Question

limit as x goes to 0:
tan(2x)/(sin(3x)cos(x))
(no l'hopital)

Answer 1

okay one of those Sinx/x problems..

Answer 2

haha, yeah. I want to see if Im calcing right without using l'hopital

Answer 3

I think you can get away just by doing some trig ...
that's what I did:
tan=sin/cos
sin3x=sin(2x+x)=sin2xconsx+cos2xsinx
& so on

Answer 4

Okay lookin for trig identities that can help you solve this.. First of all, i'd get all of this into Sinx and Cosx s

Answer 5

done. I think inik might be on the right track too

Answer 6

or not :-)

Answer 7

a lot of trig function will canceled out:
2tans/(1-tan2x)cosx (sin2xcosx+cos2x sinx)=
2sinx * cos^2x/cosx^2 (sin2x cosx +cos2x sinx)(cosx^2 - sinx^2)=
2sinx/cos2x (sin2x cosx + cos2x sinx)=

Answer 8

=2sinx/(cos2x * sinx * (2cosx^2 + cos2x) = 2/(1*(2+1)=2/3

Answer 9

that is the correct answer; just going through how you got it with trig identities

Answer 10

Okay here's what I have:
tan 2x = Sin2x / Cos2x
Sin2x = 2SinxCosx
Cos2x = Cos^2 x - Sin^2 x
Sin3x = (3Cos^2 x * Sinx - Sin^3 x)

Answer 11

phew this is getting long :-) wish we could just use l'hopital

Answer 12

why is sin(2x) 2sin(x)cos(x)?

Answer 13

tan(2x)/(sin(3x)cos(x)) = [Sin2x/Cos2x] / [Sin3xCosx] = { [2SinxCosx]/[Cos^2 x - Sin^2 x] } / {Sin3x Cosx}

Answer 14

Akileez it's a trig identity, and apparently it can be proven.. don't remember how though..

Answer 15

yeah< i didn;t recognize it off by heart

Answer 16

{ [2SinxCosx]/[Cos^2 x - Sin^2 x] } / {Sin3x Cosx} = { [2SinxCosx]/{ [Cos^2 x - Sin^2 x] Sin3x Cosx}

Answer 17

Cosines cancel:
{ [2SinxCosx]/{ [Cos^2 x - Sin^2 x] Sin3x Cosx} = { [2Sinx]/{ [Cos^2 x - Sin^2 x] Sin3x}

Answer 18

oh! the dbl angle formulas

Answer 19

i think it's easier to present sin3x as sin (2x+x)...
I got an answer - see up 2/3, which is matching answer if you use l'h rule

Answer 20

{ [2Sinx]/{ [Cos^2 x - Sin^2 x] Sin3x} = { [2Sinx]/{ [Cos^2 x - Sin^2 x] (3Cos^2 x * Sinx - Sin^3 x) }

Answer 21

inik wld you mind putting the way you worked it out into one post pls?

Answer 22

give me a min - will take a few

Answer 23

inik used different identities i believe, I will let him finish instead of confusing u with more identities.. haha

Answer 24

haha! no worries Bahrom! it is frustrating when you can simply use l'hopital - thanks to both you guys for sharing in this torturous one :-)

Answer 25

\[=2tanx/(1-\tan ^{2}x)* cosx*(\sin2x cosx+\cos2x sinx)=\]
\[=2sinx \cos ^{2} x/\cos ^{2} x (\sin2x cosx +\cos2x sinx)(\cos ^{2}x-\sin ^{2}x)=\]
\[=2sinx/ \cos2x (2sinx cosx cosx+\cos2x sinx)=\]
\[=2sinx/\cos2x sinx (2\cos ^{2}x +\cos2x)=2/\cos2x(2\cos ^{2}x+\cos2x)=2/1*(2+1)=2/3\]
=2/1*(2+1)
=2/3

Answer 26

let me know if there are questions... I think I was able to put almost all steps.... too much work!

Answer 27

Champion!!! Thnak you!

Answer 28

Ooo ye! I'm champion now! Awesome!!
& Welcome!

Answer 29

you happen to have a good link with a l'hopital review/explanation pls?

Answer 30

\[x\rightarrow0\quad\Rightarrow\quad\tan{2x}\rightarrow2x\,,\,\sin{3x}\rightarrow3x\,,\,\cos{x}\rightarrow1\]
\[\lim_{x\rightarrow0}\frac{\tan{2x}}{\sin{3x}\cos{x}}=\lim_{x\rightarrow0}\frac{2x}{3x\cdot 1}=\frac{2}{3}\]

Answer 31

thanks Nikvist; I was referring though to an entire tutorial on l'hopital - but the way you put it down makes it rather simple to deduce though - medal!