MAT1581
Assignment 3
Due 5 August 2025
, ASSIGNMENT 03
Due date: Tuesday, 5 August 2025
Question 1: Limits (6 marks)
1.1
𝑥3 − 2 (−1)3 − 2 −3 1
𝑙𝑖𝑚𝑥→−1 4
= = =
𝑥 + 4𝑥 − 3 (−1)4 + 4(−1) − 3 6 2
1.2
𝑥3 − 8 27 − 8
𝑙𝑖𝑚𝑥→3 = = 519
𝑥2 − 4 9−4
Question 2: Differentiation (22 marks)
2.1
1
1 𝑑𝑦 𝑠𝑖𝑛𝑥 + 1
𝑦= ⇒ =− 2
𝑥 − √𝑐𝑜𝑠𝑥 𝑑𝑥 (𝑥 − √cos(𝑥) )2
Answer:
sin(𝑋) /√cos(𝑥) + 2
−
2(𝑥 − √cos (𝑥)) 2
2.2
y = (x 3 + 2)(x 2 − 2x + 1)
Using the product rule:
𝑑𝑦
= (3𝑥 2 )(𝑥 2 − 2𝑥 + 1) + (𝑥 3 + 2)(2𝑥 − 2)
𝑑𝑥
Simplified:
5x 4 − 8x 3 + 3x 2 + 4x − 4
2.3
Assignment 3
Due 5 August 2025
, ASSIGNMENT 03
Due date: Tuesday, 5 August 2025
Question 1: Limits (6 marks)
1.1
𝑥3 − 2 (−1)3 − 2 −3 1
𝑙𝑖𝑚𝑥→−1 4
= = =
𝑥 + 4𝑥 − 3 (−1)4 + 4(−1) − 3 6 2
1.2
𝑥3 − 8 27 − 8
𝑙𝑖𝑚𝑥→3 = = 519
𝑥2 − 4 9−4
Question 2: Differentiation (22 marks)
2.1
1
1 𝑑𝑦 𝑠𝑖𝑛𝑥 + 1
𝑦= ⇒ =− 2
𝑥 − √𝑐𝑜𝑠𝑥 𝑑𝑥 (𝑥 − √cos(𝑥) )2
Answer:
sin(𝑋) /√cos(𝑥) + 2
−
2(𝑥 − √cos (𝑥)) 2
2.2
y = (x 3 + 2)(x 2 − 2x + 1)
Using the product rule:
𝑑𝑦
= (3𝑥 2 )(𝑥 2 − 2𝑥 + 1) + (𝑥 3 + 2)(2𝑥 − 2)
𝑑𝑥
Simplified:
5x 4 − 8x 3 + 3x 2 + 4x − 4
2.3
