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: 
Using the Taylor series:

$$ \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}\sim\_0\frac{-\sqrt{2x}x^2}{\frac{x^2}{2}\sqrt x}=-2\sqrt2\leftarrow\text{the desired limit}$$
$$\begin{aligned}\lim \_{x\to 0}\left(\frac{\sqrt{2x\left(1-x\right)}\:\log \left(1-x^2\right)}{\left(1-\cos \:\:x\right)\sqrt{\sin \:\:x}}\right)
\\& \approx\_0 \lim \_{x\to 0}\left(\frac{-\sqrt{2x\left(1-x\right)}\:x^2}{\left(1-\cos \:\:x\right)\sqrt{\sin \:\:x}}\right)
\\& \approx\_0 \lim \_{x\to 0}\left(\frac{-\sqrt{2x\left(1-x\right)}\:x^2}{\left(1-\cos \:\:x\right)\sqrt{x}}\right)
\\& \approx\_0 \lim \_{x\to 0}\left(\frac{-\sqrt{2x\left(1-x\right)}\:x^2}{\left(\frac{x^2}{2}\right)\sqrt{x}}\right)
\\& \approx\_0\lim \_{x\to 0}\left(\frac{-\sqrt{2x}\:x^2}{\left(\frac{x^2}{2}\right)\sqrt{x}}\right)
\\& = \color{red}{-2\sqrt{2}}
\end{aligned}$$