quick limit Q for math gurus :)

[MCBTunes]

lim(t -->0) sin^2(3t)/t^2

nothing too amazing..

I know lim(t-->0) sin(t)/t = 1... but I am having difficulty converting it to that form..

the first person that answers this is going to make me feel stupid... but thats ok

 

[Poop-Loops]

L'Hospital's rule?

 

[Jazzin']

Looking at it really quickly; it looks like the answer is going to be 3 or 9.

edit: seems to be 9.

 

[Minger]

Calculator shows 9.

I don't remember how to do this though; all I've done lately is derivatives.

 

[Poop-Loops]

L'HOSPITAL'S RULE!!

 

[Geoff St. Germaine]

My suggestion is the same as Poop-Loops (except I would call it "L'Hôpital's Rule" ). You should get something like lim (t->0) 3*sin(3t)/t after doing that. I'm assuming you can show that lim(t->0) sin(3t)/t = 3 (if you have to).

 

[Poop-Loops]

Nah, you can do it again and the final answer is 18cos^2(3t)/2

Obviously lim t->0 cos^2(3t) = 1, so you're left with 18/2.

 

[Dave Grossman]

lim(t -->0) sin^2(3t)/t^2

nothing too amazing..

I know lim(t-->0) sin(t)/t = 1... but I am having difficulty converting it to that form..

the first person that answers this is going to make me feel stupid... but thats ok

It's been a while but I seem to remember that for very small t, sin( t ) ~= t. Therefore, the limit as t -> 0 of sin( t )/t is 1. It's been at least 10 years since I've needed to do something like this so I may be wrong.

- Dave

 

[Poop-Loops]

That only works experimentally. For an angle T less than about 10 degrees, sinT = T. Analytically, though, that's not the case. There is still a correction to be applied, so you can't put that as your reason for them cancelling.

 

[Geoff St. Germaine]

Nah, you can do it again and the final answer is 18cos^2(3t)/2

Obviously lim t->0 cos^2(3t) = 1, so you're left with 18/2.

Sure, you can do it again, but if you're using the identity that lim (t->0) sin(t)/t is one then you don't have to. Either way is fine.

 

[jady]

Damn these threads always make me feel stupid.........

 

[Poop-Loops]

That's ok. I'm sitting in front of my Diff EQ's homework and this thread stopped me from tearing my hair out.

 

[Headroom]

Bernoulli to the rescue (again).

 

[Poop-Loops]

Bernoulli???

 

[Headroom]

Bernoulli???

http://en.wikipedia.org/wiki/L'Hôpital's_rule

http://en.wikipedia.org/wiki/Johann_Bernoulli

 

[Poop-Loops]

Ahh I see.

 

[msquared]

From Wikipedia:

The Marquis spelled his name with an 's'; however, the French language has since dropped the 's' (it was silent anyway) and added a circumflex to the preceding vowel.

It's anyone's guess as to how accurate that is, though.

 

[MCBTunes]

thanks guys, I can't believe I've never heard of "L'Hôpital's Rule"... or atleast dont remember it from high school. So the basic method to complete problems using this rule is to take n number of derivatives until you can either "See" the answer or you run into constants?

 

[Poop-Loops]

You'll run into constants. That's how people "see" the answer. You learned sin(t)/t = 1. How did they find out? Likely L'Hopital's rule.

EDIT: Well actually the main point is that you don't get something like 0/0, infinity/infinity, constant/0, etc.

 

[Geoff St. Germaine]

You'll run into constants. That's how people "see" the answer. You learned sin(t)/t = 1. How did they find out? Likely L'Hopital's rule.

EDIT: Well actually the main point is that you don't get something like 0/0, infinity/infinity, constant/0, etc.

You can show that lim (t->0) sin(t)/t = 1 with the squeeze theorem (which I'm sure you've seen). This is normally how it is shown in early calculus courses. Of course, L'Hôpital's rule works and is easier.