MAT1501 Assignment 2 ANSWERS 2025 - Due July 2025

MAT1501
ASSIGNMENT 2 2025
DUE: JULY/AUG 2025 (MEMO)

, Section A
Question 1

State if true or false; if false, correct the statement.

1.1 Statement: limx→a f (x) ​ = f (a)
False.
Correction: If f is continuous at x = a, then limx→a f (x) = f (a). ​




1.2 Statement: limx→a (f (x) − g(x)) ​ = limx→a f (x) + limx→a g(x) ​ ​




False.
Correction: limx→a (f (x) − g(x))​ = limx→a f (x) − limx→a g(x), provided both limits exist.
​ ​




f (x) limx→a f (x)−limx→a g(x)
1.3 Statement: limx→c g(x) = ​ ​




​ ​




limx→a g(x) ​
​




False.
f (x) lim f (x)
Correction: limx→a g(x) = lim x→a g(x) , provided limx→a g(x) = 0 and both limits exist. (Note: The
​




​ ​ ​ ​




x→a ​




variable inconsistency x → c vs. x → a is assumed to be a typo; corrected to x → a.)

1.4 Statement: limx→a f (x) ⋅ g(x) ​ = (limx→a f (x))(limx→a g(x))
​ ​




True, provided both limits exist.

1.5 Statement: limx→a f (x)n ​ = (n − 1) limx→a f (x)(n−1) ​




False.
n
Correction: limx→a [f (x)]n ​ = [limx→a f (x)] , provided the limit exists and is appropriate for the power
​




e.g., positive if n is fractional



Question 2
Evaluate the limits.

2.1 limx→3 (x3
​ + 2)(x2 − 5x)

Since the function is a polynomial, it is continuous everywhere. Substitute x = 3:
(33 + 2) = 27 + 2 = 29,
(32 − 5 ⋅ 3) = 9 − 15 = −6,
29 ⋅ (−6) = −174.
−174
∣x+4∣
2.2 limx→−4− x+4 ​ ​