MAT1581 Assignment 3 | Due 5 August 2025 COMPLETE ANSWERS. Ensure your succes with us

Exam (elaborations) MAT1581 Assignment 3 Memo | Due 5 August 2025

• Course • Mathematics I (MAT1581)

• Institution • University Of South Africa (Unisa)

MAT1581 Assignment 3 Memo | Due 5 August 2025. Step by Step Calculations
Provided.

Question 1: 6 Marks Determine the following limits: (1.1) lim (3) x→−1 x3 − 2 x4 + 4x −
3 (1.2) lim (3) x→3 x3 − 8 x2 − 4

1.1 lim ⁡→→ − 1 𝑥 3 − 2 𝑥 4 + 4 𝑥 − 3 x→ −1 lim

x 4 +4x−3 x 3 −2


Step 1: Substitute 𝑥
− 1 x=−1.


Numerator: ( − 1 ) 3 − 2

−1−2

− 3 Numerator: (−1) 3 −2=−1−2=−3 Denominator: ( −
1)4+4(−1)−3

1−4−3
− 6 Denominator: (−1) 4 +4(−1)−3=1−4−3=−6 Step 2: Simplify.


−3−6

1 2 −6 −3
21

, Answer:


lim ⁡𝑥 → − 1 𝑥 3 − 2 𝑥 4 + 4 𝑥 − 3
1 2 x→−1 lim


x 4 +4x−3 x 3 −2
21



1.2 lim ⁡𝑥 → 3 𝑥 3 − 8 𝑥 2 − 4 x→3 lim

x 2 −4 x 3 −8


Step 1: Substitute 𝑥
3 x=3.


Numerator: 3 3 − 8

27 − 8

19 Numerator: 3 3 −8=27−8=19 Denominator: 3 2 −
4

9−4
5 Denominator: 3 2 −4=9−4=5 Step 2: Simplify.

19 5 5 19

Answer: