MAT1512 ASSIGNMENT 6 2022

MAT1512
ASSIGNMENT 6
2022

,Solution:



𝐴𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙𝑠:


𝑑
√6 − 𝑥 − 2 𝑑𝑥 [√6 − 𝑥 − 2]
lim = lim
𝑥→2 √3 − 𝑥 − 1 𝑥→2 𝑑
− 𝑥 − 1]
𝑑𝑥 [√3
1
(− )
2√6 − 𝑥
= lim
𝑥→2 1
(− )
2√3 − 𝑥

√3 − 𝑥
= lim
𝑥→2 √6 − 𝑥

√3 − 2
=
√6 − 2
√1
=
√4
1
=
2


ANSWER:[3]




Solution:


|𝑥 + 3|
lim −
𝑥→−3 𝑥2 − 9

, (𝑥 + 3) 𝑓𝑜𝑟 𝑥 ≥ −3
|𝑥 + 3| = {
−(𝑥 + 3) 𝑓𝑜𝑟 𝑥 < −3



𝐼𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒 |𝑥 + 3| = −(𝑥 + 3)


|𝑥 + 3| −(𝑥 + 3)
lim − 2
= lim − 2
𝑥→−3 𝑥 −9 𝑥→−3 𝑥 −9
−(𝑥 + 3)
= lim −
𝑥→−3 (𝑥 + 3)(𝑥 − 3)
−1
= lim −
𝑥→−3 (𝑥 − 3)
−1
=
(−3 − 3)
1
=
6


ANSWER:[3]




Solution:



√𝑥 2 + 4𝑥 − 2𝑥 √𝑥 2 + 4𝑥
lim = lim − lim 1
𝑥→−∞ 2𝑥 𝑥→−∞ 2𝑥 𝑥→−∞




√𝑥 2 + 4𝑥 1 √𝑥 2 + 4𝑥 √𝑥 2 + 4𝑥 𝑥 2 + 4𝑥
lim = lim 𝑓𝑜𝑟 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥 = −√
𝑥→−∞ 2𝑥 2 𝑥→−∞ 𝑥 𝑥 𝑥2

1 𝑥 2 + 4𝑥
= lim −√
2 𝑥→−∞ 𝑥2