Calculus - limits: Help!

[mosende]

---

nose bleed ko ani da

[janvier]

y sub 1 + y sub 1 = y sabot.....mahilis man atong utok ani oi...

[emjee17]

hehe, pariha rata..

[emjee17]

ANOTHER PROBLEM:

Though I already know/understand Theorems of Limits, I forgot some of the basic - Algebra.

Legend:
^ - Squared
//- square root

1. lim t^2-5/2t^3+6
t---->2

2. lim 3x^2-17x+20/4x^2-25x+3
x---->4

3. 3 - index

3/ x/ - 1/ x-1
x---->1

[crush_23]

sus ma turn on jud ko basta math wiz...hihihi

na answeran naman d ay ...moanswer unta ko

[cebu.opportunities]

kakuti ba d i ani sis...gesubay man jud sa pamalaud..dugay napud ko la kagunit og libro,papel og bpen...
ambot tama ba ni...

legend: []=absolute sign


1. (x^2-1)/(x+1) = -2
(x+1)(x-1)/(x+1) = -2
x-1 = -2 ;
[x+2 = 0]
[x+2] = 0 ; consider the positive limit , thus E>0;

mao nga [x+2] < E
considering the Delta, using the limit x--> -2
x = -2
x + 2 = 0 , delta should be positive so delta>0
[x+2] < delta
compare ang E and delta;
[x+2]<E ; [x+2]<delta
[x+2]<E=delta
mao nga delta =(1/k)E , where k is a constant and in this case k is equal to 1...

.....hehehe..ambot tama ba ni, mao dili nalang nako esuwat ang no.2 kay basin mali ra japun og basin malibog rakag samot, kapoy2 type.

nalipat ka.

legend: []=absolute sign


1. (x^2-1)/(x+1) = -2
(x+1)(x-1)/(x+1) = -2
x-1 = -2 ;
[x-1+2 = 0]
[x+1] = 0 ; consider the positive limit , thus E>0;

basta x+1= 0 ky x= -1

^ i mean, amo mang gud answers kay ingon ani:


given example:

1. show that lim (4x-5) = 3
x--> 2

Proof: let E(epsilon) > 0.
Observe that: |(4x-5) - 3| = |4x-8|
= |4(x-2)|
= |4| |x-2|

Take d(delta)=E/4. Then, for all x E IR(all real nos.), such that
0<|x-2|<d, then
|(4x-5) - 3| = |4x-8|
= |4(x-2)|
= |4| |x-2|
< 4d = 4 E/4
Thus, |(4x-5) - 3| < E.
Therefore, by definition lim (4x-5) = 3.
x-->2

**ingon ani pag present sa among maestrong pwerting pagka paspasa mu discuss. tsk3.

ga lisod lisod rmn inyo teacher uy, hehe.

1. show that

lim x^2-1/x+1 = -2
x--> -1

2. prove that

lim (7-3x) = -2
x-->3

Observe that: |(x^2-1) / (x+1) +2 | = | (x+1) (x-1) / (x+1) +2 |
= |(x-1) + 2 |
= | x-1+2 |
= |1| | x +1 |

Take d(delta)=E/1. Then, for all x E IR(all real nos.), such that
0<|x+1|<d, then
| (x+1) (x-1) / (x+1) + 2 | = | x-1 + 2|
= |1 (x+1)|
= |1| |x+1|
< 1d = 1 E/1
Thus, |(x^2-1) / (x+1) + 2 | < E.
Therefore, by definition lim (x^2-1)/(x+1) = -2.
x--> -1



haha, ambot lang sad bitaw. wa kayo ko kasabot sa delta and epsilon.

1.) (x+1)(x-1)/(x+1) = -2
x-1- = -2
-1-1 = -2

2.) 7-3(3) = -2
7 - 9 = -2

^ kani man gud pinakasayon. substitute lang ba.