Return Styles: Pseud0ch, Terminal, Valhalla, NES, Geocities, Blue Moon.

Pages: 1-

Limits and trigonometry

Name: Anonymous 2007-05-29 20:40 ID:/lIhdw7k

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: Anonymous 2007-05-29 20:44 ID:TkmLk11e

squeeze theorem

Name: Anonymous 2007-05-29 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: Anonymous 2007-05-29 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

Name: Anonymous 2007-05-29 21:08 ID:Ny3Sxm+a

"Oh, by the way,
YOU'RE WELCOME!"
-Nick Burns

Name: Anonymous 2007-05-29 22:55 ID:Heaven

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: Anonymous 2007-05-29 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

Name: Anonymous 2007-05-30 0:09 ID:NdNgkq5D

>>7
oy..

Name: Anonymous 2007-05-30 0:14 ID:NdNgkq5D


(2sin(x)cos(x))/x
=(2/x)+(sin(x)/x)+(cos(x)/x)

lol wut?

Name: Anonymous 2007-05-30 0:18 ID:Atoz6iff

>>7

L'HOPITAL'S RULE, BOY.

lim    sin(2x)/x
x->0

= lim    (2cos(2x))/1
  x->0

= 2

Name: Anonymous 2007-05-30 1:31 ID:4OshKqUd

Read the post guys, he is not allowed to use L'Hopital.  >>6 has the right answer.

Name: Anonymous 2007-05-30 23:40 ID:pkwKlP+c

>>6
Why thank you. That's perfect.

Name: 4tran 2007-05-31 19:44 ID:BPfRRRZL

"Which I tried rewriting as (2/x)+(sin(x)/x)+(cos(x)/x)"
phail

You can still solve the question with the double angle method you wrote: (2sin(x)cos(x))/x = 2cos(x)[sin(x)/x] = 2*1*1 = 2

>>2 is unrelated, and >>6 has a valid answer.

Name: Anonymous 2007-06-01 21:35 ID:ADiu91Bg

suck me until my vas deferens gives out

Name: Anonymous 2007-06-02 22:22 ID:fV9oWoQI

fucking asians only board

Name: Anonymous 2007-06-03 5:51 ID:UF+L2fCN

hi maths forum

Name: Anonymous 2007-06-03 8:32 ID:PGCAJ/is

Divide by zero. Oh shi-

Name: Anonymous 2009-03-18 2:45

The word pirahna, is all I can think of that rhymes with marijuana

Marijuana MUST be legalized.

Don't change these.
Name: Email:
Entire Thread Thread List