Limits: practice problems with step-by-step solutions

Free Limits practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: direct substitution and factoring, limits at infinity, L'Hôpital's rule.

Free practice problemEasy
xy9x = 9

Find the limit L = limx→9 (√x − 3)/(x − 9).

f(x) = (√x − 3)/(x − 9)x → 9
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What you practise

  1. Easydirect substitution and factoring
  2. Mediumlimits at infinity
  3. HardL'Hôpital's rule

Key formulas

Direct substitution
try x = a first; 0/0 means simplify
Factor and cancel
xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹)
Conjugate
(√x − √a)(√x + √a) = x − a
Limits at infinity
divide by the highest power of x; equal degrees give the ratio of leading coefficients
Special limit
limx→∞ (1 + a/x)bx = eab
L'Hôpital's rule
lim f/g = lim f′/g′ (0/0 or ∞/∞)

Learn Limits from scratch

1What a limit is

A limit describes the value a function gets close to as x gets close to some number. It does not matter what happens exactly at that number.

Take f(x) = (x² − 1)/(x − 1). At x = 1 you get 0/0, which has no value. But you can try numbers close to 1.

  • x = 0.9 gives f = 1.9.
  • x = 0.99 gives f = 1.99.
  • x = 1.01 gives f = 2.01.

The values get closer and closer to 2. We write:

limx→1 (x² − 1)/(x − 1) = 2
xy12(1, 2)

The graph is a straight line with one missing point, a hole at (1, 2). The limit is the height of the hole.

Self-check A function g gives g(2.9) = 5.9, g(2.99) = 5.99 and g(3.01) = 6.01. It is not defined at x = 3. What is the limit as x → 3?
  1. 3
  2. 5.99
  3. 6
  4. It does not exist, because g(3) is not defined

Answer: 6. The values approach 6 from both sides, so the limit is 6. A limit only looks at values near x = 3, never at x = 3 itself.

2Substitution, 0/0 and factoring

Always try direct substitution first: put the number in. If you get an ordinary number, that is the limit.

limx→2 (x² + 3x) = 4 + 6 = 10

If you get 0/0, you are not finished. 0/0 is called an indeterminate form: it is a signal to simplify, not an answer.

Usually the top and the bottom share a factor that is zero at that point. Factor both, cancel the common factor, then substitute again.

(x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3

So the limit as x → 3 is 3 + 3 = 6. Useful factor patterns:

  • x² − a² = (x − a)(x + a)
  • x³ − a³ = (x − a)(x² + ax + a²)
  • x³ + a³ = (x + a)(x² − ax + a²)
  • A quadratic such as x² − x − 30 factors as (x − 6)(x + 5): find two numbers that multiply to −30 and add to −1.

With a square root, multiply top and bottom by the conjugate: (√x − 2)(√x + 2) = x − 4.

Self-check What is limx→2 (x² − 4)/(x − 2)?
  1. 0
  2. 2
  3. 4
  4. It does not exist, because you get 0/0

Answer: 4. Substitution gives 0/0, so factor: (x − 2)(x + 2)/(x − 2) = x + 2, which is 4 at x = 2. Stopping at 0/0 is the mistake behind “does not exist”.

3Limits at infinity and L’Hôpital’s rule

x → ∞ means x grows without end. For a fraction of polynomials, only the highest powers matter when x is huge.

Divide every term by the highest power of x in the bottom. Terms like 4/x and 7/x² go to 0.

(6x² + x)/(3x² − 5) = (6 + 1/x)/(3 − 5/x²) → 6/3 = 2
  • Same highest power on top and bottom: the limit is the ratio of the leading coefficients.
  • Higher power in the bottom: the limit is 0.
  • Higher power on top: the fraction grows without end, so there is no finite limit.

For √(x² + 6x) − x, multiply by the conjugate. You get 6x/(√(x² + 6x) + x), which tends to 6/(1 + 1) = 3.

L’Hôpital’s rule: if substitution gives 0/0 or ∞/∞, you may replace the top and the bottom by their derivatives, each one on its own.

limx→0 (eˣ − 1)/x = limx→0 eˣ/1 = 1

If you get 0/0 again, apply the rule once more. Trigonometric limits always use x in radians.

Self-check What is limx→∞ (4x² + 1)/(2x² + 3x)?
  1. 0
  2. 2
  3. ½
  4. ∞

Answer: 2. The highest powers are equal (x²), so the limit is the ratio of the leading coefficients: 4/2 = 2. ½ comes from dividing the wrong way round.

4The method step by step

  1. Substitute the number. An ordinary number is the answer.
  2. If you get 0/0, factor the top and the bottom, or multiply by the conjugate if there is a root.
  3. Cancel the common factor and substitute again.
  4. If x → ∞, divide every term by the highest power of x and let terms like a/x go to 0.
  5. If factoring does not work (eˣ, sin, cos, ln), use L’Hôpital’s rule on the 0/0 or ∞/∞ form.
  6. Check with a number very close to the point, for example x = 1.001.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Find L = limx→−2 (x³ + 8)/(x + 2).

1

Try substitution

((−2)³ + 8)/(−2 + 2) = (−8 + 8)/0 = 0/0

This is the indeterminate form 0/0, so you must simplify first.

2

Factor the top

8 = 2³, so the top is a sum of cubes with a = 2.

x³ + 8 = (x + 2)(x² − 2x + 4)
3

Cancel the common factor

(x + 2)(x² − 2x + 4)/(x + 2) = x² − 2x + 4

This is allowed because x only comes close to −2, so x + 2 is never exactly 0.

4

Substitute again

(−2)² − 2·(−2) + 4 = 4 + 4 + 4 = 12
5

Check the answer

Try x = −1.99: (−7.880599 + 8)/0.01 = 11.94, which is close to 12.

xy−212

The graph is the parabola x² − 2x + 4 with a hole at (−2, 12).

Answer

L = 12

6Common mistakes

  • Writing 0/0 as the answer, or saying the limit does not exist. 0/0 means: simplify first.
  • Using L’Hôpital’s rule when substitution does not give 0/0 or ∞/∞.
  • Using the quotient rule in L’Hôpital’s rule. Differentiate the top and the bottom separately.
  • At infinity, comparing the first terms written instead of the highest powers.
  • Using degrees instead of radians in limits with sin or cos.
Self-check What is limx→1 (x² + 1)/(x + 1)?
  1. 0
  2. 1
  3. 2
  4. It does not exist

Answer: 1. Substitution gives 2/2 = 1, an ordinary number, so that is the limit. Using L’Hôpital here (2x/1 = 2) is wrong, because the form was not 0/0.

7How exam questions look

  • Easy: direct substitution and factoring. Substitution gives 0/0, then you factor and cancel, as in the example.
  • Medium: limits at infinity. Divide by the highest power, or use the conjugate for a root such as √(x² + 4x) − x.
  • Hard: L’Hôpital’s rule, for example (eˣ − 1 − x)/x² or (1 − cos 3x)/x² as x → 0. You may need the rule twice.

A quick check: put in a number very close to the point with a calculator. The result should be close to your answer.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
xyx = 1

Find the limit L = limx→1 (x4 − 1)/(x − 1).

f(x) = (x4 − 1)/(x − 1)x → 1

Find L

Show hints
  1. Substitution gives 0/0. Factor the numerator.
  2. xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹).
  3. x4 − 1 = (x − 1)(x3 + x2 + x + 1).
Show the answer and the step-by-step solution

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Try direct substitution

numerator → 0 and denominator → 0, so the form is 0/0.

Factor and cancel

(x4 − 1)/(x − 1) = x3 + x2 + x + 1

Substitute

L = 1·13 + 1·12 + 1·1 + 1·1 = 4

Answer

L = 4

This is the derivative of xⁿ at x = a: n·aⁿ⁻¹.

Example 2Medium
xyy = L

Find the limit L = limx→∞ (1 + 1/x)x. Give a decimal value.

f(x) = (1 + 1/x)xx → ∞

Find L

Show hints
  1. The form is 1^∞. Write L = e^(ln L) and find ln L.
  2. ln L = limx→∞ x·ln(1 + 1/x).
  3. For small u, ln(1 + u) ≈ u. Here u = 1/x.
Show the answer and the step-by-step solution

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Take the logarithm

ln L = limx→∞ x·ln(1 + 1/x)

Evaluate

ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.

ln L = 1·1 = 1

Undo the logarithm

L = e1 = 2.718

Answer

L = 2.718
Example 3Hard
xyx = 0

Find the limit L = limx→0 ln(1 + 2x)/sin(3x), with x in radians.

f(x) = ln(1 + 2x)/sin(3x)x → 0

Find L

Show hints
  1. At x = 0: ln 1 = 0 and sin 0 = 0, so the form is 0/0.
  2. (ln(1 + 2x))′ = 2/(1 + 2x), (sin 3x)′ = 3·cos 3x.
  3. Substitute x = 0 into f′/g′.
Show the answer and the step-by-step solution

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

L'Hôpital's rule

L = limx→0 [2/(1 + 2x)]/[3·cos 3x]

Substitute x = 0

L = 2/3 = 0.6667

Answer

L = 0.6667

Questions and answers

What is Limits?

A limit describes the value a function gets close to as x gets close to some number. It does not matter what happens exactly at that number.

Take f(x) = (x² − 1)/(x − 1). At x = 1 you get 0/0, which has no value. But you can try numbers close to 1.

How do I solve Limits problems step by step?

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

What are common mistakes in Limits?
  • Keep every term of the second factor: xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹), and at x = a all n terms are equal.
  • The form 1^∞ is indeterminate, not 1. Take the natural logarithm first.
  • Keep the numerator's derivative on top: lim f/g = lim f′/g′.
What do the Limits exercises cover?
  • Easy: direct substitution and factoring
  • Medium: limits at infinity
  • Hard: L'Hôpital's rule

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Limits worksheet with answers
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