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.

Just added as a too long comment.
Many ways to get the answer have been given in answers. Using Taylor expansions with one more term, you could even know how is approached the limit since $$x^4 \cos ^2(x)+2 x^2 \sin ^2(2 x)=9 x^4-\frac{35 x^6}{3}+O\left(x^8\right)$$ making $$x^4 \cos ^2(x)+2 x^2 \sin ^2(2 x)-x^4=8 x^4-\frac{35 x^6}{3}+O\left(x^8\right)$$
$$\sqrt{x^4 \cos ^2(x)+2 x^2 \sin ^2(2 x)-x^4}=2 \sqrt{2}\, x^2-\frac{35 }{12 \sqrt{2}}x^4+O\left(x^6\right)$$ $$\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}=2 \sqrt{2}-\frac{35 x^2}{12 \sqrt{2}}+O\left(x^4\right)$$ which shows the limit and how it is approached.
Moreover, this provides an easy estimation of the function "close" to $x=0$. For example $$x=\frac \pi 6 \implies \text{exact} \approx 2.28503 \qquad  \text{approximation} \approx 2.26301$$ 
$$x=\frac \pi 4 \implies \text{exact} \approx 1.65598 \qquad  \text{approximation} \approx 1.55624$$
