UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
MATHEMATICS I (ENGINEERING)
Assignment 03 — Differentiation (Study Guide 2) — 2026
⋄
Module Code: MAT1581
Module Name: Mathematics I (Engineering)
Assignment No.: Assignment 03
Due Date: Thursday, 6 August 2026, 23:59
Semester: Year Module — 2026
Submitted in partial fulfilment of the requirements for Mathematics I (Engineering)
at the University of South Africa.
, UNISA | MAT1581 Assignment 03 — Differentiation
Question 1: Evaluating Limits by Direct Substitution
Both limits below are evaluated at a point where the denominator does not vanish, so direct
substitution applies without any need for factorisation or L’Hospital’s rule.
1.1 Limit of a Rational Function as x → −2
The limit to evaluate is
x3 − x + 1
lim .
x→−2 x4 − 4x + 3
Substituting x = −2 into the numerator and denominator gives
(−2)3 − (−2) + 1 −8 + 2 + 1 −5
4
= = .
(−2) − 4(−2) + 3 16 + 8 + 3 27
Since the denominator is non-zero at x = −2, this substitution is valid and
x3 − x + 1 5
lim 4
=− .
x→−2 x − 4x + 3 27
1.2 Limit of a Rational Function as x → 1
The limit to evaluate is
2x3 + 16
lim .
x→1 3x4 − 243
Substituting x = 1 gives
2(1)3 + 16 2 + 16 18
= = .
3(1)4 − 243 3 − 243 −240
Dividing the numerator and denominator by their common factor of 6 simplifies this to
18 3
=− .
−240 40
Therefore
2x3 + 16 3
lim =− .
x→1 3x4 − 243 40
Page 1 of 8
College of Science, Engineering and Technology
⋄
MATHEMATICS I (ENGINEERING)
Assignment 03 — Differentiation (Study Guide 2) — 2026
⋄
Module Code: MAT1581
Module Name: Mathematics I (Engineering)
Assignment No.: Assignment 03
Due Date: Thursday, 6 August 2026, 23:59
Semester: Year Module — 2026
Submitted in partial fulfilment of the requirements for Mathematics I (Engineering)
at the University of South Africa.
, UNISA | MAT1581 Assignment 03 — Differentiation
Question 1: Evaluating Limits by Direct Substitution
Both limits below are evaluated at a point where the denominator does not vanish, so direct
substitution applies without any need for factorisation or L’Hospital’s rule.
1.1 Limit of a Rational Function as x → −2
The limit to evaluate is
x3 − x + 1
lim .
x→−2 x4 − 4x + 3
Substituting x = −2 into the numerator and denominator gives
(−2)3 − (−2) + 1 −8 + 2 + 1 −5
4
= = .
(−2) − 4(−2) + 3 16 + 8 + 3 27
Since the denominator is non-zero at x = −2, this substitution is valid and
x3 − x + 1 5
lim 4
=− .
x→−2 x − 4x + 3 27
1.2 Limit of a Rational Function as x → 1
The limit to evaluate is
2x3 + 16
lim .
x→1 3x4 − 243
Substituting x = 1 gives
2(1)3 + 16 2 + 16 18
= = .
3(1)4 − 243 3 − 243 −240
Dividing the numerator and denominator by their common factor of 6 simplifies this to
18 3
=− .
−240 40
Therefore
2x3 + 16 3
lim =− .
x→1 3x4 − 243 40
Page 1 of 8
