $.

In an exam, justify continuity by stating all three conditions in order: the value, the limit, and their equality.

An economics example: a bulk-discount cliff. A supplier charges \$5 per unit for orders of up to 100 units, but \$4 per unit on the whole order once it exceeds 100. Total cost is $ C(q) = 5q $ for $ q \le 100 $ and $ C(q) = 4q $ for $ q > 100 $. At $ q = 100 $ the left limit is \$500 but the right limit is \$400: a jump discontinuity. Buying 101 units costs \$404, less than buying 100. Firms facing such a cliff bunch their orders just above the threshold, which is why many tariffs charge the lower rate only on units beyond the threshold. Then $ C(q) = 500 + 4(q-100) $ for $ q > 100 $, both one-sided limits are \$500, and the cost is continuous.

Exam tipAP free-response questions often ask whether a piecewise function is continuous at the join. Write three short lines: “$ f(2) = \ldots $”, “$ \lim_{x\to2^-} f(x) = \ldots $ and $ \lim_{x\to2^+} f(x) = \ldots $, so $ \lim_{x\to2} f(x) = \ldots $”, then “since the limit equals $ f(2) $, $ f $ is continuous at $ x = 2 $”. A bare “yes, the pieces meet” earns nothing.

Infinite limits and vertical asymptotes

When substitution gives $ \frac{k}{0} $ with $ k \neq 0 $, the function divides by something tiny, so its size explodes. The only question is the sign on each side. Take $ \frac{3}{x-2} $ as $ x \to 2 $. Just below 2, $ x-2 $ is small and negative, so the fraction is large and negative: $ \lim_{x\to2^-} \frac{3}{x-2} = -\infty $. Just above 2 it is large and positive: $ \lim_{x\to2^+} \frac{3}{x-2} = +\infty $.

Key termVertical asymptote: the line $ x = a $ is a vertical asymptote of $ f $ if at least one of $ \lim_{x\to a^-} f(x) $ and $ \lim_{x\to a^+} f(x) $ is $ \infty $ or $ -\infty $.

Worked example

$ f(x) = \dfrac{x^2-x-6}{x^2-4x + 3} $. Find all discontinuities of $ f $, classify each, and describe the behaviour of $ f $ near them.

Step 1Factorise: $ f(x) = \dfrac{(x-3)(x + 2)}{(x-3)(x-1)} $. The denominator is zero at $ x = 1 $ and $ x = 3 $.

Step 2For $ x \neq 3 $, $ f(x) = \dfrac{x + 2}{x-1} $. At $ x = 3 $ the factor cancels and $ \lim_{x\to3} f(x) = \tfrac52 $: a removable discontinuity, a hole at $ \left(3, \tfrac52\right) $.

Step 3At $ x = 1 $ nothing cancels and the top tends to 3. Below 1 the bottom is negative, so $ f \to -\infty $; above 1, $ f \to +\infty $.

Step 4So $ x = 1 $ is an infinite discontinuity, with vertical asymptote $ x = 1 $.

Figure 1 shows both features, plus the horizontal asymptote $ y = 1 $ from the next section.

y = (x² − x − 6)/(x² − 4x + 3)

simplifies to (x + 2)/(x − 1), x ≠ 3

Figure 1. The factor x − 3 cancels, leaving a hole at (3, 5/2): a removable discontinuity. The factor x − 1 does not cancel, so x = 1 is a vertical asymptote. Equal degrees on top and bottom give the horizontal asymptote y = 1.

Common mistakeAssuming every zero of the denominator gives a vertical asymptote. Factorise first. If the factor cancels completely, you have a hole, not an asymptote. Only a zero of the simplified denominator gives an asymptote.

Limits at infinity and horizontal asymptotes

Now let $ x $ itself grow without bound. The basic fact is that a fixed number divided by a huge one is tiny: $ \lim_{x\to\pm\infty} \frac{1}{x^n} = 0 $ for $ n > 0 $, and $ \lim_{x\to\infty} e^{-x} = 0 $. If $ f(x) \to L $ as $ x \to \infty $ or as $ x \to -\infty $, the line $ y = L $ is a horizontal asymptote.

For a rational function, the highest powers on the top and bottom dominate everything else. Divide top and bottom by the highest power of $ x $ in the denominator and every lower term becomes a constant over a power of $ x $, which tends to 0.

Degree of top $ m $, bottom $ n $	Limit as $ x \to \pm\infty $	Example
$ m < n $	0	$ \frac{5x + 2}{x^3-1} \to 0 $
$ m = n $	ratio of leading coefficients	$ \frac{6x^2-x}{3x^2 + 4} \to 2 $
$ m > n $	unbounded; no horizontal asymptote	$ \frac{x^2 + 1}{x} $

Worked example

Find (a) $ \displaystyle\lim_{x\to\infty} \frac{6x^2-x}{3x^2 + 4} $ and (b) the horizontal asymptotes of $ y = \dfrac{\sqrt{9x^2 + 1}}{2x-5} $.

Step 1(a) Divide by $ x^2 $: $ \dfrac{6-\frac1x}{3 + \frac{4}{x^2}} \to \dfrac{6}{3} = 2 $.

Step 2(b) For large $ |x| $, $ \sqrt{9x^2 + 1} \approx \sqrt{9x^2} = 3|x| $. Careful: $ |x| = x $ for $ x > 0 $ but $ -x $ for $ x < 0 $.

Step 3As $ x \to \infty $, divide by $ x $: $ \dfrac{\sqrt{9 + 1/x^2}}{2-5/x} \to \dfrac32 $.

Step 4As $ x \to -\infty $, divide by $ |x| = -x $: $ \dfrac{\sqrt{9 + 1/x^2}}{-2 + 5/x} \to -\dfrac32 $.

So there are two horizontal asymptotes, $ y = \tfrac32 $ and $ y = -\tfrac32 $. Square roots are the usual way to get different behaviour at the two ends.

An economics example: average cost and economies of scale. A firm has fixed cost \$1,200 and a constant \$15 cost for each unit. Its average cost is $ AC(Q) = \frac{1200 + 15Q}{Q} = 15 + \frac{1200}{Q} $. At $ Q = 100 $, $ AC = \$27 $; at $ Q = 500 $, \$17.40; at $ Q = 1000 $, \$16.20. As $ Q \to \infty $, $ \frac{1200}{Q} \to 0 $, so $ AC \to \$15 $: the fixed cost is spread ever thinner, and average cost sinks towards the cost of one more unit without ever reaching it (Figure 2). At the other end, $ AC \to \infty $ as $ Q \to 0^+ $, because a handful of units carry the whole \$1,200. The same kind of limit defines continuous compounding: \$2,000 at 6% compounded $ n $ times a year grows to $ 2000\left(1 + \frac{0.06}{n}\right)^n $, which is \$2,120.00 for $ n = 1 $, \$2,123.36 for $ n = 12 $, and tends to $ 2000e^{0.06} \approx \$2{,}123.67 $ as $ n \to \infty $.

Average cost AC = 15 + 1200/Q $ per unit

AC = (1200 + 15Q)/Q

Figure 2. As output grows, the $1,200 fixed cost is spread over more units and average cost falls towards the $15 cost of each extra unit: a horizontal asymptote.

The Intermediate Value Theorem

If you walk continuously from sea level to a 1,000 m summit, you must pass every height in between. You cannot skip 731 m without jumping. That is the Intermediate Value Theorem.

Key termIntermediate Value Theorem (IVT): if $ f $ is continuous on the closed interval $ [a, b] $ and $ N $ is any number between $ f(a) $ and $ f(b) $, then $ f(c) = N $ for at least one $ c $ in $ [a, b] $. In particular, if $ f(a) $ and $ f(b) $ have opposite signs, $ f $ has a zero in $ (a, b) $.

Worked example

Show that the equation $ \cos x = x $ has a solution between 0 and 1, and narrow it to an interval of width 0.1.

Step 1Let $ g(x) = \cos x-x $. It is a difference of continuous functions, so it is continuous on $ [0, 1] $.

Step 2$ g(0) = 1 > 0 $ and $ g(1) = \cos 1-1 \approx -0.460 < 0 $ (radians).

Step 3Since 0 lies between $ g(1) $ and $ g(0) $, the IVT gives a $ c $ in $ (0, 1) $ with $ g(c) = 0 $, that is $ \cos c = c $.

Step 4Narrow it: $ g(0.7) \approx 0.065 > 0 $ and $ g(0.8) \approx -0.103 < 0 $, so the root lies in $ (0.7, 0.8) $.

The theorem says a root exists; it does not say where. Repeated sign checks close in on it, which is the idea behind the numerical methods of Step 31.

g(x) = cos x − x changes sign on [0, 1]

g(x) = cos x − x

Figure 3. g is continuous, positive at 0 and negative at 1, so its graph must cross the x-axis somewhere in between. The IVT guarantees the root; it does not say where it is.

Common mistakeForgetting continuity. $ f(x) = \frac{1}{x-2} $ has $ f(1) = -1 $ and $ f(3) = 1 $, yet $ \frac{1}{x-2} = 0 $ has no solution, because $ f $ is not continuous on $ [1, 3] $. Always write “$ f $ is continuous on $ [a, b] $” before you quote the IVT, and remember it promises at least one value of $ c $, not exactly one.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. State the vertical and horizontal asymptotes of $ y = \dfrac{4x-1}{x + 3} $. Show worked answer

The denominator is zero at $ x = -3 $, where the top is $ -13 \neq 0 $, so $ x = -3 $ is a vertical asymptote.
Top and bottom both have degree 1, so $ y \to \frac41 = 4 $ as $ x \to \pm\infty $: horizontal asymptote $ y = 4 $.

EasyQ2. Find $ \displaystyle\lim_{x\to\infty} \frac{3x^3-x}{6x^3 + 2x^2} $ and $ \displaystyle\lim_{x\to\infty} \frac{2x + 1}{x^2 + 5} $. Show worked answer

Equal degrees: the limit is the ratio of leading coefficients, $ \frac36 = \frac12 $.
Top degree 1 is below bottom degree 2, so the limit is 0.

MediumQ3. $ f(x) = \dfrac{x^2-7x + 10}{x-5} $ for $ x \neq 5 $, and $ f(5) = k $. Find $ k $ so that $ f $ is continuous at $ x = 5 $, and name the type of discontinuity for any other $ k $. Show worked answer

$ x^2-7x + 10 = (x-5)(x-2) $, so $ f(x) = x-2 $ for $ x \neq 5 $.
$ \lim_{x\to5} f(x) = 3 $, so $ k = 3 $.
For any other $ k $ the limit exists but differs from $ f(5) $: a removable discontinuity.

MediumQ4. Find $ \lim_{x\to2^-} \dfrac{x + 1}{(x-2)^2} $ and $ \lim_{x\to2^+} \dfrac{x + 1}{(x-2)^2} $. What does this tell you about the graph? Show worked answer

The top tends to 3, a positive number.
The bottom $ (x-2)^2 $ tends to 0 and is positive on both sides.
So both one-sided limits are $ +\infty $.
The line $ x = 2 $ is a vertical asymptote, and the graph rises to $ +\infty $ on both sides of it.

MediumQ5. Show that $ x^3-4x + 1 = 0 $ has a root between 1 and 2, and find an interval of width 0.1 that contains it. Show worked answer

$ f(x) = x^3-4x + 1 $ is a polynomial, so it is continuous on $ [1, 2] $.
$ f(1) = -2 < 0 $ and $ f(2) = 1 > 0 $, so by the IVT there is a root in $ (1, 2) $.
$ f(1.8) = 5.832-7.2 + 1 = -0.368 < 0 $ and $ f(1.9) = 6.859-7.6 + 1 = 0.259 > 0 $.
So the root lies in $ (1.8, 1.9) $.

ExamQ6. (AP style) Let $ f(x) = \sqrt{x + 3} $ for $ -3 \le x < 1 $ and $ f(x) = k-x $ for $ x \ge 1 $. (a) Find the value of $ k $ for which $ f $ is continuous at $ x = 1 $. Justify your answer. [3] (b) For this value of $ k $, explain why there must be a value $ c $ with $ 1 < c < 5 $ and $ f(c) = -1 $. [2] (c) Find $ \displaystyle\lim_{x\to\infty} \frac{f(x)}{2x + 1} $. [2] Show worked answer

(a) $ \lim_{x\to1^-} f(x) = \sqrt{4} = 2 $ [1]. $ \lim_{x\to1^+} f(x) = f(1) = k-1 $ [1]. Continuity needs $ k-1 = 2 $, so $ k = 3 $; then the limit exists and equals $ f(1) $ [1].
(b) $ f(x) = 3-x $ is continuous on $ [1, 5] $, with $ f(1) = 2 $ and $ f(5) = -2 $ [1]. Since $ -1 $ lies between $ -2 $ and 2, the IVT gives $ c $ in $ (1, 5) $ with $ f(c) = -1 $ [1].
(c) For large $ x $, $ \frac{3-x}{2x + 1} = \frac{3/x-1}{2 + 1/x} $ [1] $ \to -\frac12 $ [1].
Total: 7 marks.

Check yourself

1. What kind of discontinuity does $ f(x) = \dfrac{x-1}{x^2-1} $ have at $ x = 1 $?

$ f(x) = \frac{1}{x + 1} $ for $ x \neq \pm1 $, so $ \lim_{x\to1} f(x) = \frac12 $ exists but $ f(1) $ is undefined: a hole. The infinite discontinuity is at $ x = -1 $.

2. What is the horizontal asymptote of $ y = \dfrac{2x^2 + 1}{x^2-9} $?

Equal degrees, so $ y \to \frac21 = 2 $. The lines $ x = \pm3 $ are the vertical asymptotes, not horizontal ones.

3. Which condition must you state before using the Intermediate Value Theorem on $ [a, b] $?

The IVT needs continuity on the closed interval. It does not need $ f $ to be increasing, and for a root you want $ f(a) $ and $ f(b) $ to have opposite signs.

Your score: 0 out of 3

The short version

Continuity at $ a $ needs three things: $ f(a) $ defined, the limit existing, and the two being equal.
Discontinuities are removable (a hole, fixed by redefining one value), jump (one-sided limits differ) or infinite (a vertical asymptote).
$ \frac{k}{0} $ with $ k \neq 0 $ means an infinite limit; check the sign on each side. Factorise first: a cancelled factor gives a hole, not an asymptote.
As $ x \to \pm\infty $, compare degrees: lower on top gives 0, equal gives the ratio of leading coefficients, higher gives no horizontal asymptote.
The IVT: a function continuous on $ [a, b] $ takes every value between $ f(a) $ and $ f(b) $, so a sign change means a root.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Degree of top \( m \), bottom \( n \)	Limit as \( x \to \pm\infty \)	Example
\( m < n \)	0	\( \frac{5x + 2}{x^3-1} \to 0 \)
\( m = n \)	ratio of leading coefficients	\( \frac{6x^2-x}{3x^2 + 4} \to 2 \)
\( m > n \)	unbounded; no horizontal asymptote	\( \frac{x^2 + 1}{x} \)

Step 10: Continuity, Asymptotes and the Intermediate Value Theorem

What continuity means

Removing a discontinuity and joining pieces

Infinite limits and vertical asymptotes

Limits at infinity and horizontal asymptotes

The Intermediate Value Theorem

Practise: 6 questions

Check yourself

The short version

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