How to find $\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}$ without L'Hopital's Rule. 
How would you find $\displaystyle\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}$ without L'Hopital's Rule?

The way the problem is set up, it makes me think I would try and use the fact that
$\displaystyle\lim_{x \to 0}\frac{1-\cos(x)}{x}=0$ or $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}=1$
So one idea I did was to multiply the top and bottom by $2x$ like so:
$\displaystyle\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}\cdot\frac{2x}{2x}$.
Then I would let $\theta=2x$:
$\displaystyle\lim_{\theta \to 0}\frac{1-\cos(\theta)}{\sin^2{(\frac{3}{2}\theta)}}\cdot\frac{\theta}{\theta}$
which would let me break it up:
$\displaystyle\lim_{\theta \to 0}\frac{1-\cos(\theta)}{\theta}\cdot \frac{\theta}{\sin^2(\frac{3}{2}\theta)}$
So I was able to extract a trigonometric limit that is zero or at least see it. The second part needed more work. My immediate suspicion was maybe the whole limit will go to zero, but when checking with L'Hopital's Rule, I get $\frac{2}{9}$..... :/
 A: Also try this one: with 
$$1-\cos2t=2\sin^2t$$
then
$$\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}=\lim_{x \to 0}\frac{2\sin^2x}{\sin^23x}\times\dfrac{(3x)^2}{2(x)^2}\times\dfrac{2}{9}=\lim_{x \to 0}\frac{2\sin^2x}{2(x)^2}\times\dfrac{(3x)^2}{\sin^23x}\times\dfrac{2}{9}=\color{blue}{\dfrac{2}{9}}$$
A: Notice that 
$$
\frac{1 - \cos t}{t^2} = \frac{1 - \cos t}{t^2} \cdot \frac{1 + \cos t}{1 + \cos t} = \frac{1 - \cos^2 t}{t^2(1 + \cos t)} = \frac{\sin^2 t}{t^2} \cdot \frac{1}{1 + \cos t}.
$$
Hence
$$
\lim_{t \to 0}\frac{1 - \cos t}{t^2} = \frac{1}{2}.
$$
Returning back to your problem:
$$
\frac{1 - \cos(2x)}{\sin^2 (3x)} = \frac{1 - \cos(2x)}{x^2} \cdot \frac{x^2}{\sin^2(3x)} = 4 \cdot \frac{1 - \cos(2x)}{(2x)^2} \cdot \frac{1}{9} \cdot \frac{(3x)^2}{\sin^2(3x)}.
$$
Now take the limit...
A: As in the other solutions,
$$\frac{1-\cos2x}{\sin^23x}=2\left(\frac{\sin x}{\sin 3x}\right)^2.$$
But
$$\sin 3x=3\sin x-4\sin^3x.$$
Therefore
$$\frac{1-\cos2x}{\sin^23x}=2\left(\frac{1}{3-4\sin^2x}\right)^2\to\frac29$$
as $x\to0$.
A: $\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots \approx x ,\text{as }  x \to 0$
that is $\displaystyle\lim_{x \to 0} \sin x = x \implies \displaystyle\lim_{x \to 0} \dfrac{\sin x}{x}=1$
$\displaystyle\lim_{x \to 0} \dfrac{1−\cos(2x)}{\sin^2(3x)}=\displaystyle\lim_{x \to 0} \dfrac{2\sin^2x }{\sin^2(3x)}=\dfrac29\displaystyle\lim_{x \to 0} \dfrac{\dfrac{\sin^2x}{x^2} }{\dfrac{\sin^2(3x)}{(3x)^2}}=\dfrac29\dfrac{\displaystyle\lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)^2 }{\displaystyle\lim_{3x \to 0}\left( \dfrac{\sin(3x)}{3x}\right)^2}=\dfrac29$
A: Taylor expansion:
$$
\lim_{x\to0}\frac{1-\cos2x}{\sin^23x}=
\lim_{x\to0}\frac{1-\left(1-\dfrac{(2x)^2}{2!}+o(x^2)\right)}{\left(\dfrac{3x}{1!}+o(x)\right)^2}=
\lim_{x\to0}\frac{2x^2+o(x^2)}{9x^2+o(x^2)}=\frac{2}{9}
$$
In a different way, use $1-\cos2x=2\sin^2x$, so you get
$$
\lim_{x\to0}2\left(\frac{\sin x}{\sin3x}\right)^2
$$
Now
$$
\lim_{x\to0}\frac{\sin x}{\sin 3x}=
\lim_{x\to0}\frac{1}{3}\frac{\sin x}{x}\frac{3x}{\sin 3x}=\frac{1}{3}
$$
A: You have unnecessarily complicated the problem .
Answer:
$$\lim_{x\to 0}\ \frac{1-\cos2x}{\sin^2{3x}} $$
Using  :
$$1 - \cos2x = {2\sin^2{x}}$$
$$\lim_{x\to 0}\ \cfrac{2\sin^2{x}}{\sin^2{3x}} $$
$$2\  \lim_{x\to 0}\ \cfrac{\sin^2{(x)}}{(x)^2} \cdot \ \cfrac{(x)^2}{\sin^2(3x)} $$
Now multiply the numerator and denominator by $9$ i.e $(3)^2$ to obtain 
$$\cfrac{2}{(3)^2}\  \lim_{x\to 0}\ \cfrac{\sin^2{(x)}}{(x)^2} \cdot \ \cfrac{(3)^2(x)^2}{\sin^2(3x)} $$
$$\cfrac{2}{9}\  \lim_{x\to 0}\ \cfrac{\sin^2{(x)}}{(x)^2} \cdot \ \cfrac{(3x)^2}{\sin^2(3x)} $$
$$\cfrac{2}{9}\  \lim_{x\to 0}\ \biggl(\cfrac{\sin{x}}{x}\biggr)^2 \cdot \ \biggl(\cfrac{3x}{\sin{3x}}\biggr)^2 $$
Using $$\lim_{x\to 0} \cfrac{\sin x}{x} = 1$$
And   $$\lim_{3x\to 0} \cfrac{\sin (3x)}{(3x)} = 1$$
It Becomes:-
$$\cfrac{2}{9}\  \lim_{x\to 0}\ 1\cdot \ 1$$
And
$$\cfrac{2}{9}\  \lim_{x\to k}\ 1 = 1$$
$$ Answer : \cfrac{2}{9}\ $$
