Mathematics · Simulation and virtual lab Free
Limits: approaching a value from the left and the right
Move x closer and closer to a point and watch what number f(x) approaches, from the left and from the right.
- You will learn to
- Estimate limits numerically and graphically from both sides.
- Decide whether a limit exists using one-sided limits.
- Confirm limits by factorising, using the conjugate or dividing by the highest power.
- You will measure
- Values of f(x) as x approaches a from the left and the right (or as x grows).
- Time needed
- About 25–35 min
- Equipment
- Nothing extra. Real version: scientific calculator, graph paper.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To estimate limits numerically and graphically by approaching a point from both sides, to recognise when a limit exists, and to confirm limits algebraically by factorising, using the conjugate, or dividing by the highest power.
What to do
- Choose (x² − 4) ÷ (x − 2). Tap Move closer several times, watching both orange and purple points. Record the values.
- What number do the values approach? Is f(2) defined? Tap Show the algebra.
- Repeat for (x³ − 1) ÷ (x − 1) and (√(x + 4) − 2) ÷ x. Write each limit.
- Choose 1 ÷ (x − 1). What happens to f(x) from the left and from the right?
- Choose (x² − 1) ÷ |x − 1|. Compare the two one-sided limits. Does the limit exist?
- Choose (2x² + 3) ÷ (x² + 1) as x → ∞. Move x to 10, 100, 1 000… What value does f(x) approach?
Results
Questions
What does lim x→a f(x) = L mean?
f(x) can be made as close to L as we like by taking x close enough to a (but not equal to a), from both sides.
Find lim x→3 (x² − 9) ÷ (x − 3).
(x − 3)(x + 3) ÷ (x − 3) = x + 3 → 6.
Find lim x→0 (√(x + 9) − 3) ÷ x.
Multiply by the conjugate: 1 ÷ (√(x + 9) + 3) → 1 ÷ 6.
When does a limit not exist?
When the left-hand and right-hand limits are different, or when f(x) grows without bound (infinite limit).
Find lim x→∞ (3x² − x) ÷ (2x² + 5).
Divide by x²: (3 − 1/x) ÷ (2 + 5/x²) → 3 ÷ 2 = 1.5.
Why can the limit at x = 2 exist even though f(2) is undefined?
A limit describes the values near x = 2, not at x = 2 itself.
For teachers: expected results and how the model works
Six functions: (x² − 4) ÷ (x − 2) → 4 (removable hole); (x³ − 1) ÷ (x − 1) → 3; (√(x + 4) − 2) ÷ x → 0.25 (conjugate); 1 ÷ (x − 1) at 1 (infinite, no limit, vertical asymptote); (x² − 1) ÷ |x − 1| at 1 (left −2, right +2, no limit); (2x² + 3) ÷ (x² + 1) → 2 as x → ∞ (horizontal asymptote). The slider sets h = 10⁻ⁿ (x = a ± h), or x = 10ⁿ for the limit at infinity, up to n = 6; values are shown to 7 significant figures. The table on the picture keeps the last five steps so learners can see the pattern. Point out rounding: very small h can give rounding errors on calculators.
Exam practice Exam set
National-exam style questions on this simulation (15 marks). Try each one on paper first, then open the marking scheme.
Evaluate (a) lim x→1 (x² + x − 2) ÷ (x² − 1), (b) lim x→4 (√x − 2) ÷ (x − 4). (6 marks)
Marking scheme: (a) (x + 2)(x − 1) ÷ ((x + 1)(x − 1)) (2) = (x + 2) ÷ (x + 1) → 3 ÷ 2 (1). (b) x − 4 = (√x − 2)(√x + 2) (1), so the expression = 1 ÷ (√x + 2) (1) → 1 ÷ 4 (1).
Find the horizontal asymptote of y = (4x² − 1) ÷ (2x² + x). (3 marks)
Marking scheme: Divide by x²: (4 − 1/x²) ÷ (2 + 1/x) (1); as x → ∞ this → 4 ÷ 2 (1) = 2, so y = 2 (1).
f(x) = x + 1 for x < 2 and f(x) = 5 − x for x ≥ 2. Find the left and right limits at x = 2 and say whether the limit exists. (3 marks)
Marking scheme: Left: 2 + 1 = 3 (1); right: 5 − 2 = 3 (1); they are equal, so lim x→2 f(x) = 3 (1).
Complete a table of values of f(x) = (x² − 4) ÷ (x − 2) for x = 1.9, 1.99, 2.01, 2.1 and use it to estimate the limit as x → 2. (3 marks)
Marking scheme: 3.9, 3.99, 4.01, 4.1 (2); the limit is 4 (1).
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Limits: approaching a value from the left and the right
Aim: To estimate limits numerically and graphically by approaching a point from both sides, to recognise when a limit exists, and to confirm limits algebraically by factorising, using the conjugate, or dividing by the highest power.
Apparatus (real laboratory): scientific calculator, graph paper.
What to do (with the simulation on www.donatstem.co.rw)
- Choose (x² − 4) ÷ (x − 2). Tap Move closer several times, watching both orange and purple points. Record the values.
- What number do the values approach? Is f(2) defined? Tap Show the algebra.
- Repeat for (x³ − 1) ÷ (x − 1) and (√(x + 4) − 2) ÷ x. Write each limit.
- Choose 1 ÷ (x − 1). What happens to f(x) from the left and from the right?
- Choose (x² − 1) ÷ |x − 1|. Compare the two one-sided limits. Does the limit exist?
- Choose (2x² + 3) ÷ (x² + 1) as x → ∞. Move x to 10, 100, 1 000… What value does f(x) approach?
Results
| # | x (left) | f(x) | x (right) | f(x) |
|---|
| 1 | | | | |
| 2 | | | | |
| 3 | | | | |
| 4 | | | | |
| 5 | | | | |
| 6 | | | | |
| 7 | | | | |
| 8 | | | | |
Graph
No graph: sketch each function near x = a, marking holes, asymptotes and the limit.
Questions
- What does lim x→a f(x) = L mean? [2]
- Find lim x→3 (x² − 9) ÷ (x − 3). [2]
- Find lim x→0 (√(x + 9) − 3) ÷ x. [2]
- When does a limit not exist? [2]
- Find lim x→∞ (3x² − x) ÷ (2x² + 5). [2]
- Why can the limit at x = 2 exist even though f(2) is undefined? [2]
- Evaluate (a) lim x→1 (x² + x − 2) ÷ (x² − 1), (b) lim x→4 (√x − 2) ÷ (x − 4). [6]
- Find the horizontal asymptote of y = (4x² − 1) ÷ (2x² + x). [3]
- f(x) = x + 1 for x < 2 and f(x) = 5 − x for x ≥ 2. Find the left and right limits at x = 2 and say whether the limit exists. [3]
- Complete a table of values of f(x) = (x² − 4) ÷ (x − 2) for x = 1.9, 1.99, 2.01, 2.1 and use it to estimate the limit as x → 2. [3]
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Limits: approaching a value from the left and the right
Expected results
Six functions: (x² − 4) ÷ (x − 2) → 4 (removable hole); (x³ − 1) ÷ (x − 1) → 3; (√(x + 4) − 2) ÷ x → 0.25 (conjugate); 1 ÷ (x − 1) at 1 (infinite, no limit, vertical asymptote); (x² − 1) ÷ |x − 1| at 1 (left −2, right +2, no limit); (2x² + 3) ÷ (x² + 1) → 2 as x → ∞ (horizontal asymptote). The slider sets h = 10⁻ⁿ (x = a ± h), or x = 10ⁿ for the limit at infinity, up to n = 6; values are shown to 7 significant figures. The table on the picture keeps the last five steps so learners can see the pattern. Point out rounding: very small h can give rounding errors on calculators.
Answers
- What does lim x→a f(x) = L mean? [2]
f(x) can be made as close to L as we like by taking x close enough to a (but not equal to a), from both sides.
- Find lim x→3 (x² − 9) ÷ (x − 3). [2]
(x − 3)(x + 3) ÷ (x − 3) = x + 3 → 6.
- Find lim x→0 (√(x + 9) − 3) ÷ x. [2]
Multiply by the conjugate: 1 ÷ (√(x + 9) + 3) → 1 ÷ 6.
- When does a limit not exist? [2]
When the left-hand and right-hand limits are different, or when f(x) grows without bound (infinite limit).
- Find lim x→∞ (3x² − x) ÷ (2x² + 5). [2]
Divide by x²: (3 − 1/x) ÷ (2 + 5/x²) → 3 ÷ 2 = 1.5.
- Why can the limit at x = 2 exist even though f(2) is undefined? [2]
A limit describes the values near x = 2, not at x = 2 itself.
- Evaluate (a) lim x→1 (x² + x − 2) ÷ (x² − 1), (b) lim x→4 (√x − 2) ÷ (x − 4). [6]
(a) (x + 2)(x − 1) ÷ ((x + 1)(x − 1)) (2) = (x + 2) ÷ (x + 1) → 3 ÷ 2 (1). (b) x − 4 = (√x − 2)(√x + 2) (1), so the expression = 1 ÷ (√x + 2) (1) → 1 ÷ 4 (1).
- Find the horizontal asymptote of y = (4x² − 1) ÷ (2x² + x). [3]
Divide by x²: (4 − 1/x²) ÷ (2 + 1/x) (1); as x → ∞ this → 4 ÷ 2 (1) = 2, so y = 2 (1).
- f(x) = x + 1 for x < 2 and f(x) = 5 − x for x ≥ 2. Find the left and right limits at x = 2 and say whether the limit exists. [3]
Left: 2 + 1 = 3 (1); right: 5 − 2 = 3 (1); they are equal, so lim x→2 f(x) = 3 (1).
- Complete a table of values of f(x) = (x² − 4) ÷ (x − 2) for x = 1.9, 1.99, 2.01, 2.1 and use it to estimate the limit as x → 2. [3]
3.9, 3.99, 4.01, 4.1 (2); the limit is 4 (1).