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?

$$\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}$$
$$=\sqrt2\sqrt{\lim_{x\to0_+}(1-x)}\cdot\frac1{\sqrt{\lim_{x\to0_+}\dfrac{\sin x}x}}\cdot\lim_{x\to0_+}(1+\cos x)\cdot\lim_{x\to0_+}\frac{\ln(1-x^2)}{(-x^2)}\cdot\lim_{x\to0_+}\frac{-x^2}{(1-\cos^2x)} $$
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 $$\lim_{h\to0}\frac{\sin h}h=1=\lim_{h\to0}\frac{\ln(1+h)}h$$
