How to solve this limit without using L'Hopital's Rule: $\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}$? 
Find:
  $$\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}\cdot$$

I tried to simplify the expression, but I kept getting stuck. I also tried to make a substitution $u=\frac{1}{x}$, but I got stuck again. Please give me hint or explain what to do. Thanks.

$$\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2} = \frac{\sqrt{x^4\cos^2{x}+8x^2 \sin^2x \cos^2x-x^4}}{x^2}=
\frac{\sqrt{x^4(\cos^2{x}-1)+8x^2 \sin^2x \cos^2x}}{x^2}=
\frac{\sqrt{-x^4\sin^2 x+8x^2 \sin^2x \cos^2x}}{x^2} =
\frac{\sqrt{x^2\sin^2 x(8 \cos^2x - x^2)}}{x^2} =
\frac{\sqrt{\sin^2 x(8 \cos^2x - x^2)}}{x} =
\sqrt{ \left(\frac{\sin x}{x}\right)^2 (8 \cos^2x - x^2)}
$$
Now, take the limit
$$\lim_{x\rightarrow 0^{-}}\sqrt{ \left(\frac{\sin x}{x}\right)^2 (8 \cos^2x - x^2)} = \lim_{x\rightarrow 0^{-}}\sqrt{ \left(\frac{\sin x}{x}\right)^2} \lim_{x\rightarrow 0^{-}}\sqrt{(8 \cos^2x - x^2)} = 1 \times \sqrt{8} = 2\sqrt{2}$$
