Find the limit $$L=\lim_{x\to0}\dfrac{\ln(2x^2-2x+1)+\arcsin{2x}}{(\sqrt{1-4x}-1)(1-\cos(2x))}$$
First let's get rid of the square root.
$$L=\lim_{x\to0}\dfrac{\left[\ln(2x^2-2x+1)+\arcsin(2x)\right]\left(\sqrt{1-4x}+1\right)}{(1-4x-1)(1-\cos(2x))}$$
I thought it would be a good idea to write the limit as a sum of the following limits $$L\color{red}{\stackrel{?}{\color{red}{=}}}-\dfrac12\lim_{x\to0}\dfrac{\ln(2x^2-2x+1)}{x(1-\cos(2x))}-\dfrac12\dfrac{\arcsin(2x)}{x(1-\cos(2x))}=-\dfrac12L_1-\dfrac12L_2$$ Note that the first limit is $-\infty$ though, so the equality does not actually hold: $$L_1=
\lim_{x\to0}\dfrac{\ln(2x^2-2x+1)}{x(1-\cos(2x))}\cdot\dfrac{2x^2-2x}{2x^2-2x}=2\lim_{x\to0}\dfrac{x-1}{1-\cos(2x)}\\=\lim_{x\to0}\dfrac{x-1}{\sin^2x}\cdot\dfrac{x^2}{x^2}=-\infty.$$ As a matter of fact, because of the negative coefficient $\left(-\dfrac12\right),$ it would become $+\infty$. This does not change anything of course. I just decided to mention it.
I don't see how to approach the problem if we are not supposed to somehow "split" the limit.
I'm looking for a solution that doesn't rely on any fancy split ideas, but rather on a general approach and intuitive ideas. Thanks!
Best Answer
As $u\to0$, $\sqrt{1+u}=1+\frac u2+o(u)$ and $\cos u=1-\frac{u^2}2+o(u^2)$ hence as $x\to0$, $$(\sqrt{1-4x}-1)(1-\cos(2x))\sim(-2x)(2x^2)=-4x^3.$$ This incites us to expand the numerator at the order $3$. As $u\to0$, $$\begin{align} \ln(1+u)&=u-\frac{u^2}2+\frac{u^3}3+o(u^3),\\ \arcsin u&=u+\frac{u^3}6+o(u^3), \end{align}$$ hence as $x\to0$, $$\begin{align} \ln(1-2x+2x^2)&=(-2x+2x^2)-\frac{4x^2-8x^3}2+\frac{(-2x)^3}3+o(x^3)\\ &=-2x+\frac{4x^3}3+o(x^3),\\ \arcsin(2x)&=2x+\frac{4x^3}3+o(x^3) \end{align}$$ and therefore, $$\lim_{x\to0} \frac{\ln(2x^2-2x+1)+\arcsin{2x}}{(\sqrt{1-4x}-1)(1-\cos(2x))}=\lim_{x\to0}\frac{\frac{8x^3}3}{-4x^3}=-\frac23.$$