Limits and continuity: 7 exam questions solved step by step

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What this lesson covers

  • A function is continuous where its graph has no break, and differentiable where it also has no sharp corner.
  • Reading these from a piecewise graph is Achieved-level work that is often left blank.
Key idea

7 questions, one checklist

  • 1. Limit: both sides head to one value.
  • 2. Continuous: limit = value.
  • 3. Differentiable: continuous, one gradient.
  • 4. A reason with every answer.
Your turn

Question 1 · Achieved

  • The graph of y = f(x) is shown.
  • Find the limit of f(x) as x → 2
  • from the left, and from the right.
  • Find lim(x→2) f(x), and f(2).
Question 1 · Achieved
Worked example

Q1: limits at x = 2

  • Follow each piece in.
  • Value: the closed dot.
  1. from the left: f(x) → 3along the line
  2. from the right: f(x) → 4along the curve
  3. 3 ≠ 4the sides disagree
  4. lim(x→2) f(x) does not exist
  5. f(2) = 4the closed dot
  6. f(2) = 3open dot: not on it
Your turn

Question 2 · Achieved

  • The same graph.
  • Find lim(x→4) f(x), and f(4).
  • Is f continuous at x = 4? Give a reason.
  • Careful: two dots sit above x = 4.
Question 2 · Achieved
Worked example

Q2: at x = 4

  • Limit: where it heads.
  • Value: the closed dot.
  1. from the left: f(x) → 4
  2. from the right: f(x) → 4
  3. lim(x→4) f(x) = 4both sides agree
  4. f(4) = 4open dot: a hole
  5. f(4) = 2the closed dot
  6. 4 ≠ 2limit ≠ value
  7. not continuous at x = 4
Your turn

Question 3 · Merit

  • The same graph.
  • For 0 < x < 7, where is f
  • not differentiable? Give reasons.
Question 3 · Merit
Worked example

Q3: not differentiable where?

  • Continuous first.
  • Then one gradient.
  1. x = 2: a jumpnot continuous
  2. x = 4: a holenot continuous
  3. x = 5: lim f(x) = 1 = f(5)continuous
  4. left of 5: falling, f′ < 0
  5. right: (5 − 1) ÷ (7 − 5) = 2rise over run
  6. f′ < 0 and f′ = 2: a cornergradients differ
  7. x = 3: smooth, f′(3) = 0differentiable
  8. not differentiable at x = 2, 4, 5
Your turn

Question 4 · Merit

  • The same graph.
  • For which values of x is f′(x) > 0?
  • Where is f′(x) < 0, and f′(x) = 0?
  • Careful at x = 2 and x = 4.
Question 4 · Merit
Worked example

Q4: the sign of f′(x)

  • Rising: f′ > 0.
  • Falling: f′ < 0.
  1. 0 < x < 2: rising, gradient 1
  2. 2 < x < 3: rising
  3. x = 3: maximum, f′(3) = 0
  4. 3 < x < 5, x ≠ 4: falling
  5. 5 < x < 7: rising, gradient 2
  6. f′(x) > 0: 0 < x < 2, 2 < x < 3
  7. and 5 < x < 7
  8. f′(x) < 0: 3 < x < 4, 4 < x < 5
  9. f′(x) = 0: x = 3
Your turn

Question 5 · Merit, in context

  • A drone's height h m after t s is
  • h = t² + 3 for 0 ≤ t ≤ 1,
  • h = 6t − 2 for 1 < t ≤ 4.
  • Is h continuous and differentiable at t = 1?
Worked example

Q5: the drone at t = 1

  • Do the heights meet?
  • Do the rates match?
  1. left: h(1) = 1² + 3 = 4the first piece
  2. right: 6(1) − 2 = 4as t → 1 from above
  3. lim h(t) = 4 = h(1)continuous
  4. left: h′(t) = 2t, h′(1) = 2
  5. right: h′(t) = 6
  6. 2 ≠ 6the rates differ
  7. not differentiable at t = 1
  8. height 4 m both sides: no jump
  9. climb rate jumps 2 to 6 m/s
Your turn

Question 6 · Excellence

  • f(x) = a√x for 0 ≤ x ≤ 4,
  • f(x) = x + b for x > 4.
  • f is differentiable at x = 4.
  • Find a and b.
Worked example

Q6: find a and b

  • The pieces meet.
  • The gradients match.
  1. a√4 = 4 + bpieces meet at 4
  2. 2a = 4 + b
  3. d/dx a√x = a/(2√x)a√x = ax^(1/2)
  4. d/dx (x + b) = 1
  5. a/(2√4) = 1gradients match
  6. a/4 = 1, so a = 4
  7. 2(4) = 4 + b, so b = 4substitute a
  8. a = 4, b = 4
  9. 4√4 = 8 = 4 + 4
Your turn

Question 7 · Excellence

  • The graph of f(x) = |x² − 4x| is shown.
  • Find where f is not differentiable,
  • and where f′(x) = 0. Justify both.
Question 7 · Excellence
Worked example

Q7: write f in pieces

  • Where is x² − 4x negative?
  • There, change its sign.
  1. x² − 4x = x(x − 4)
  2. x(x − 4) < 0 for 0 < x < 4between the roots
  3. f(x) = x² − 4x, x ≤ 0 or x ≥ 4
  4. f(x) = 4x − x², 0 < x < 4sign changed
  5. f′(x) = 2x − 4 outsidedifferentiate
  6. f′(x) = 4 − 2x inside
  7. f is continuous everywherepieces meet at 0, 4
Worked example

Q7: corners, and f′(x) = 0

  • Gradients at 0 and at 4.
  • Reject a root off its piece.
  1. x = 0: left 2(0) − 4 = −4
  2. right 4 − 2(0) = 4
  3. x = 4: left 4 − 2(4) = −4
  4. right 2(4) − 4 = 4
  5. −4 ≠ 4 at both: cornersgradients differ
  6. not differentiable at x = 0, 4
  7. 2x − 4 = 0: x = 2not in x ≤ 0, x ≥ 4
  8. 4 − 2x = 0: x = 2, f(2) = 4the inside piece
  9. f′(x) = 0 only at (2, 4)
Summary

The checklist that solved all seven

  • Both sides, then the value, then the gradients, with a reason each time.
StepHow
Limitboth sides agree
Valuethe closed dot
Continuouslimit equals value
Cornergradients differ
Sign of f′rising +, falling −
Constantsmeet, then match

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