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March 4, 2011 @ 05:18 PM
CraftySince90

Post: 901

Join Date: Dec 2006

Location: YB house, SF Bay Are...

Find LIM x-->2 : square root of : x^2-4 / x^2-3x+2

How the heck do i even get this started...
March 4, 2011 @ 05:39 PM
exhale

Post: 1023

Join Date: Jul 2009

Location: In your heart.

graph that shit. see where it's going at x=2.
March 4, 2011 @ 06:40 PM
Imco0l

Post: 767

Join Date: Dec 2008

Location: Canada

no limit.
March 4, 2011 @ 06:41 PM
Nuhhhice

Post: 141

Join Date: Jul 2010

no limit.


yup, limit does not exist
March 4, 2011 @ 09:40 PM
Virungă

Post: 2790

Join Date: Jan 2010

Location: Connecticut

March 4, 2011 @ 11:13 PM
jaded

moderator

Post: 7992

Join Date: Feb 2008

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Find LIM x-->2 : square root of : x^2-4 / x^2-3x+2

How the heck do i even get this started...


edit: the work i typed out was wrong

i'm the best mayne, i deed it

March 4, 2011 @ 11:32 PM
jaded

moderator

Post: 7992

Join Date: Feb 2008

Location: ಠ______ಠ

Today we had to do:

If the series of numbers 123456789123456789123456789 is recurring, what is the 2011th digit.

i'm the best mayne, i deed it

March 5, 2011 @ 12:15 AM
Carlton C-Note Banks

Post: 1715

Join Date: Jan 2009

The limit's 4...
If the function is (x^2-4)/(x^2-3x+2) then it factors to ( (x-2) (x+2) )/( (x-1) (x-2) )
The (x-2)s cancel, so it's the function (x+2)/(x-1). Plug in 2 and you get 4/1, which is 4.
I tried doing this in my head earlier and was canceling the wrong parts after factoring but used online calculators to help since I don't have a piece of paper to write this down in front of me.
March 5, 2011 @ 12:19 AM
jaded

moderator

Post: 7992

Join Date: Feb 2008

Location: ಠ______ಠ

The limit's 4...
If the function is (x^2-4)/(x^2-3x+2) then it factors to ( (x-2) (x+2) )/( (x-1) (x-2) )
The (x-2)s cancel, so it's the function (x+2)/(x-1). Plug in 2 and you get 4/1, which is 4.
I tried doing this in my head earlier and was canceling the wrong parts after factoring but used online calculators to help since I don't have a piece of paper to write this down in front of me.


you forgot the square root part, so i think it'd be two.

i'm the best mayne, i deed it

March 5, 2011 @ 12:22 AM
Carlton C-Note Banks

Post: 1715

Join Date: Jan 2009

lol my bad, you're right. Just disregard my post then, didn't even catch the square root part. In that case you all are likely right about it not having a limit, haven't checked though.

Edit, checked again and it is 2. lol Right again about that part. Been away from limits and stuff too long lol
March 5, 2011 @ 12:26 AM
jaded

moderator

Post: 7992

Join Date: Feb 2008

Location: ಠ______ಠ

lol my bad, you're right. Just disregard my post then, didn't even catch the square root part. In that case you all are likely right about it not having a limit, haven't checked though.

Edit, checked again and it is 2. lol Right again about that part. Been away from limits and stuff too long lol


yeah, me too. my work was wrong so i deleted it. i didn't factor. i wasn't a big fan of limits. i did really well with derivatives.

i'm the best mayne, i deed it

March 5, 2011 @ 12:27 AM
Carlton C-Note Banks

Post: 1715

Join Date: Jan 2009

BTW, your math was right I think, but when you saw that 2.001 and 1.999 gave answers that were very close to 2 (not exactly 2 of course), that actually meant that the limit was 2, since the discrepancy was minimal. There'd be an issue if for instance 2.001 gave something like 1 and 1.999 gave something like 3, then there would be a hole. So your math was right
March 5, 2011 @ 12:30 AM
jaded

moderator

Post: 7992

Join Date: Feb 2008

Location: ಠ______ಠ

BTW, your math was right I think, but when you saw that 2.001 and 1.999 gave answers that were very close to 2 (not exactly 2 of course), that actually meant that the limit was 2, since the discrepancy was minimal. There'd be an issue if for instance 2.001 gave something like 1 and 1.999 gave something like 3, then there would be a hole. So your math was right


i'm too lazy to type it back out, it was the long way of doing it anyhow.

i'm the best mayne, i deed it

March 5, 2011 @ 05:46 PM
CraftySince90

Post: 901

Join Date: Dec 2006

Location: YB house, SF Bay Are...

Thats what i figured.. do all the work to 4, then it would just square to a limit of 2?
March 5, 2011 @ 09:33 PM
isitnyu

Post: 165

Join Date: Aug 2010

you could also go to WOLFRAMALPHA.com and enter that limit in exactly like you did in your intial post and they'll show you how to do it and give you the answer.
March 5, 2011 @ 10:27 PM
jaded

moderator

Post: 7992

Join Date: Feb 2008

Location: ಠ______ಠ

you could also go to WOLFRAMALPHA.com and enter that limit in exactly like you did in your intial post and they'll show you how to do it and give you the answer.


that's a neat site.

i'm the best mayne, i deed it

March 5, 2011 @ 10:35 PM
jaded

moderator

Post: 7992

Join Date: Feb 2008

Location: ಠ______ಠ

I looked at it again today and actually wrote it out on paper and it's 2, just like we figured.

When you initially plug 2 in to the limit, you should end up with sqrt(0/0), so you know you have to do more work. When you have to do more work, the first thing you want to do is factor something like this to see if you can cancel anything out.

The top can factor to: (x+2)(x-2)
The bottom can factor to: (x-1)(x-2)

The (x-2)'s cancel out, so you are left with sqrt((x+2)/(x-1))

Plug 2 back into the function and you get sqrt((2+2)/(2-1)) which is sqrt(4) which equals 2.

So your limit as you approach 2 from both sides is 2.

i'm the best mayne, i deed it


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