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?

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.
