Question

Evaluate lim θ->0 (sin(3θ)cot(2θ))/(2θcot(θ))

Answer 1

I think you first have to re-write it as:

\[\lim_{\theta \rightarrow 0}\frac{\frac{1}{2\theta}cot(2\theta)}{\frac{1}{sin(3\theta)}cot(\theta)}\]

Which is of the form \(\frac{\infty}{\infty}\) and use l'hopitals.

Answer 2

\[\lim_{\theta \rightarrow 0}\frac{\sin(3 \theta)*\frac{\cos(2 \theta)}{\sin(2 \theta)}}{2 \theta*\frac{\cos( \theta)}{\sin(\theta)}}\]
\[=\lim_{\theta \rightarrow 0}\frac{3 \theta \frac{\sin3 \theta}{3 \theta}*\cos(\theta)*\frac{2 \theta}{\sin(2 \theta)}}{2*\cos(\theta)*\frac{\theta}{\sin(\theta)}}\]
\[=\frac{3*0*1*1*1}{2*1*1}\]

Answer 3

oops i made a mistake

Answer 4

do you see in the numerator that I multiply by 2 theta but i didn't divide by 2 theta

Answer 5

so actually the limit is not 0

Answer 6

so if we make that fix we should get 3/4 if i didnt make another mistake
\[=\frac{3*1*1*1*\frac{1}{2}}{2*1*1}=\frac{\frac{3}{2}}{2}=\frac{3}{4}\]

Answer 7

yep looks good
you can also compare your answer graphically if you wish

Answer 8

Hrm.. I'm assuming you used l'hopital's in that first step there, but I thought we had to have it in one of the specific forms. Where originally it's of the form \(\frac{0\times \infty}{0 \times \infty}\). I suppose you can split them up and take the derivative of each though rather than inverting the 0's as I did.

Answer 9

no i didn't use l'hopital
i just rewrote the function

Answer 10

i wrote the cot as cos/sin

Answer 11

and then i used the property that
as x->0, sinx/x->1

Answer 12

or i mean i put it in terms of sinx and cosx

Answer 13

and cosx->1 as x->0

Answer 14

oh I see. Bleh.

Answer 15

\[\lim_{\theta \rightarrow 0}\frac{\sin(3 \theta)*\frac{\cos(2 \theta)}{\sin(2 \theta)}}{2 \theta \frac{\cos(\theta)}{\sin(\theta)}}\]
\[=\lim_{\theta \rightarrow 0}\frac{3 \theta*\frac{\sin(3 \theta)}{3 \theta}*\cos(2 \theta)*\frac{2 \theta}{\sin(2 \theta)}*\frac{1}{2 \theta}}{2 \cos(\theta) \frac{\theta}{\sin(\theta)}}\]

Answer 16

there maybe that looks better

Answer 17

i went back in and put the thing i was missing lol

Answer 18

perhaps lhopital way is better since this looks messey but i don't know lhopital could be messey too

Answer 19

it probably is what with all those chain rules

Answer 20

and product rules

Answer 21

maybe this is the best way for this one

Answer 22

probably.

Answer 23

wow thank you guys :)