UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
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MAT1581: Mathematics I (Engineering)
Assignment 03 | Year Module, 2026
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MAT1581
Module Code:
Mathematics I (Engineering)
Module Name:
Differentiation, from Study Guide 2
Essay Topic:
Assignment 03
Assignment Number:
06 August 2026 (23:59)
Due Date:
47
Total Marks:
Submitted in partial fulfilment of the requirements
for Mathematics I (Engineering), UNISA 2026
, UNISA | MAT1581 Assignment 03: Differentiation
Question 1: Limits
Both limits below are evaluated at a point where the denominator is non-zero, so direct substitution
applies without needing to resolve an indeterminate form.
x3 − x + 1
1.1 lim
x→−2 x4 − 4x + 3
Substituting x = −2 into the numerator and denominator gives
(−2)3 − (−2) + 1 −8 + 2 + 1
=
(−2)4 − 4(−2) + 3 16 + 8 + 3
−5
= .
27
x3 − x + 1 5
lim 4
=−
x→−2 x − 4x + 3 27
2x3 + 16
1.2 lim
x→1 3x4 − 243
Substituting x = 1 gives
2(1)3 + 16 2 + 16
4
=
3(1) − 243 3 − 243
18
= .
−240
Dividing the numerator and denominator by their common factor of 6 simplifies this to
3
− .
40
2x3 + 16 3
lim 4
=−
x→1 3x − 243 40
Page 2 of 9
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT1581: Mathematics I (Engineering)
Assignment 03 | Year Module, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
MAT1581
Module Code:
Mathematics I (Engineering)
Module Name:
Differentiation, from Study Guide 2
Essay Topic:
Assignment 03
Assignment Number:
06 August 2026 (23:59)
Due Date:
47
Total Marks:
Submitted in partial fulfilment of the requirements
for Mathematics I (Engineering), UNISA 2026
, UNISA | MAT1581 Assignment 03: Differentiation
Question 1: Limits
Both limits below are evaluated at a point where the denominator is non-zero, so direct substitution
applies without needing to resolve an indeterminate form.
x3 − x + 1
1.1 lim
x→−2 x4 − 4x + 3
Substituting x = −2 into the numerator and denominator gives
(−2)3 − (−2) + 1 −8 + 2 + 1
=
(−2)4 − 4(−2) + 3 16 + 8 + 3
−5
= .
27
x3 − x + 1 5
lim 4
=−
x→−2 x − 4x + 3 27
2x3 + 16
1.2 lim
x→1 3x4 − 243
Substituting x = 1 gives
2(1)3 + 16 2 + 16
4
=
3(1) − 243 3 − 243
18
= .
−240
Dividing the numerator and denominator by their common factor of 6 simplifies this to
3
− .
40
2x3 + 16 3
lim 4
=−
x→1 3x − 243 40
Page 2 of 9
