<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] Note that, as $x \to 0$, we have
$$
\sin x \sim x \implies\sin^2(2x)\approx4x^2 
\\ \cos x \sim1 \implies\cos^2(x)\approx1
$$
We can use these to greatly simplify the limit at hand:
$$\lim_{x \to 0}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}=\lim_{x \to 0}\frac{\sqrt{8x^4}}{x^2} = \color{red}{\sqrt 8}$$
If you want a mor rigorous proof, we can instead use squeeze theorem or $O$ notation. Using the latter, we can write $\sin^2(2x) \in 4x^2 + O(x^6)$ and $\cos^2(x) \in 1+O(x^4)$ as $x \to 0$, giving us
$$\begin{align}
\lim_{x \to 0}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2} &=\lim_{x \to 0}\frac{\sqrt{x^4[1+O(x^4)]+2x^2[4x^2+O(x^6)]-x^4}}{x^2}
\\&=\lim_{x \to 0}\frac{\sqrt{x^4+8x^4+O(x^8)-x^4}}{x^2}
\\&=\lim_{x \to 0}\sqrt{\frac{8x^4+O(x^8)}{x^4}}
\\&=\lim_{x \to 0}\sqrt{8+O(x^4)}
\\&=\sqrt{8+\lim_{x \to 0}O(x^4)}
\\&= \color{red}{\sqrt 8}
\end{align}$$
</s>