Question

Limits:

Answer 1

idea is this: if you replace x by -3 in the denominator you get 0. so the only way you have a hope of having a limit that is not undefined is if the numerator is 0 also when
x=-3
so replace x by -3 in the numerator, set = 0 and solve

Answer 2

\[2x^2+ax+a+14\] put x = -3 to get
\[18-3a+a+14=0\]

Answer 3

i get a = 16

Answer 4

so numerator is
\[2x^2+16x+30\]

Answer 5

Great..it was right

Answer 6

ok what did you get for the limit?

Answer 7

one sec..let me do that part

Answer 8

ok factor and cancel

Answer 9

-2

Answer 10

wait

Answer 11

\[\frac{2x^2+16x+30}{x^2+2x-3}=\frac{2(x+3)(x+5)}{(x+3)(x-1)}=\]
\[\frac{2(x+5)}{(x-1)}\]

Answer 12

-1

Answer 13

i dunno. i got the above and if you replace x by -3 you get what?

Answer 14

\[\frac{2(-3+5)}{-3-1}\]

Answer 15

oh yeah -1 ! good work

Answer 16

I got the answer already....i got -1...it was right

Answer 17

yeah i saw it. good going

Answer 18

I have 2 more limit questions i need help with..do you think you can assist me

Answer 19

why not

Answer 20

cant see it that well. if you replace x by 4 on the left and right do you get the same number?

Answer 21

actually it doesn't look like it. you get 16-13=3 on the left yes?

Answer 22

oh and you get 3 on the right as well! so the limit is 3

Answer 23

the old "squeeze theorem " or whatever it is called

Answer 24

ok the answer was right but can you please explain it some more so can understand better

Answer 25

well you have an inequality that goes like this
\[3\leq \lim_{x->4} f(x)\leq 3\]

Answer 26

which pretty much forces
\[lim_{x->4}f(x)=3\]

Answer 27

left and right side of inequality just plug in 4 to get the limit. since you get 3 both on the left and right that means the thing in the middle must also be 3

Answer 28

ok I get it..THANKS

Answer 29

yw

Answer 30

ok one more and then I am done bothering you

Answer 31

having a hard time reading this open office stuff. i see the denominator is
\[\sqrt{5-x}-1\] yes? so when you plug in 4 you get 0. what is the numerator?

Answer 32

ah i bet
\[\sqrt{8-x}-2\]

Answer 33

yes

Answer 34

got it. you want
\[\lim_{x->4}\frac{\sqrt{8-x}-1}{\sqrt{5-x}-1}\]

Answer 35

oops numerator is wrong

Answer 36

\[\frac{\sqrt{8-x}-2}{\sqrt{5-x}-1}\]

Answer 37

correct!

Answer 38

gimmick is to multiply numerator and denominator by "conjugate" of the denominator.

Answer 39

yes i figured that! but it became so complex

Answer 40

\[\frac{\sqrt{8-x}-2}{\sqrt{5-x}-1}\times \frac{\sqrt{5-x}+1}{\sqrt{5-x}+1}\]

Answer 41

yeah it pretty much sucks

Answer 42

let me see if i can figure out a trick. no l'hopitals rule yet i am sure

Answer 43

Sorry i am back! I had to put my son to bed...

Answer 44

ok let me try to figure it out because i have to get this last problem done by 11pm

Answer 45

on line course?

Answer 46

Nope..online homework! It is an in class course at Purdue University

Answer 47

What times are you usually on this website because I have an exam on Tuesday and it would be great to have some assistance from you to study for it if you have some free time to answer some questions.

Answer 48

well if you just want the answer... it is\[\frac{1}{2}\]

Answer 49

if you want to know how to get it without using l'hopital's rule i will have to do some heavy algebra

Answer 50

lol the answer is great but I still need to know how to do it because I want to pass this exam

Answer 51

What is l'hopital's rule?

Answer 52

if you haven't gotten to derivatives yet you cannot use it

Answer 53

oh ok yea we are not on derivatives yet...but did u get my question from above

Answer 54

are you still there?

Answer 55

yeah i am trying to figure out the algebra for this and it is not coming to me

Answer 56

maybe multiplying by the conjugate is not the trick, although you will get 4 - x in the denominator

Answer 57

problem is you get a bloody mess in the numerator

Answer 58

I will ask my professor tomorrow...

Answer 59

What times are you usually on this website because I have an exam on Tuesday and it would be great to have some assistance from you to study for it if you have some free time to answer some questions.

Answer 60

usually in the evenings around this time.

Answer 61

the problem is that l'hopital is so easy, but you have to take derivatives

Answer 62

got one in a book similar?

Answer 63

let me call in the reinforcements

Answer 64

LOL! yes i have one similar

Answer 65

hey thanks for the invitation

Answer 66

there you are. thanks
i cannot do this algebra.

Answer 67

maybe multiplying by the conjugate is not the right thing to do. and no cheating by using l'hopital

Answer 68

@ taylor
my myininaya is a ringer!

Answer 69

\[\lim_{x \rightarrow 2} \sqrt{6-x}-2/\sqrt{3-x}-1\]

Answer 70

so i'm sorry there is too much too read what are we doing

Answer 71

trying to compute this limit
\[lim_{x->4}\frac{\sqrt{8-x}-2}{\sqrt{5-x}-1}\]

Answer 72

oh okay i think i can do that give me a sec

Answer 73

easy answer of 1/2 using l'hopital. multiplying by the conjugate leads to algebra from hell

Answer 74

i will be back in a few guys...I have a computer assignment due by 11:59 and still have 16 problems left so let me get that completed and then come back to this..

THANKS FOR ALL THE HELP

Answer 75

well just put in 1/2 but i am eagerly awaiting my myininaya's solution

Answer 76

i got it...

Answer 77

i am sure you would!

Answer 78

\[\frac{\sqrt{8-x}-2}{\sqrt{5-x}-1}*\frac{\sqrt{5-x}+1}{\sqrt{5-x}+1}*\frac{\sqrt{8-x}+2}{\sqrt{8-x}+2}\]
\[\frac{8-x-4}{5-x-1}*\frac{\sqrt{5-x}+1}{\sqrt{8-x}+2}=\frac{\sqrt{5-x}+1}{\sqrt{8-x}+2}\]
just plug in 4 now!

Answer 79

lorda mercy conjugate twice! shoulda known. you are a superstar

Answer 80

lol

Answer 81

oh of course. you make it look so easy. saw that i should get a factor of 4-x top and bottom.

Answer 82

:)

Answer 83

so is taylor gone i guess

Answer 84

i see what u did..THANKS

Answer 85

np