need help on a calculus problem

stickybytes

Golden Member
Sep 3, 2003
1,043
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0
I've been working on this problem all night and i can't seem to find the answer.

Find the limit of f(x) as x approaches 6

f(x)= y+6
----------------
y squared-36

Please also explain your steps.
Thanks.
 

EyeMWing

Banned
Jun 13, 2003
15,670
1
0
Uhhhhhhhhh. 12, because f(6) is 12.
Pay attention in class.

Oh, wait. WTF. Nevermind. Skip that problem, it's too much for me to think about tonight.
 
Nov 3, 2004
10,491
22
81
Originally posted by: stickybytes
I've been working on this problem all night and i can't seem to find the answer.

Find the limit of f(x) as x approaches 6

f(x)= y+6
----------------
y squared-36

Please also explain your steps.
Thanks.

are yous ure you copied it down right?
theer's no x in that equation
 

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
Originally posted by: stickybytes
I've been working on this problem all night and i can't seem to find the answer.

Find the limit of f(x) as x approaches 6

f(x)= y+6
----------------
y squared-36

Please also explain your steps.
Thanks.



I'm almost positive your problem was supposed to be.

lim (x-->6) of (x-6)/(x^2-36)

Either that or:

lim (x-->-6) of (x+6)/(x^2-36)
 

ruffilb

Diamond Member
Feb 6, 2005
5,096
1
0
Originally posted by: stickybytes
I've been working on this problem all night and i can't seem to find the answer.

Find the limit of f(x) as x approaches 6

f(x)= y+6
----------------
y squared-36

Please also explain your steps.
Thanks.

uh, 6/-36 = -1/6

if it's ((x+6)/(x^2-36)) then

((x+6)/(x^2-36)) =
((x+6)/((x+6)(x-6)))

lim (x+6)/
x => 6 (x+6)(x-6)

=

lim 1/(x-6)
x => 6

= 1/0 indicating a vertical asymptote. If the top is (x-6), it comes out to be


((x-6)/(x^2-36)) =
((x-6)/((x+6)(x-6)))

lim (x-6)/
x => 6 (x+6)(x-6)

=

lim 1/(x+6)
x => 6

= 1/(6+6) = 1/12
 
Nov 3, 2004
10,491
22
81
Originally posted by: ruffilb
Originally posted by: stickybytes
I've been working on this problem all night and i can't seem to find the answer.

Find the limit of f(x) as x approaches 6

f(x)= y+6
----------------
y squared-36

Please also explain your steps.
Thanks.

uh, 6/-36 = -1/6

wtf?
 

ruffilb

Diamond Member
Feb 6, 2005
5,096
1
0
Originally posted by: IAteYourMother
Originally posted by: ruffilb
Originally posted by: stickybytes
I've been working on this problem all night and i can't seem to find the answer.

Find the limit of f(x) as x approaches 6

f(x)= y+6
----------------
y squared-36

Please also explain your steps.
Thanks.

uh, 6/-36 = -1/6

wtf?


A constant function and e^x are walking on Broadway. Then suddenly the constant function sees a differential operator approaching and runs away. So e^x follows him and asks why the hurry. "Well, you see, there's this differential operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...!" "Ah," says e^x, "he won't bother ME, I'm e to the x!" and he walks on. Of course he meets the differential operator after a short distance.
e^x: "Hi, I'm e^x"
diff.op.: "Hi, I'm d/dy"

XD
 

ruffilb

Diamond Member
Feb 6, 2005
5,096
1
0
Originally posted by: IAteYourMother
Originally posted by: ruffilb
Originally posted by: Auryg
Jus' type it in to yer graphing calc!

Ti-89 ftw.

But 83-86's don't have a limit function.

so you just graph it and find y at x=6... not that hard

f(x)= y+6 doesn't graph.

And the OP probably want's to do it the legit way, although I must admit I use my graphing calculator like, 99% of the time ;)