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?
 A: Using the Taylor series:
$$ \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}\sim_0\frac{-\sqrt{2x}x^2}{\frac{x^2}{2}\sqrt x}=-2\sqrt2\leftarrow\text{the desired limit}$$
A: $$\lim_{x\to0_+} \frac{\sqrt{2x(1-x)} \log(1-x^2)}{(1-\cos x)\sqrt{\sin x}}$$
$$=\sqrt2\sqrt{\lim_{x\to0_+}(1-x)}\cdot\frac1{\sqrt{\lim_{x\to0_+}\dfrac{\sin x}x}}\cdot\lim_{x\to0_+}(1+\cos x)\cdot\lim_{x\to0_+}\frac{\ln(1-x^2)}{(-x^2)}\cdot\lim_{x\to0_+}\frac{-x^2}{(1-\cos^2x)} $$
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 $$\lim_{h\to0}\frac{\sin h}h=1=\lim_{h\to0}\frac{\ln(1+h)}h$$
A: 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.
A: $$\begin{aligned}\lim _{x\to 0}\left(\frac{\sqrt{2x\left(1-x\right)}\:\log \left(1-x^2\right)}{\left(1-\cos \:\:x\right)\sqrt{\sin \:\:x}}\right)
\\& \approx_0 \lim _{x\to 0}\left(\frac{-\sqrt{2x\left(1-x\right)}\:x^2}{\left(1-\cos \:\:x\right)\sqrt{\sin \:\:x}}\right)
\\& \approx_0 \lim _{x\to 0}\left(\frac{-\sqrt{2x\left(1-x\right)}\:x^2}{\left(1-\cos \:\:x\right)\sqrt{x}}\right)
\\& \approx_0 \lim _{x\to 0}\left(\frac{-\sqrt{2x\left(1-x\right)}\:x^2}{\left(\frac{x^2}{2}\right)\sqrt{x}}\right)
\\& \approx_0\lim _{x\to 0}\left(\frac{-\sqrt{2x}\:x^2}{\left(\frac{x^2}{2}\right)\sqrt{x}}\right)
\\& = \color{red}{-2\sqrt{2}}
\end{aligned}$$
