<s>[INST] 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.
 [/INST] The expression in square roots can be simplified as $$8x^{2}\sin^{2}x\cos^{2}x-x^{4}\sin^{2}x$$ and hence the numerator becomes $$x\sin x\sqrt{8\cos^{2}x - x^{2}} $$ and thus we can see that the given expression can be written as $$\frac{\sin x} {x} \cdot\sqrt{8\cos^{2}x-x^{2}}\to 1\cdot \sqrt{8\cdot 1^{2}-0^{2}}=2\sqrt{2}$$ and thus the desired limit is $2\sqrt{2}$.
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