⋆ ⋆ˆ ⋆
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MAT1512: CALCULUS A
OCT/NOV Examination 2026 Preparation
Comprehensive Revision Guide — Covers OCT/NOV 2023 to OCT/NOV 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
University of South Africa (UNISA)
Exam Revision Guide
MAT1512
Module Code:
Calculus A
Module Name:
OCT/NOV 2023, 2024 & 2025
Papers Covered:
OCT/NOV 2026
Target Exam:
100 per paper
Total Marks:
2 Hours
Duration:
Prof Z.I. Ali / Mr S. Blose
Examiner(s):
All calculations must be shown. Focus on understanding, not memorisation.
Exam Revision Notes | MAT1512 | 2026
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
PAPER 1 — OCTOBER/NOVEMBER 2023
Examiners: Mr S. Blose & Dr Z.A. Idriss | 100 Marks | 2 Hours
Page 2 of 31
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
Question 1 (2023) [30 marks]
(a) Limits [18 marks]
Question: Determine the following limits (if they exist):
x2 − 4
(i) lim (3)
x→2 x − 2
x3 + 1
(ii) lim (3)
x→−1 x + 1
x2 + 4x
(iii) lim (3)
x→∞ 2x − 1
x
(iv) lim √ (3)
x→∞ 1 + x
x2 − 9
(v) lim 2 (3)
x→3 x − 9
sin t + tan 2t
(vi) lim (3)
t→0 t
Answer:
x2 − 4
(i) lim
x→2 x − 2
Factor the numerator: x2 − 4 = (x − 2)(x + 2).
(x − 2)(x + 2)
lim = lim (x + 2) = 2 + 2 = 4
x→2 x−2 x→2
x3 + 1
(ii) lim
x→−1 x + 1
Factor numerator using sum of cubes: x3 + 1 = (x + 1)(x2 − x + 1).
(x + 1)(x2 − x + 1)
lim = lim (x2 − x + 1) = 1 + 1 + 1 = 3
x→−1 x+1 x→−1
x2 + 4x
(iii) lim
x→∞ 2x − 1
Divide numerator and denominator by x (highest power in denominator):
x+4 ∞
lim 1 = 2 = +∞
x→∞ 2− x
The limit does not exist (diverges to +∞).
x
(iv) lim √
x→∞ 1 + x
Page 3 of 31
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
√
Divide numerator and denominator by x:
√
x ∞
lim = = +∞
x→∞ √1 +1 0+1
x
The limit diverges.
x2 − 9
(v) lim
x→3 x2 − 9
The expression simplifies to 1 for all x ̸= 3, so the limit is 1 .
sin t + tan 2t
(vi) lim
t→0 t
Split the limit and use the standard results limt→0 sin t
t = 1 and limt→0 tan θ
θ = 1:
sin t tan 2t tan 2t
lim + lim = 1 + 2 lim = 1 + 2(1) = 3
t→0 t t→0 t t→0 2t
Exam Tip
0
For forms: always try factoring first. For limits involving sin or tan as argument
0
→ 0, use the fundamental limits limθ→0 sinθ θ = 1 and limθ→0 tanθ θ = 1. Multiply and
divide by the correct factor to match the argument.
(b) Squeeze Theorem [5 marks]
1 2
Question: Use the Squeeze Theorem to determine lim x sin .
x→0 x
Answer:
1
We know −1 ≤ sin ≤ 1 for all x ̸= 0.
x
Multiplying through by x2 ≥ 0:
2 12
−x ≤ x sin ≤ x2
x
Since limx→0 (−x2 ) = 0 and limx→0 (x2 ) = 0, by the Squeeze Theorem:
2 1
lim x sin = 0
x→0 x
Page 4 of 31
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
Key Concept
Squeeze Theorem: If g(x) ≤ f (x) ≤ h(x) near x = a and limx→a g(x) =
limx→a h(x) = L, then limx→a f (x) = L.
(c) Continuity of a Piecewise Function [7 marks]
Question: Let
x+2 if x > −1
h(x) = x2 if − 1 ≥ x < 1
3−x if x ≥ 1
Determine at which value(s) of x is h(x) continuous. (4+3)
Answer:
A function is continuous at x = c if: (1) h(c) is defined; (2) limx→c h(x) exists; (3) limx→c h(x) =
h(c).
Check potential break-points x = −1 and x = 1:
At x = −1:
• h(−1) = (−1)2 = 1 (using x2 branch, since −1 ≥ x < 1 includes −1)
• limx→−1+ h(x) = limx→−1+ (x + 2) = 1
• limx→−1− h(x) = limx→−1− (x2 ) = 1
• Both one-sided limits equal h(−1) = 1, so h is continuous at x = −1.
At x = 1:
• h(1) = 3 − 1 = 2 (using 3 − x branch since x ≥ 1)
• limx→1− h(x) = limx→1− (x2 ) = 1
• limx→1+ h(x) = limx→1+ (3 − x) = 2
• Left limit ̸= right limit, so h is discontinuous at x = 1 (jump discontinuity).
Therefore h(x) is continuous for all x ∈ R except at x = 1.
Page 5 of 31
, MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
Question 2 (2023) [25 marks]
(a) Differentiation from First Principles [8 marks]
Question: Using the first principle of differentiation, determine the derivative of f (x) =
2x2 − 3x + 5 at x = 2.
Answer:
The derivative from first principles is:
f (x + h) − f (x)
f ′ (x) = lim
h→0 h
Compute f (x + h):
f (x + h) = 2(x + h)2 − 3(x + h) + 5 = 2x2 + 4xh + 2h2 − 3x − 3h + 5
So:
f (x + h) − f (x) = 4xh + 2h2 − 3h
4xh + 2h2 − 3h
f ′ (x) = lim = lim (4x + 2h − 3) = 4x − 3
h→0 h h→0
At x = 2: f ′ (2) = 4(2) − 3 = 5
(b) Differentiation Rules [12 marks]
Question: Find the derivatives of the following functions using the appropriate rules of
differentiation:
x2 + x
(i) f (x) = (4)
sin x cos x
(ii) g(x) = cos(5x) + sin(x2 ) (4)
(iii) y = (x3 + 1)4 ln(x) (4)
Answer:
u ′ u′ v − uv ′
(i) Using the Quotient Rule: =
v v2
Let u = x2 + x, v = sin x cos x = 1
2 sin 2x.
Page 6 of 31
]]
MAT1512: CALCULUS A
OCT/NOV Examination 2026 Preparation
Comprehensive Revision Guide — Covers OCT/NOV 2023 to OCT/NOV 2025
⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆ ⋄ ⋆
University of South Africa (UNISA)
Exam Revision Guide
MAT1512
Module Code:
Calculus A
Module Name:
OCT/NOV 2023, 2024 & 2025
Papers Covered:
OCT/NOV 2026
Target Exam:
100 per paper
Total Marks:
2 Hours
Duration:
Prof Z.I. Ali / Mr S. Blose
Examiner(s):
All calculations must be shown. Focus on understanding, not memorisation.
Exam Revision Notes | MAT1512 | 2026
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
PAPER 1 — OCTOBER/NOVEMBER 2023
Examiners: Mr S. Blose & Dr Z.A. Idriss | 100 Marks | 2 Hours
Page 2 of 31
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
Question 1 (2023) [30 marks]
(a) Limits [18 marks]
Question: Determine the following limits (if they exist):
x2 − 4
(i) lim (3)
x→2 x − 2
x3 + 1
(ii) lim (3)
x→−1 x + 1
x2 + 4x
(iii) lim (3)
x→∞ 2x − 1
x
(iv) lim √ (3)
x→∞ 1 + x
x2 − 9
(v) lim 2 (3)
x→3 x − 9
sin t + tan 2t
(vi) lim (3)
t→0 t
Answer:
x2 − 4
(i) lim
x→2 x − 2
Factor the numerator: x2 − 4 = (x − 2)(x + 2).
(x − 2)(x + 2)
lim = lim (x + 2) = 2 + 2 = 4
x→2 x−2 x→2
x3 + 1
(ii) lim
x→−1 x + 1
Factor numerator using sum of cubes: x3 + 1 = (x + 1)(x2 − x + 1).
(x + 1)(x2 − x + 1)
lim = lim (x2 − x + 1) = 1 + 1 + 1 = 3
x→−1 x+1 x→−1
x2 + 4x
(iii) lim
x→∞ 2x − 1
Divide numerator and denominator by x (highest power in denominator):
x+4 ∞
lim 1 = 2 = +∞
x→∞ 2− x
The limit does not exist (diverges to +∞).
x
(iv) lim √
x→∞ 1 + x
Page 3 of 31
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
√
Divide numerator and denominator by x:
√
x ∞
lim = = +∞
x→∞ √1 +1 0+1
x
The limit diverges.
x2 − 9
(v) lim
x→3 x2 − 9
The expression simplifies to 1 for all x ̸= 3, so the limit is 1 .
sin t + tan 2t
(vi) lim
t→0 t
Split the limit and use the standard results limt→0 sin t
t = 1 and limt→0 tan θ
θ = 1:
sin t tan 2t tan 2t
lim + lim = 1 + 2 lim = 1 + 2(1) = 3
t→0 t t→0 t t→0 2t
Exam Tip
0
For forms: always try factoring first. For limits involving sin or tan as argument
0
→ 0, use the fundamental limits limθ→0 sinθ θ = 1 and limθ→0 tanθ θ = 1. Multiply and
divide by the correct factor to match the argument.
(b) Squeeze Theorem [5 marks]
1 2
Question: Use the Squeeze Theorem to determine lim x sin .
x→0 x
Answer:
1
We know −1 ≤ sin ≤ 1 for all x ̸= 0.
x
Multiplying through by x2 ≥ 0:
2 12
−x ≤ x sin ≤ x2
x
Since limx→0 (−x2 ) = 0 and limx→0 (x2 ) = 0, by the Squeeze Theorem:
2 1
lim x sin = 0
x→0 x
Page 4 of 31
,MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
Key Concept
Squeeze Theorem: If g(x) ≤ f (x) ≤ h(x) near x = a and limx→a g(x) =
limx→a h(x) = L, then limx→a f (x) = L.
(c) Continuity of a Piecewise Function [7 marks]
Question: Let
x+2 if x > −1
h(x) = x2 if − 1 ≥ x < 1
3−x if x ≥ 1
Determine at which value(s) of x is h(x) continuous. (4+3)
Answer:
A function is continuous at x = c if: (1) h(c) is defined; (2) limx→c h(x) exists; (3) limx→c h(x) =
h(c).
Check potential break-points x = −1 and x = 1:
At x = −1:
• h(−1) = (−1)2 = 1 (using x2 branch, since −1 ≥ x < 1 includes −1)
• limx→−1+ h(x) = limx→−1+ (x + 2) = 1
• limx→−1− h(x) = limx→−1− (x2 ) = 1
• Both one-sided limits equal h(−1) = 1, so h is continuous at x = −1.
At x = 1:
• h(1) = 3 − 1 = 2 (using 3 − x branch since x ≥ 1)
• limx→1− h(x) = limx→1− (x2 ) = 1
• limx→1+ h(x) = limx→1+ (3 − x) = 2
• Left limit ̸= right limit, so h is discontinuous at x = 1 (jump discontinuity).
Therefore h(x) is continuous for all x ∈ R except at x = 1.
Page 5 of 31
, MAT1512 | Calculus A — Exam Revision OCT/NOV 2023–2025
Question 2 (2023) [25 marks]
(a) Differentiation from First Principles [8 marks]
Question: Using the first principle of differentiation, determine the derivative of f (x) =
2x2 − 3x + 5 at x = 2.
Answer:
The derivative from first principles is:
f (x + h) − f (x)
f ′ (x) = lim
h→0 h
Compute f (x + h):
f (x + h) = 2(x + h)2 − 3(x + h) + 5 = 2x2 + 4xh + 2h2 − 3x − 3h + 5
So:
f (x + h) − f (x) = 4xh + 2h2 − 3h
4xh + 2h2 − 3h
f ′ (x) = lim = lim (4x + 2h − 3) = 4x − 3
h→0 h h→0
At x = 2: f ′ (2) = 4(2) − 3 = 5
(b) Differentiation Rules [12 marks]
Question: Find the derivatives of the following functions using the appropriate rules of
differentiation:
x2 + x
(i) f (x) = (4)
sin x cos x
(ii) g(x) = cos(5x) + sin(x2 ) (4)
(iii) y = (x3 + 1)4 ln(x) (4)
Answer:
u ′ u′ v − uv ′
(i) Using the Quotient Rule: =
v v2
Let u = x2 + x, v = sin x cos x = 1
2 sin 2x.
Page 6 of 31
