MAT1512 Assignment 9 2021 (Oct/Nov 2020 exam)

MAT1512 ASSIGNMENT 9 2021



Question 1



(𝑎)
(𝑖)



𝑀𝑒𝑡ℎ𝑜𝑑 1 (𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑠𝑎𝑡𝑖𝑜𝑛)

𝑥 2 + 𝑥 − 20
𝐿 = lim
𝑥→−5 3(𝑥 + 5)

𝑥 2 + 5𝑥 − 4𝑥 − 20
𝐿 = lim
𝑥→−5 3(𝑥 + 5)
𝑥(𝑥 + 5) − 4(𝑥 + 5)
𝐿 = lim
𝑥→−5 3(𝑥 + 5)
(𝑥 + 5)(𝑥 − 4)
𝐿 = lim
𝑥→−5 3(𝑥 + 5)
(𝑥 − 4)
𝐿 = lim
𝑥→−5 3
(−5 − 4)
𝐿=
3
−9
𝐿=
3
𝐿 = −3



𝑀𝑒𝑡ℎ𝑜𝑑 2 (𝐿′ 𝐻𝑜𝑝𝑖𝑡𝑎𝑙 ′ 𝑠 𝑟𝑢𝑙𝑒)
𝑥 2 + 𝑥 − 20
𝐿 = lim
𝑥→−5 3(𝑥 + 5)

(−5)2 + (−5) − 20
𝐿=
3(−5 + 5)
25 − 5 − 20
𝐿=
3(0)
0
𝐿=
0
0
𝑊𝑒 𝑔𝑒𝑡 𝑎𝑛 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 . 𝑊𝑒 𝑚𝑎𝑦 𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑝𝑖𝑡𝑎𝑙 ′ 𝑠 𝑅𝑢𝑙𝑒.
0

, 𝑥 2 + 𝑥 − 20
𝐿 = lim
𝑥→−5 3(𝑥 + 5)

𝑥 2 + 𝑥 − 20
𝐿 = lim
𝑥→−5 3𝑥 + 15

𝑑 2
(𝑥 + 𝑥 − 20)
𝐿 = lim 𝑑𝑥
𝑥→−5 𝑑
(3𝑥 + 15)
𝑑𝑥
2𝑥 + 1 − 0
𝐿 = lim
𝑥→−5 3+0
2𝑥 + 1
𝐿 = lim
𝑥→−5 3
2(−5) + 1
𝐿=
3
−10 + 1
𝐿=
3
−9
𝐿=
3
𝐿 = −3


(𝑖𝑖)
sin(5𝑡)
𝐿 = lim
𝑡→0 𝑡 2 + 4𝑡
sin(5 × 0)
𝐿=
(0)2 + 4 × 0
sin(0)
𝐿=
0+0
0
𝐿=
0
0
𝑊𝑒 𝑔𝑒𝑡 𝑎𝑛 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚 𝑜𝑓 . 𝑊𝑒 𝑚𝑎𝑦 𝑎𝑝𝑝𝑙𝑦 𝐿′ 𝐻𝑜𝑝𝑖𝑡𝑎𝑙 ′ 𝑠 𝑅𝑢𝑙𝑒.
0
sin(5𝑡)
𝐿 = lim
𝑡→0 𝑡 2 + 4𝑡

𝑑
(sin(5𝑡))
𝐿 = lim 𝑑𝑡
𝑡→0 𝑑
(𝑡 2 + 4𝑡)
𝑑𝑡
cos(5𝑡) × 5
𝐿 = lim
𝑡→0 2𝑡 + 4
5 cos(5𝑡)
𝐿 = lim
𝑡→0 2𝑡 + 4

, 5 cos(5 × 0)
𝐿=
2×0+4
5 × cos(0)
𝐿=
0+4
5×1
𝐿=
4
5
𝐿=
4


(𝑖𝑖𝑖)
3 − |𝑥|
𝐿 = lim
𝑥→−∞ 2|𝑥| + 1




𝑥 𝑖𝑓 𝑥 ≥ 0
|𝑥| = {
−𝑥 𝑖𝑓 𝑥 < 0



𝑆𝑖𝑛𝑐𝑒 𝑥 𝑖𝑠 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑖𝑛𝑔 − ∞, 𝑤𝑒 𝑚𝑎𝑦 𝑎𝑠𝑠𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑥 < 0.
∴ |𝑥| = −𝑥
3 − |𝑥|
𝐿 = lim
𝑥→−∞ 2|𝑥| + 1

3 − (−𝑥)
𝐿 = lim
𝑥→−∞ 2(−𝑥) + 1

3+𝑥
𝐿 = lim
𝑥→−∞ −2𝑥 + 1

𝑥+3
𝐿 = lim
𝑥→−∞ −2𝑥 + 1

𝑥 3
+
𝐿 = lim 𝑥 𝑥
𝑥→−∞ 2𝑥 1
− 𝑥 +𝑥

3
1+𝑥
𝐿 = lim
𝑥→−∞ 1
−2 + 𝑥

1+0
𝐿=
−2 + 0
1
𝐿=−
2