Question

lim x.sin3x/1-cos2x
x->0

Answer 1

Hey! I'm here

Answer 2

\[\lim_{x->0} \frac{\sin(3x)}{1-\cos(2x)}\]

Answer 3

Ok we will probably be using either or both of these limits:
\[\lim_{u->0} \frac{\sin(u)}{u}=1\]
OR
\[\lim_{u->0} \frac{\cos(u)-1}{u}=0\]

Answer 4

I think it would cool if there was someway to write the bottom as one term but we won't be able to do it without one factor
Do you know what factor we need?

Answer 5

Like we need to multiply top and bottom by ----?-----

Answer 6

look!
\[\lim_{x \rightarrow 0} x \sin3x/2\sin2x = xsin3x/2sinxsinx = \lim x \times \sin3x/x/2sinx \times sinx/x = 3/2\]

Answer 7

\[1-\cos(2x)=1-(\cos^2(x)-\sin^2(x))=1-\cos^2(x)+\sin^2(x)\]
\[=1-(1-\sin^2(x))+\sin^2(x)\]
\[2 \sin^2(x)=2 \sin(x) \sin(x)\]
Yay we can do it that was we write the bottom! :)

Answer 8

\[\lim_{x \rightarrow 0}\frac{\sin(3x)}{2 \sin(x) \sin(x)}\]

Answer 9

Ok yea I see that is what you have on the bottom at one part
Good Job

Answer 10

ok so...
in order to use \[\lim_{u->0} \frac{\sin(u)}{u}=1\]
We need to have sin(3x)/(3x) or even x/(sin(x)) <-we actually prove this also goes to 1 on the way to proving the first
so we have
\[\frac{1}{2}\lim_{x \rightarrow 0}\frac{\sin(3x)}{1} \frac{1}{\sin(x)} \frac{1}{\sin(x)}\]
=
\[\frac{3}{2}\lim_{x \rightarrow 0}\frac{\sin(3x)}{3x} \frac{x}{\sin(x)} \frac{1}{\sin(x)}\]
the first thing I did was multiply by 3x/3x
Do you see that?

Answer 11

yes thanks

Answer 12

ok so we have
\[\frac{3}{2}(1)(1) \cdot \lim_{x->0}\frac{1}{\sin(x)}\]

Answer 13

But there is no way to any manipulation so that csc(x) is continuous at x=0
The limit does not exist

Answer 14

no problem
may you solve it?
\[\lim_{x \rightarrow 0} 2x^{2}/1-cosx\]