How to Evaluate Limits: Substitution, 0/0, Conjugates and Limits at Infinity

In 30 seconds

A limit describes the value a function approaches as its input gets close to some number, or grows without bound, even if the function never actually gets there.
Always try substitution first. If it gives a normal number, that is the limit.
If substitution gives $ \tfrac{0}{0} $, the expression needs reshaping: factorise and cancel, or rationalise a square root, then substitute again.
For limits as $ x \to \infty $, divide every term by the highest power in the denominator. Terms like $ \tfrac{12\,000}{q} $ shrink to zero.
Exam point: write $ \lim $ on every line until the final substitution, and say which technique you used.

The case: what does one phone stand really cost?

Two friends start a small business 3D-printing phone stands. Before they sell a single stand they spend $12,000 on printers, software and a website. After that, each stand costs $4 in filament and packaging. So the total cost of making $ q $ stands is $ C(q) = 12\,000 + 4q $ dollars, and the average cost per stand is

\[ AC(q) = \frac{12\,000 + 4q}{q} \]

Their first 1,000 stands cost $16 each on average. They ask two questions. As sales grow, what does the average cost per stand settle down to? And how many stands must they make before the average cost drops below $4.50? Later, when overtime pushes costs up, they will also want the cost of one extra stand at a given output, which leads to a $ \tfrac{0}{0} $ limit.

The business and its numbers are illustrative, but the shape of the problem is common to every firm with fixed costs, and it is exactly the kind of limit that economics and calculus exams like to set.

The concept: where is the function heading?

Look at some values. At $ q = 1000 $, $ AC = 12 + 4 = 16 $. At $ q = 10\,000 $, $ AC = 1.2 + 4 = 5.20 $. At $ q = 100\,000 $, $ AC = 4.12 $. At a million stands, $ AC = 4.012 $. The numbers are heading towards 4, and they get as close to 4 as we like if $ q $ is large enough. They never reach it, because some of the $12,000 is always shared out across each stand.

That is what a limit captures: not the value at a point, but the value the function is heading towards. We write $ \displaystyle\lim_{q \to \infty} AC(q) = 4 $, read “the limit of AC as $ q $ tends to infinity is 4”. On a graph it is a horizontal asymptote.

The same idea works at a finite point. A function can be undefined at $ x = a $, for example because it would mean dividing by zero, and still head cleanly towards a single value as $ x $ approaches $ a $ from either side.

Key termLimit: $ \displaystyle\lim_{x \to a} f(x) = L $ means that $ f(x) $ can be made as close to $ L $ as we like by taking $ x $ close enough to $ a $ (but not equal to it). As $ x \to \infty $, it means $ f(x) $ gets as close to $ L $ as we like once $ x $ is large enough^[1].

The formula

Two results do most of the work. The first is the reason substitution works for polynomials, rational functions and roots wherever they are defined:

\[ \lim_{x \to a} f(x) = f(a) \]

when $ f $ is continuous at $ a $. The second handles limits at infinity of a ratio of polynomials. If the top has degree $ n $ with leading coefficient $ a_n $, and the bottom has degree $ m $ with leading coefficient $ b_m $, then

\[ \lim_{x \to \infty} \frac{a_n x^n + \dots}{b_m x^m + \dots} = \frac{a_n}{b_m} \]

when $ n = m $. The limit is 0 if $ n < m $, and there is no finite limit if $ n > m $. Both rest on one simple fact: $ \dfrac{k}{x^p} \to 0 $ as $ x \to \infty $ for any constant $ k $ and any $ p > 0 $.

Chapter 9 of Mathematics Made Simple teaches limits and continuity from tables and graphs to algebra.Get the textbook · $12

Why these tools, and when they fail

For the phone stands, $ AC(q) $ is a ratio of two degree-1 polynomials, $ 4q + 12\,000 $ over $ q $. Both grow forever, so “substitute infinity” gives $ \tfrac{\infty}{\infty} $, which is meaningless. Dividing top and bottom by $ q $ turns the expression into something with a single shrinking term, and the limit becomes obvious.

At a finite point, substitution is the first tool and usually the last. It fails only when it produces an indeterminate form like $ \tfrac{0}{0} $^[5]. That result does not mean the limit is 0, or undefined. It means top and bottom share a factor that is zero at that point, and the factor must be removed. If substitution gives a non-zero number over zero, such as $ \tfrac{5}{0} $, there is no shared factor to cancel: the function shoots off towards $ \pm\infty $ and the limit does not exist.

Setting it up

Identify where $ x $ is heading: a finite number $ a $, or $ \infty $.
Substitute. A normal number? Done.
$ \tfrac{0}{0} $? Factorise top and bottom and cancel the common factor, or, if there is a square root, multiply top and bottom by the conjugate. Then substitute again.
$ x \to \infty $? Divide every term by the highest power of $ x $ in the denominator, then let each $ \tfrac{k}{x^p} $ go to zero.
Interpret the answer in context, with units.

In the case, $ q $ is the number of stands (positive), AC is in dollars per stand, and the limit is taken as $ q \to \infty $.

Solving it

Worked example

For $ AC(q) = \dfrac{12\,000 + 4q}{q} $, find $ \displaystyle\lim_{q \to \infty} AC(q) $, and find how many stands are needed for the average cost to fall below $4.50.

Step 1Divide every term by $ q $, the highest power in the denominator: $ AC(q) = \dfrac{12\,000/q + 4}{1} = \dfrac{12\,000}{q} + 4 $.

Step 2As $ q \to \infty $, $ \dfrac{12\,000}{q} \to 0 $.

Step 3So $ \displaystyle\lim_{q \to \infty} AC(q) = 0 + 4 = 4 $ dollars per stand.

Step 4Below $4.50 means $ \dfrac{12\,000}{q} + 4 < 4.5 $, so $ \dfrac{12\,000}{q} < 0.5 $.

Step 5Since $ q > 0 $, multiply through: $ 12\,000 < 0.5q $, so $ q > 24\,000 $.

The average cost tends to $4, the variable cost of one stand. To get it below $4.50, the friends must make more than 24,000 stands; at exactly 24,000 it equals $4.50.

Average cost per phone stand, AC = (12 000 + 4q)/q $ per unit

average cost AC(q)

Figure 1. Average cost falls fast at first, then flattens towards the horizontal asymptote AC = 4. It gets as close to $4 as you like, but never reaches it, because some fixed cost is always spread over each unit.

Now the $ \tfrac{0}{0} $ techniques. Each part below uses a different method, so that you can see when each one is needed.

Worked example

Find (a) $ \displaystyle\lim_{x \to 3} \frac{2x^2-1}{x + 2} $ (b) $ \displaystyle\lim_{x \to -3} \frac{x^2 + x-6}{x^2-9} $ (c) $ \displaystyle\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} $.

Step 1(a) Substitute: $ \dfrac{2(9)-1}{3 + 2} = \dfrac{17}{5} $. A normal number, so the limit is $ \dfrac{17}{5} = 3.4 $.

Step 2(b) Substitute: top $ 9-3-6 = 0 $, bottom $ 9-9 = 0 $. This is $ \tfrac{0}{0} $, so factorise.

Step 3$ \dfrac{(x + 3)(x-2)}{(x + 3)(x-3)} = \dfrac{x-2}{x-3} $ for $ x \ne -3 $.

Step 4Substitute again: $ \dfrac{-3-2}{-3-3} = \dfrac{-5}{-6} = \dfrac{5}{6} $.

Step 5(c) Substitute: $ \dfrac{2-2}{0} = \tfrac{0}{0} $. There is a square root, so multiply by the conjugate $ \sqrt{x} + 2 $.

Step 6$ \dfrac{(\sqrt{x}-2)(\sqrt{x} + 2)}{(x-4)(\sqrt{x} + 2)} $ $ = \dfrac{x-4}{(x-4)(\sqrt{x} + 2)} $ $ = \dfrac{1}{\sqrt{x} + 2} $

Step 7Substitute: $ \dfrac{1}{2 + 2} = \dfrac{1}{4} $.

The answers are $ 3.4 $, $ \tfrac{5}{6} $ and $ \tfrac{1}{4} $. In (b) and (c) the original function is undefined at the point, but its limit exists: the graph has a hole, not a wall (Figure 2).

Worked example

(Exam style.) Overtime makes the friends’ cost rise faster at high volume: $ C(q) = 12\,000 + 4q + 0.0001q^2 $. The cost per extra stand near $ q = 10\,000 $ is $ \displaystyle\lim_{q \to 10\,000} \frac{C(q)-C(10\,000)}{q-10\,000} $. Find it.

Step 1Substituting $ q = 10\,000 $ gives $ \tfrac{0}{0} $: the top is $ C(10\,000)-C(10\,000) = 0 $.

Step 2The 12,000 cancels: $ C(q)-C(10\,000) $ $ = 4(q-10\,000) $ $ {} + 0.0001(q^2-10\,000^2) $.

Step 3Factorise the difference of squares: $ q^2-10\,000^2 = (q-10\,000)(q + 10\,000) $.

Step 4Cancel $ q-10\,000 $: the fraction becomes $ 4 + 0.0001(q + 10\,000) $.

Step 5Substitute: $ 4 + 0.0001(20\,000) = 4 + 2 = 6 $.

At 10,000 stands an extra stand costs about $6. This limit is the marginal cost, $ C'(10\,000) $: the derivative is defined as exactly this kind of $ \tfrac{0}{0} $ limit.

The Practice Questions book drills every limit technique: substitution, factorising, conjugates and infinity.Get 1,565 questions · $9

What the answers tell us

The $4 limit is the variable cost. As output grows, the fixed $12,000 is spread so thinly that each stand’s share of it vanishes, leaving only the $4 of filament and packaging. That is a general result: for any cost $ C = F + vq $, $ \displaystyle\lim_{q \to \infty} AC = v $. Economists call the falling part spreading the overhead, and it is one source of economies of scale.

Reasonableness and sensitivity. The limit cannot be below $4, since each stand costs at least that to make, and the answer passes that check. Halving the fixed cost to $6,000 would not change the limit at all; it would only change how fast AC gets there (12,000 stands instead of 24,000 to reach $4.50). The quadratic cost in the third example behaves differently: there $ AC = \dfrac{12\,000}{q} + 4 + 0.0001q $, and the last term grows, so AC has no finite limit as $ q \to \infty $. Rising costs eventually beat the spreading of overheads.

Limits are not values. In part (b) the function is not defined at $ x = -3 $, yet the limit is $ \tfrac{5}{6} $. In the case, AC never equals 4. A limit tells you where a function is heading, which is often more useful than where it is.

y = (x² + x − 6)/(x² − 9) has a hole, not a wall, at x = −3

y = (x² + x − 6)/(x² − 9)

Figure 2. At x = −3 both top and bottom are zero (0/0), but the graph just has a missing point: the limit is 5/6. At x = 3 only the bottom is zero, so the function blows up and no limit exists. The dashed line y = 1 is the limit as x → ±∞.

For the full chapter on limits and continuity, including one-sided limits and the squeeze theorem, see the Mathematics Made Simple textbook.

Common mistakeWriting $ \tfrac{0}{0} = 0 $ or “undefined, so no limit”. A $ \tfrac{0}{0} $ result says only that you need to reshape the expression. In part (b) the limit turned out to be $ \tfrac{5}{6} $, not 0 and not undefined.

Common mistakeDropping the $ \lim $ sign too early. Writing $ \dfrac{x-2}{x-3} = \dfrac{5}{6} $ is false, since the left side depends on $ x $. Keep $ \displaystyle\lim_{x \to -3} $ on every line until you substitute.

Exam tipIn AP Calculus AB and BC, Unit 1 covers limits by algebraic manipulation (topics 1.5–1.6) and limits at infinity and horizontal asymptotes (topic 1.15)^[2]. IB Mathematics: analysis and approaches introduces limits in SL topic 5.1, linked to the derivative^[3]. Cambridge 9709 does not set limit techniques as a separate topic, but Paper 1 differentiation treats the gradient of a curve as the limit of gradients of chords, which is the same idea as the third worked example^[4].

Get both Mathematics Made Simple books for $16: the limits chapter plus hundreds of fully worked questions.Both books · $16

Practice questions

Try these on paper first. Each worked answer names the technique. For many more, with full solutions, see the Mathematics Made Simple practice questions book.

EasyQ1. Find $ \displaystyle\lim_{x \to 2} \frac{x^3-3x}{x + 1} $. Show worked answer

Substitution works: $ \dfrac{8-6}{2 + 1} = \dfrac{2}{3} $.

EasyQ2. Find $ \displaystyle\lim_{x \to \infty} \frac{7x-3}{2x + 5} $. Show worked answer

Divide every term by $ x $: $ \dfrac{7-3/x}{2 + 5/x} $.
As $ x \to \infty $, $ 3/x \to 0 $ and $ 5/x \to 0 $, so the limit is $ \dfrac{7}{2} = 3.5 $.

MediumQ3. Find $ \displaystyle\lim_{x \to 2} \frac{x^2-3x + 2}{x^2-4} $. Show worked answer

Substitution gives $ \tfrac{0}{0} $, so factorise: $ \dfrac{(x-1)(x-2)}{(x-2)(x + 2)} = \dfrac{x-1}{x + 2} $ for $ x \ne 2 $.
Substitute: $ \dfrac{1}{4} $.

MediumQ4. Find $ \displaystyle\lim_{x \to \infty} \left( \sqrt{x^2 + 6x}-x \right) $. Show worked answer

This is $ \infty-\infty $, so multiply and divide by the conjugate $ \sqrt{x^2 + 6x} + x $.
Top: $ (x^2 + 6x)-x^2 = 6x $, so the expression is $ \dfrac{6x}{\sqrt{x^2 + 6x} + x} $.
Divide by $ x $ (for $ x > 0 $, $ \sqrt{x^2 + 6x}/x = \sqrt{1 + 6/x} $): $ \dfrac{6}{\sqrt{1 + 6/x} + 1} $.
As $ x \to \infty $: $ \dfrac{6}{1 + 1} = 3 $.

ExamQ5. A bakery’s weekly cost of making $ q $ loaves is $ C(q) = 900 + 15q $ cents. (a) Find $ \displaystyle\lim_{q \to \infty} \frac{C(q)}{q} $ and interpret it. [2] (b) Find the smallest whole number of loaves for which the average cost is less than 15.1 cents. [2] (c) A new cost model is $ C(q) = 900 + 15q + 0.02q^2 $. Show that $ \displaystyle\lim_{q \to 50} \frac{C(q)-C(50)}{q-50} = 17 $. [3] (d) Explain why, for the new model, the average cost has no finite limit as $ q \to \infty $. [1] Show worked answer

(a) $ \dfrac{C(q)}{q} = \dfrac{900}{q} + 15 $ [1]. As $ q \to \infty $, $ 900/q \to 0 $, so the limit is 15 cents: the average cost approaches the variable cost per loaf [1].
(b) $ \dfrac{900}{q} < 0.1 $ gives $ q > 9000 $ [1], so the smallest whole number is 9001 loaves [1].
(c) $ C(q)-C(50) = 15(q-50) + 0.02(q^2-2500) $ [1] $ = (q-50)\left[15 + 0.02(q + 50)\right] $ [1]. Cancel and let $ q \to 50 $: $ 15 + 0.02(100) = 17 $ [1].
(d) $ \dfrac{C(q)}{q} = \dfrac{900}{q} + 15 + 0.02q $, and $ 0.02q \to \infty $, so the average cost grows without bound [1].

Frequently asked questions

What is the first thing to try when evaluating a limit?

Direct substitution. For polynomials, and for rational and root functions at points where they are defined, the limit is simply the function’s value. Only if substitution gives an indeterminate form such as $ \tfrac{0}{0} $ do you need factorising or rationalising.

What does 0/0 mean in a limit?

It means the top and bottom both vanish at that point, usually because they share a factor. It does not tell you the answer. Cancel the shared factor (or rationalise) and substitute again.

How do I evaluate a limit as x tends to infinity?

For a fraction, divide every term by the highest power of $ x $ in the denominator. Every term of the form $ k/x^p $ then tends to zero, and what is left is the limit. If the top has the higher degree, there is no finite limit.

Is a limit the same as the function’s value?

Only when the function is continuous there. A limit describes where the function is heading, so it can exist where the function is undefined (a hole) and can differ from the value at a jump or removed point.

References

Strang, G. and Herman, E. (2016) Calculus Volume 1, sections 2.2 “The Limit of a Function”, 2.3 “The Limit Laws” and 4.6 “Limits at Infinity and Asymptotes”. Houston, TX: OpenStax. https://openstax.org/details/books/calculus-volume-1
College Board (2020) AP Calculus AB and AP Calculus BC Course and Exam Description, Unit 1 “Limits and Continuity”. New York: College Board.
International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide (first assessment 2021). Geneva: IBO.
Cambridge University Press & Assessment. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. cambridgeinternational.org
Stewart, J. (2016) Calculus: Early Transcendentals, 8th edn, chapter 2 “Limits and Derivatives”. Boston, MA: Cengage Learning.

How to Evaluate Limits: Substitution, 0/0, Conjugates and Limits at Infinity

The case: what does one phone stand really cost?

The concept: where is the function heading?

The formula

Why these tools, and when they fail

Setting it up

Solving it

What the answers tell us

Practice questions

Frequently asked questions

What is the first thing to try when evaluating a limit?

What does 0/0 mean in a limit?

How do I evaluate a limit as x tends to infinity?

Is a limit the same as the function’s value?

References

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