0
(Jan/Feb)
MAT1512 Exam
2026
, The A+Math Chicken
QUESTION 1
a. (i)
Evaluate the following limit (state if the limit does not exist):
𝑥2 − 4
lim
𝑥→2 𝑥 − 2
[3 marks]
ANSWER:
𝑥2 − 4 (𝑥 + 2)(𝑥 − 2)
lim = lim
𝑥→2 𝑥 − 2 𝑥→2 𝑥−2
= lim (𝑥 + 2)
𝑥→2
= (2) + 2
=4
a. (ii)
Evaluate the following limit (state if the limit does not exist):
2𝑥 3 + 2𝑥 + 1
lim
𝑥→∞ 𝑥3 + 1
[4 marks]
ANSWER:
2𝑥 3 2𝑥 1
3 + 𝑥3 + 𝑥3
3
2𝑥 + 2𝑥 + 1 𝑥
lim = lim [ ]
𝑥→∞ 𝑥3 + 1 𝑥→∞ 𝑥3 1
+
𝑥3 𝑥3
𝑥 1
2+ 2+ 3
= lim [ 𝑥 𝑥 ]
𝑥→∞ 1
1+ 3
𝑥
2 + 2(0) + 0
=
1+0
=2
, The A+Math Chicken
a. (iii)
Evaluate the following limit (state if the limit does not exist):
√𝑥 − 1
lim
𝑥→1 𝑥 − 1
[4 marks]
ANSWER:
√𝑥 − 1 √𝑥 − 1 √𝑥 + 1
lim = lim [ ∙ ]
𝑥→1 𝑥 − 1 𝑥→1 𝑥 − 1 √𝑥 + 1
2
(√𝑥) + √𝑥 − √𝑥 − 1
= lim [ ]
𝑥→1 (𝑥 − 1)(√𝑥 + 1)
𝑥−1
= lim
𝑥→1 (𝑥 − 1)(√𝑥 + 1)
1
= lim
𝑥→1 √𝑥 +1
1
=
√1 + 1
1
=
2
a. (iv)
Evaluate the following limit (state if the limit does not exist):
𝑥
lim
𝑥→0 √1 +𝑥−1
[4 marks]
ANSWER:
𝑥 𝑥 √1 + 𝑥 − 1
lim = lim [ ∙ ]
𝑥→0 √1 +𝑥−1 𝑥→0 √1 + 𝑥 − 1 √1 + 𝑥 − 1
, The A+Math Chicken
𝑥(√1 + 𝑥 − 1)
= lim [ 2 ]
𝑥→0
(√1 + 𝑥) − 1
𝑥(√1 + 𝑥 − 1)
= lim
𝑥→0 1+𝑥−1
𝑥(√1 + 𝑥 − 1)
= lim
𝑥→0 𝑥
= lim (√1 + 𝑥 − 1)
𝑥→0
= √1 + 0 − 1
=2
(b)
Use the Squeeze (Sandwich) Theorem to evaluate:
1
lim 𝑛2 sin ( )
𝑛→0 𝑛
[3 marks]
ANSWER:
The given sin graph oscillates between −1 and 1 .
1
−1 ≤ sin ( ) ≤ 1
𝑛
1
𝑛2 × −1 ≤ 𝑛2 × sin ( ) ≤ 1 × 𝑛2
𝑛
1
−𝑛2 ≤ 𝑛2 sin ( ) ≤ 𝑛2
𝑛
lim 𝑛2 = 0 and lim (−𝑛2 ) = 0
𝑥→0 𝑥→0
1
∴ lim 𝑛2 sin ( ) = 0
𝑥→0 𝑛
(Jan/Feb)
MAT1512 Exam
2026
, The A+Math Chicken
QUESTION 1
a. (i)
Evaluate the following limit (state if the limit does not exist):
𝑥2 − 4
lim
𝑥→2 𝑥 − 2
[3 marks]
ANSWER:
𝑥2 − 4 (𝑥 + 2)(𝑥 − 2)
lim = lim
𝑥→2 𝑥 − 2 𝑥→2 𝑥−2
= lim (𝑥 + 2)
𝑥→2
= (2) + 2
=4
a. (ii)
Evaluate the following limit (state if the limit does not exist):
2𝑥 3 + 2𝑥 + 1
lim
𝑥→∞ 𝑥3 + 1
[4 marks]
ANSWER:
2𝑥 3 2𝑥 1
3 + 𝑥3 + 𝑥3
3
2𝑥 + 2𝑥 + 1 𝑥
lim = lim [ ]
𝑥→∞ 𝑥3 + 1 𝑥→∞ 𝑥3 1
+
𝑥3 𝑥3
𝑥 1
2+ 2+ 3
= lim [ 𝑥 𝑥 ]
𝑥→∞ 1
1+ 3
𝑥
2 + 2(0) + 0
=
1+0
=2
, The A+Math Chicken
a. (iii)
Evaluate the following limit (state if the limit does not exist):
√𝑥 − 1
lim
𝑥→1 𝑥 − 1
[4 marks]
ANSWER:
√𝑥 − 1 √𝑥 − 1 √𝑥 + 1
lim = lim [ ∙ ]
𝑥→1 𝑥 − 1 𝑥→1 𝑥 − 1 √𝑥 + 1
2
(√𝑥) + √𝑥 − √𝑥 − 1
= lim [ ]
𝑥→1 (𝑥 − 1)(√𝑥 + 1)
𝑥−1
= lim
𝑥→1 (𝑥 − 1)(√𝑥 + 1)
1
= lim
𝑥→1 √𝑥 +1
1
=
√1 + 1
1
=
2
a. (iv)
Evaluate the following limit (state if the limit does not exist):
𝑥
lim
𝑥→0 √1 +𝑥−1
[4 marks]
ANSWER:
𝑥 𝑥 √1 + 𝑥 − 1
lim = lim [ ∙ ]
𝑥→0 √1 +𝑥−1 𝑥→0 √1 + 𝑥 − 1 √1 + 𝑥 − 1
, The A+Math Chicken
𝑥(√1 + 𝑥 − 1)
= lim [ 2 ]
𝑥→0
(√1 + 𝑥) − 1
𝑥(√1 + 𝑥 − 1)
= lim
𝑥→0 1+𝑥−1
𝑥(√1 + 𝑥 − 1)
= lim
𝑥→0 𝑥
= lim (√1 + 𝑥 − 1)
𝑥→0
= √1 + 0 − 1
=2
(b)
Use the Squeeze (Sandwich) Theorem to evaluate:
1
lim 𝑛2 sin ( )
𝑛→0 𝑛
[3 marks]
ANSWER:
The given sin graph oscillates between −1 and 1 .
1
−1 ≤ sin ( ) ≤ 1
𝑛
1
𝑛2 × −1 ≤ 𝑛2 × sin ( ) ≤ 1 × 𝑛2
𝑛
1
−𝑛2 ≤ 𝑛2 sin ( ) ≤ 𝑛2
𝑛
lim 𝑛2 = 0 and lim (−𝑛2 ) = 0
𝑥→0 𝑥→0
1
∴ lim 𝑛2 sin ( ) = 0
𝑥→0 𝑛
