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.
 A: $$\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}$$
$$=\lim_{x\to0^-}\sqrt{\cos^2{x}+\frac{8\sin^2{2x}}{(2x)^2}-1}$$
$$=\sqrt{8}$$
$$=2\sqrt{2}$$
A: Factor by $x^4$ inside the square root and simplify by $x^2$.
then using the classical limit
$$\lim_0\frac {\sin (X)}{X}=1$$
we get $$2\sqrt {2} $$
A: 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
$$\begin{align}
\lim_{x \to 0}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2} &=\lim_{x \to 0}\frac{\sqrt{x^4[1+O(x^4)]+2x^2[4x^2+O(x^6)]-x^4}}{x^2}
\\&=\lim_{x \to 0}\frac{\sqrt{x^4+8x^4+O(x^8)-x^4}}{x^2}
\\&=\lim_{x \to 0}\sqrt{\frac{8x^4+O(x^8)}{x^4}}
\\&=\lim_{x \to 0}\sqrt{8+O(x^4)}
\\&=\sqrt{8+\lim_{x \to 0}O(x^4)}
\\&= \color{red}{\sqrt 8}
\end{align}$$
A: $$\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2} = \frac{\sqrt{x^4\cos^2{x}+8x^2 \sin^2x \cos^2x-x^4}}{x^2}=
\frac{\sqrt{x^4(\cos^2{x}-1)+8x^2 \sin^2x \cos^2x}}{x^2}=
\frac{\sqrt{-x^4\sin^2 x+8x^2 \sin^2x \cos^2x}}{x^2} =
\frac{\sqrt{x^2\sin^2 x(8 \cos^2x - x^2)}}{x^2} =
\frac{\sqrt{\sin^2 x(8 \cos^2x - x^2)}}{x} =
\sqrt{ \left(\frac{\sin x}{x}\right)^2 (8 \cos^2x - x^2)}
$$
Now, take the limit
$$\lim_{x\rightarrow 0^{-}}\sqrt{ \left(\frac{\sin x}{x}\right)^2 (8 \cos^2x - x^2)} = \lim_{x\rightarrow 0^{-}}\sqrt{ \left(\frac{\sin x}{x}\right)^2} \lim_{x\rightarrow 0^{-}}\sqrt{(8 \cos^2x - x^2)} = 1 \times \sqrt{8} = 2\sqrt{2}$$
A: 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 $$x=\frac \pi 6 \implies \text{exact} \approx 2.28503 \qquad  \text{approximation} \approx 2.26301$$ 
$$x=\frac \pi 4 \implies \text{exact} \approx 1.65598 \qquad  \text{approximation} \approx 1.55624$$
A: 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 $2\sqrt{2}$.
