MAT1512 Assignment 2 solutions 2026
UNISA
• Total Marks: 50
Date: Due: Monday, 22 June 2026, 8:00 PM
DO NOT FORGET TO GIVE ME FEED BACK ON HOW MUCH YOU SCORED IN THIS
ASSIGNMENT, YOU FEEDBACK HELPS US IMPROVE
, MAT1512 – Assignment 02 Solutions
Show all steps and calculations
QUESTION 1
(a)
𝑥2 − 9
lim 2
𝑥→3 𝑥 − 5𝑥 + 6
Factorise numerator and denominator
𝑥 2 − 9 = (𝑥 − 3)(𝑥 + 3)
𝑥 2 − 5𝑥 + 6 = (𝑥 − 3)(𝑥 − 2)
Therefore
𝑥2 − 9 (𝑥 − 3)(𝑥 + 3)
=
𝑥 2 − 5𝑥 + 6 (𝑥 − 3)(𝑥 − 2)
Cancel (𝑥−3):
𝑥+3
=
𝑥−2
Substitute 𝒙 = 𝟑
𝑥+3 3+3
lim =
𝑥→3 𝑥 − 2 3−2
6
=
1
6
UNISA
• Total Marks: 50
Date: Due: Monday, 22 June 2026, 8:00 PM
DO NOT FORGET TO GIVE ME FEED BACK ON HOW MUCH YOU SCORED IN THIS
ASSIGNMENT, YOU FEEDBACK HELPS US IMPROVE
, MAT1512 – Assignment 02 Solutions
Show all steps and calculations
QUESTION 1
(a)
𝑥2 − 9
lim 2
𝑥→3 𝑥 − 5𝑥 + 6
Factorise numerator and denominator
𝑥 2 − 9 = (𝑥 − 3)(𝑥 + 3)
𝑥 2 − 5𝑥 + 6 = (𝑥 − 3)(𝑥 − 2)
Therefore
𝑥2 − 9 (𝑥 − 3)(𝑥 + 3)
=
𝑥 2 − 5𝑥 + 6 (𝑥 − 3)(𝑥 − 2)
Cancel (𝑥−3):
𝑥+3
=
𝑥−2
Substitute 𝒙 = 𝟑
𝑥+3 3+3
lim =
𝑥→3 𝑥 − 2 3−2
6
=
1
6
