How do I evaluate $\lim _{x\to 0}\left(\csc^2\left(2x\right)-\frac{1}{4x^2}\right)$ without using L'Hopital?

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I've tried it until this step: $\lim _{x\to 0}\left(\frac{\left(2x-\sin\left(2x\right)\right)\left(2x+\sin\left(2x\right)\right)}{\left(2x\cdot \: \sin\left(2x\right)\right)^2}\right)$ and got confused. I also tried expanding the limit into $\lim _{2x\to 0}\left(\frac{\left(2x-\sin\left(2x\right)\right)}{\sin\left(2x\right)\cdot 2x}\right)\cdot \lim _{x\to 0}\left(\frac{\left(2x+\sin\left(2x\right)\right)}{\sin\left(2x\right)\cdot 2x}\right)$ but the right-side limit doesn't exist.