29/09/2014
Lecture and Home Assignment
Find the indicated limit.
| Q:01 Lim(x à 0) (3x2 – 5x + 3) | Q:02 Lim(x à 2) (2x2 – 10x) |
| Q:03 Lim(x à 2) {(x2/3) – 7x2} | Q:04 Lim(x à 3) (2x – 8)/(x + 4) |
| Q:05 Lim(x à 2) (2x2 + 5)/(x3 + 4x - 2) | Q:06 Lim(x à -4) 250 |
| Q:07 Lim(x à 0) 175 | Q:08 Lim(x à -10) (x2 + 25)/x |
| Q:09 Lim(x à 1) (3x3 + 2x2 x) | Q:10 Lim(x à -2) (-x4 + 8x2) |
| Q:11 Lim(x à -2) (- x2 + 5x2 +10) | Q:12 Lim(x à -2) (5x3 + 10x2) |
| Q:13 Lim(x à 5) (4x - 20)/(3x2 - 7x +5) | Q:14 Lim(x à -2) (10 + x2)/(5 – 8x + 2x2) |
| Q:15 Lim(x à -3) (6x3 + 2x)(5x + 10) | Q:16 Lim(x à 5) [{(x + 3)/(x + 6)}{x2 - 12}] |
| Q:17 Lim(x à -4) (x2 + 8x – 14)/(2x + 7) | Q:18 Lim(x à 1) (x2 + 3x – 24)/(x – 3) |
| Q:19 Lim(x à -4) (x2 - 16)/(x + 4) | Q:20 Lim(x à 9) (81 – x2)/(9 + x) |
| Q:21 Lim(x à a) (4x3 – 5x2 + 10) | Q:22 Lim(x à -d) (x2 – 2x + 3) |
Definite state of function: example:
f(x) = x2 + 2x è = f(2) = (2)2 + 2(2) è = 8 -: so 8 is a real/definite number so function is definite:-
Horizontal asymptotes: when answer is in definite number or real number.
Example: lim(x à ∞) f(x) then lim(x à ∞) (3x2 + 5x)/(4x2 5) è = ¾
Vertical asymptotes: opposite to horizontal asymptotes.
(Q: 25 Page683) Solution:
H.A V.A
lim(x à ∞) -3x/(5x + 100) by putting x value -20
lim(x à ∞) x (-3)/ x {5+ (100/x)} lim(x à -20) -3x/(5x + 100)
lim(x à ∞) -3/{5 + (100/x)} -3(-20)/{5(-20) + 100)
-3/{5 + (100/∞)} = -3/5 60/(-100 +100) = ∞
Find the indicated limit & comment on the existence of any asymptotes.
| Q:23 Lim(x à ∞) 4 /x2 | Q:24 Lim(x à ∞) (5x – 3)/(x + 10) |
| :25 Lim(x à ∞) -3x/(5x+100) | Q:26 Lim(x à ∞) (8x + 10)/- 4x |
| Q:27 Lim(x à ∞) 2x | Q:28 Lim(x à ∞) (x2 + 3)/(x – 4) |
| Q:29 Lim(x à ∞) -8x/(4x + 1000) | Q:30 Lim(x à ∞) (5x + 10000)/(2x – 5000) |
| Q:31 Lim(x à ∞) (100 – 3x3)/-x3 | Q:32 Lim(x à ∞) (3x3 – 500)/(5000 – x3) |
Continuity: the continuity exists when 2 terms are validated the terms are below:
Ø Function is definite at x = a at x = 2
Ø Lim(x à a) f(x) = F(a)
1. Example of continues: f(x) = X3
Solution: f(a) = f(x) so
F(a) f(x)
Put x = 2 lim(x à 2) x3
F(2) = (2)3 = 8 answer (2)3 = 8 answer
So function is continues
2. Example of discontinues No. 01: f(x) = 1 / (x3 – x)
Solution: Check where function is discontinues
x3 Рx = 0 ̬ = x (x2 Р1) = 0 ̬ = x (x Р1) (x + 1) = 0 ̬ x = 0, x = 1 , x = -1
f(x) at 0 è = f(0) = 1/{(0)3 – 0} è = 1/0 ∞ answer
f(x) at 1 è = f(1) = 1/{(1)3 – 1} è = 1/0 ∞ answer
f(x) at -1 è = f(-1) = 1/{(-1)3 - (-1)} è = 1/0 ∞ answer
3. Example of discontinues No. 02: f(x) = 3x2/ (x + 5)
Solution: Check where function is discontinues
X + 5 = 0 è = x = -5
F(x) at -5 è = f(-5) = 3(-5)2/{(-5) + 5} è = 28/0 ∞ answer
Find discontinuity
| Q:33 f(x) = 3x2 + 2x +1 | Q:34 f(x) = 1/(x + 2) |
| Q:35 f(x) = x6/5 | Q:36 f(x) = 3x2/(x + 5) |
| Q:37 f(x) = 1/(8-2x) | Q:38 f(x) = |x| |
| Q:39 f(x) = (2x + 3)/(x2 + 4x – 21) | Q:40 f(x) = (x – 2)/(x2 – 8x) |
| Q:41 f(x) = (4x – 3)/(x3 – x2 – 6x) | Q:42 f(x) = (5/x)/(2x2 + 7x – 15) |
| Q:43 f(x) = 20/(x2 – 3x – 10) | Q44 f(x) = (4/x)/(18 + 3x – x2) |
| Q:45 f(x) = {10/(5 – x)}/ (4 – x2) | Q46 f(x) = {5/(3 – x)}/(x2 – 16) |
| Q:47 f(x) = (3x – 5)/(x4 – 27x) | Q:48 f(x) = {3/(x2 -1)}/{2/(x2 – 4)} |
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