Okay. So I have this homework problem here and I can't really figure out how to do it. The problem is...
The limit of sin(2x)/x as x approaches 0.
I know how to do it using my TI-83, but I'm supposed to be able to do it by hand using these four facts:
The limit of sin(x)/x as x approaches zero = 1
The limit of (cos(x)-1)/x as x approaches zero = 0
The limit of sin(x) as x approaches zero = 0
The limit of cos(x) as x approaches zero = 1
I've tried rewriting it using the double angle formula thing to get...
(2sin(x)cos(x))/x
Which I tried rewriting as (2/x)+(sin(x)/x)+(cos(x)/x)
But then I get all confused and I don't know what to do. Help?
Name:
Anonymous2007-05-29 20:44 ID:TkmLk11e
squeeze theorem
Name:
Anonymous2007-05-29 20:59 ID:/lIhdw7k
Uh...
What?
I looked it up on wikipedia, and I understand what it is, but I have no clue how to apply it to this. And I'm sure we don't have to use it because we haven't learned it at all yet.
Name:
Anonymous2007-05-29 21:00 ID:Ny3Sxm+a
The limit is 2.
How about this argument:
lim x->0 Sin(2x)/x
= lim x->0 2Cos(2x) (by L'Hopital's rule)
= (lim x->0 2)(lim x->0 Re(e^(2ix))) (where Re(z) denotes the real part of z and e is the base of the natural log)
= 2(lim x->0 Cos(x)^2 - Sin(x)^2)
= 2((lim x->0 Cos(x))^2 - (lim x->0 Sin(x))^2)
= 2((1)^2 - (0)^2)
= 2
The limit of sin(2x)/2x as x goes to 0 is 1, which can easily be seen from the first fact you're given. sin(2x)/x = 2 * sin(2x)/2x, so the limit as x goes to 0 of sin(2x)/x is equal to 2.
Name:
Anonymous2007-05-29 23:25 ID:icWvRjw5
if x on the bottom have to divide by zero and u cant so trick question cant answer this or id say 0