- HELP- calculus problem :(

[Omegachi]

Find the following limit:

lim tan2x
(x->0) x

 

[jamesave]

change to the cos and sin term from tan., distribute.

 

[OS]

ugh, nm did it wrong and I don't feel like finding my mistake.

 

[TripleJ]

2.25?

 

[Soybomb]

Thanks guys, you just made the nightmares return! I"ll send you my copy of mathematica, god knows I dont want to ever use it again

 

[Omegachi]

i think the answer is suppose to be 2

lim 2sin2x
(x->0) 2xcos2x

= 2

??????
i don't get how they go from the original equation to

lim 2sin2x
(x->0) 2xcos2x

 

[Dandaman]

I think it's 0. Try L'Hopital's Rule. Then you should get zero. I think...it's been awhile

 

[thEnEuRoMancER]

It's what Dandaman said. I think the answer is 2.

 

[Omegachi]

what is the L hopital rule?

 

[iLoveDivX]

anyone with a ti-89 can give you a definite answer. I have one but forgot how to use it.

 

[TripleJ]

Euromancer, Dandaman said it was 0.

I don't know, calculus was just so hard, I forgot how to do it quickly.

 

[trek]

according to the ti-89 it is 2.

-Trek

 

[Soybomb]

Perhaps I'm having a dumb moment here (I did drop calculus afterall) but can't you plug in a few sample numbers as x approaches 0 and see the trend for an answer?

 

[thEnEuRoMancER]

<< what is the L hopital rule? >>



if lim f(x) = 0
.. (x->0)

and lim g(x) = 0
.... (x->0)

then lim f(x)/g(x) = lim f'(x)/g'(x)
(x->0)....................(x->0)

 

[Aelus]

The answer is 2

with lim sin(2x) / xcos(2x) = lim 2cos2x / (cos2x - 2xsin2x)

then, if you fill in 0; you get 2x1 / 1 = 2

btw, if you need a ti89 to solve this, you should be ashamed of yourself

Aelus

 

[AznMaverick]

Aelus: how'd you get (cos2x - 2xsin2x) in the denominator?

 

[Aelus]

originally you have xcos2x, this is a multiplication of 2 functions, so you have to use the f'(x)g(x)+g'(x)f(x) rule.

Aelus

 

[AznMaverick]

i see it!...thanks

 

[Omegachi]

ahhhhhhhhhhhhhhhh..........i get it now.....aelus, you're the best

 

[Aelus]

You might want to study your derivation rules by heart, they are very valuable

i did my high school with a Ti-40, and now i have a Ti83, and i only use it for matrix math in electricity and all, for calculations, and for tetris, everythign else is a job for the brains, and i can recommend it, because it's great for showing off

Aelus

 

[Omegachi]

haha... cool

where did you get the programing for tetris

 

[Aelus]

Downloaded it off a friends machine, i don't have the computer kit, so i'm afraid i can't help you

Aelus

 

[Dandaman]

yeah, forgot about that chain rule in the denominator, oh well

 

[blahblah99]

lim x-> 0 tan(2x) / x

using L'hopital's rule, u get

lim x->0 2*sec^2(2x)/1

sec^2(2x) = 1/cos^2(2x)

lim x->0 of cos(2x) is 1,

so you get 2/1 = 2;

 

[mofo888]

Damn Calculus, it's been a really...I can't believe I don't remember any of this stuff.
Based on Aelus solution, here is another way to explain it. (basically just wording)

We all know --> lim tan (2x) / x --> yields sin(2x) / x cos (2x)

Using L'hopital's rule, you take the derivatinve of above, the top part becomes

lim 2 cos (2x) and the bottom, using the product rule f(x)g(x)= f(x) g'(x) + g(x) f'(x)

gives cos (2x) - 2x sin (2x) therefore combine top and bottom

= lim 2cos2x / (cos2x - 2xsin2x)

thus using x = 0 you will get 2!

thanks Aelus for reminding me all this crazy stuff.