Evaluating trigonometric limit $\csc^2(2x) - \frac{1}{4x^2}$ 
$$\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right]$$

I've tried to use l'Hôpital's rules but still can't find the answer. Here's my approach:
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-4\sin(2x)\cos(2x)}{8x\sin^2(2x)+16x^2\sin(2x)\cos(2x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-2\sin(4x)}{8x\sin^2(2x)+8x^2\sin(4x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x-\sin(4x)}{4x\sin^2(2x)+4x^2\sin(4x)} \right] \\
\end{aligned}
$$
Am I using the correct way? How to solve it correctly?
P. S. I tried to use calculator and it outputs one third $1/3$.
 A: hint: replace the denominator on the third line of your proof $\sin^2(2x)$ by $(2x)^2$ and apply L'hopitale rule three times to the expression: $\dfrac{4x^2-\sin^2(2x)}{16x^4}$. I did it and it works. Try it. Note that "replace" here means you write: $\sin^2 (2x) = 4x^2\cdot \left(\dfrac{\sin(2x)}{2x}\right)^2$, and the second factor approaches $1$ when $x \to 0$.
A: Based on DeepSea's hint, I've managed to solve by myself.
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2\times 4x^2\left(\frac{\sin(2x)}{2x}\right)^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2\times 4x^2} \right] \times  \lim_{x\rightarrow 0} \left[ \frac{1}{\left(\frac{\sin(2x)}{2x}\right)^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{16x^4} \right] \times  1 \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-2\sin(4x)}{64x^3} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8-8\cos(4x)}{192x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1-\cos(4x)}{24x^2} \right] \\
=& \frac{1}{24} \times \lim_{x\rightarrow 0} \left[ \frac{1-\cos(4x)}{x^2} \right] \\
=& \frac{1}{24} \times \frac{1}{2} \times 4^2 \\
=& \frac{1}{3} \\
\end{aligned}
$$
A: Almost.
I'll edit your answer.
$\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin(2x)} - \frac{1}{2x} \right]\left[ \frac{1}{\sin(2x)} + \frac{1}{2x} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{2x-(2x)^3/6+O(x^5))} - \frac{1}{2x} \right]\left[ \frac1{2x} + \frac{1}{2x} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{2x-(2x-(2x)^3/6+O(x^5))}{2x(2x-(2x)^3/6+O(x^5)))} \right]\left[ \frac1{x}\right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{(2x)/6+O(x^3))}{(1-(2x)^2/6+O(x^4)))} \right]\left[ \frac1{x}\right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1/3+O(x^2))}{1-(2x)^2/6+O(x^4))} \right] \\
=&\frac13\\
\end{aligned}
$
