MAT1581 Assignment 3 2026 Solutions Due 06 August 2026

UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology



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MATHEMATICS I (ENGINEERING)
Assignment 03 — Differentiation (Study Guide 2) — 2026


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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




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