Evaluating limit $\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}$ 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?

Using the Taylor series:
$$ \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}\sim_0\frac{-\sqrt{2x}x^2}{\frac{x^2}{2}\sqrt x}=-2\sqrt2\leftarrow\text{the desired limit}$$
