MAT1512 Assignment 2 Due 22 June 2026 |Calculus A|

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


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MAT1512: Calculus I, Assignment 02

Worked Solutions in Question and Answer Format

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MAT1512
Module Code:
Calculus I
Module Name:
Tutorial Letter 102/3/2026
Reference:
Assignment 02
Assignment:
50
Total Marks:
22 June 2026
Date:




Worked solutions prepared for revision purposes, UNISA

, UNISA | MAT1512 Assignment 02, Limit Evaluations



Question 1: Evaluating Limits by Factorisation and Standard Forms

Part (a)

Question. Evaluate
x2 − 9
lim .
x→3 x2 − 5x + 6

Solution.
The numerator factors as
x2 − 9 = (x − 3)(x + 3).

The denominator factors as
x2 − 5x + 6 = (x − 3)(x − 2).

Substituting these factors, the limit becomes
(x − 3)(x + 3)
lim .
x→3 (x − 3)(x − 2)

The factor (x − 3) cancels, since x ̸= 3 throughout the limiting process, leaving
x+3
lim .
x→3 x − 2


Substituting x = 3 gives
3+3 6
= = 6.
3−2 1

x2 − 9
lim = 6
x→3 x2 − 5x + 6



Part (b)

Question. Evaluate √
1 + 2x − 1
lim .
x→0 x
Solution.
√
The expression is multiplied by the conjugate of the numerator, 1 + 2x + 1, over itself:
√ √
1 + 2x − 1 1 + 2x + 1
×√ .
x 1 + 2x + 1

The numerator becomes a difference of squares:
(1 + 2x) − 1 2x
= √
= √
.
x 1 + 2x + 1 x 1 + 2x + 1

The factor x cancels, since x ̸= 0 throughout the limiting process, leaving
2
√ .
1 + 2x + 1

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