Question: Find the limit $$\lim\_{x\to0\_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}.$$

First I've rewritten it as $$\sqrt{ \frac{2x(1-x)}{\sin x}} \frac{\log(1-x^2)}{1-\cos x}$$

Now we can use the approximation $\sin x = x + O(x^3)$ to see that the left factor will converge to $\sqrt{2}$, right? Now I've used L'Hospital and the same approximation for $\sin x$ to find that the right factor converges to $-2$. So the original limit is $-2 \sqrt{2}$.

Is this correct? Is there a better way to do this?

Answer: 
$$\lim\_{x\to0\_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}$$

$$=\sqrt2\sqrt{\lim\_{x\to0\_+}(1-x)}\cdot\frac1{\sqrt{\lim\_{x\to0\_+}\dfrac{\sin x}x}}\cdot\lim\_{x\to0\_+}(1+\cos x)\cdot\lim\_{x\to0\_+}\frac{\ln(1-x^2)}{(-x^2)}\cdot\lim\_{x\to0\_+}\frac{-x^2}{(1-\cos^2x)} $$

For the last limit $$\lim\_{x\to0\_+}\frac{-x^2}{(1-\cos^2x)}=-\frac1{\left(\lim\_{x\to0\_+}\dfrac{\sin x}x\right)^2}=\cdots $$

The rest can be managed using $$\lim\_{h\to0}\frac{\sin h}h=1=\lim\_{h\to0}\frac{\ln(1+h)}h$$
You've found the right approach. I would use your knowlege about the Taylor approximations to intelligently split the limit up into 2 simpler limits then be more formal about solving it.

$$\begin{align}
\lim\_{x\to0\_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}} &= \lim\_{x\to0\_+} \left(\sqrt{ \frac{2x(1-x)}{\sin x}} \frac{\log(1-x^2)}{1-\cos x}\right)\\
&= \lim\_{x\to0\_+} \sqrt{ \frac{2x(1-x)}{\sin x}} \cdot \lim\_{x\to0\_+}\frac{\log(1-x^2)}{1-\cos x}\\
&= \sqrt{\lim\_{x\to0\_+} \frac{2x(1-x)}{\sin x}} \cdot \lim\_{x\to0\_+}\frac{\log(1-x^2)}{1-\cos x}\\
\text{L'Hopital on both limits}\\
&= \sqrt{\lim\_{x\to0\_+} \frac{2 - 4x}{\cos x}} \cdot \lim\_{x\to0\_+}\frac{\dfrac{-2x}{1 - x^2}}{\sin x}\\
&= \sqrt{2} \cdot \lim\_{x\to0\_+}\frac{-2x}{\sin x(1-x^2)}\\
\text{L'Hopital again}\\
&= \sqrt{2} \cdot \lim\_{x\to0\_+}\frac{-2}{\cos x - 2x\sin x - x^2 \cos x}\\
&= -2\sqrt{2}
\end{align}
$$

Now that I look at your question again I realize this is pretty much exactly how you explained your solution. Oh well.