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limits

Ok, here's an interesting limit question. I thought that when you have a radical, you multiply top and bottom by the conjugate,true? How about this one?

lim sqrt(9+x^2)/(x-3)
x-->3

So what happens here? do you square top and bottom to get rid of the radical or???
I don't know what theorems you've been taught but putting mathematical rigour aside for a moment, this limit is clearly non-existent. When you approach 3 from above you get a different answer from when you approach it from below. You could try bounding the function by some other (easier to work with) function to prove this.
(edited 12 years ago)
Reply 2
Avatar for DFranklin
DFranklin
18
Original post by Snazarina
Ok, here's an interesting limit question. I thought that when you have a radical, you multiply top and bottom by the conjugate,true? How about this one?

lim sqrt(9+x^2)/(x-3)
x-->3

So what happens here? do you square top and bottom to get rid of the radical or???
The limit doesn't exist. (Suppose it did, and the limit = L, then lim⁡x→39+x2=Llim⁡x→3(x−3)\lim_{x\to 3} \sqrt{9+x^2} = L \lim_{x \to 3} (x-3) (as long as the limits in this equation exist, which they clearly do).

Since the RHS = 0 and the LHS = 3 sqrt 2, this is not possible).
You're welcome.
Reply 4
Avatar for Hasufel
Hasufel
16
True - no limit exists.
You can also see this (assuming you want to) by using a combination of limit rules: Limit[Sqrt(something)] = Sqrt[Limit(that something)], together with L`Hop`s rule to read:

Sqrt[Limitx->3{L`Hops rule for 9+x^2/(x-3)^2}]= undefined - with the assumption that x>0 w.l.o.g
(edited 12 years ago)
Reply 5
Avatar for DFranklin
DFranklin
18
Original post by Hasufel
True - no limit exists.
You can also see this (assuming you want to) by using a combination of limit rules: Limit[Sqrt(something)] = Sqrt[Limit(that something)], together with L`Hop`s rule to read:

Sqrt[Limitx->3{L`Hops rule for 9+x^2/(x-3)^2}]= undefined - with the assumption that x>0 w.l.o.g
You can't apply L'Hopital unless the original limit is indeterminate.

(e.g. applying L'Hopital to (9+(x-3))/(x-3) at x = 3 would tell you the limit = 1, which it isn't...)
Reply 6
Avatar for Snazarina
Snazarina
OP
2
Oh gee thanks everyone. So it's simply non existent. And how do u tell if it's existent or not? I mean they all look the same
Original post by Snazarina
Oh gee thanks everyone. So it's simply non existent. And how do u tell if it's existent or not? I mean they all look the same


If you have one variable then you just need to show that the right hand limit and the left hand limit approach the same number.
Reply 8
Avatar for Hasufel
Hasufel
16
Original post by DFranklin
You can't apply L'Hopital unless the original limit is indeterminate.

(e.g. applying L'Hopital to (9+(x-3))/(x-3) at x = 3 would tell you the limit = 1, which it isn't...)


WRONG! - 2/3rds - but i take your point
(edited 12 years ago)
Reply 9
Avatar for nuodai
nuodai
17
Original post by Hasufel
WRONG!


Wut? L'Hôpital doesn't apply here. Just because it gives you the right answer doesn't mean the method is right.
Reply 10
Avatar for Hasufel
Hasufel
16
Original post by DFranklin
You can't apply L'Hopital unless the original limit is indeterminate.

(e.g. applying L'Hopital to (9+(x-3))/(x-3) at x = 3 would tell you the limit = 1, which it isn't...)


True! i bow to thee! (silly me!)

EDIT: apologies! wasn`t concentrating (my brain is fried - i`m studying from different books at once, you know: arse....elbow...)
(edited 12 years ago)

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