Evaluating $\lim_{x \to 0} \frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)}$ without L'Hopital Could you help me with this limit please:
$$
\lim_{x \to 0}  \frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)}
$$
I know that with L'Hopital is easy but I want to do it without that theorem. I have already tried converting it  to:
$$
 \frac{2\sin^2(x^2/2+x)}{2\sin^2(x^2/2+x/2)}
$$
or expanding the sin of the sum of two angles:
$$
 \frac{\sin(x^2/2)\cos(x)+\sin(x)\cos(x^2/2)}{\sin(x^2/2)\cos(x/2)+\sin(x/2)\cos(x^2/2)}
$$
but I can not advance further than that and eliminate the indetermination 0/0. I would really appreciate if you could help me please.

\begin{align}
\lim_{x\to0}\frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)} &= \lim_{x\to0}\frac{2\sin^2(\frac{x^2+2x}{2})}{2\sin^2(\frac{x^2+x}{2})} \\
&= \lim_{x\to0}\frac{\sin^2(\frac{x^2+2x}{2})}{\left(\frac{x^2+2x}{2}\right)^2}\cdot\frac{\left(\frac{x^2+x}{2}\right)^2}{\sin^2(\frac{x^2+x}{2})}\cdot\frac{\left(\frac{x^2+2x}{2}\right)^2}{\left(\frac{x^2+x}{2}\right)^2} \\
&= 1\cdot1\cdot\lim_{x\to0}{\frac{\left(\frac{x^2+2x}{2}\right)^2}{\left(\frac{x^2+x}{2}\right)^2}}\\
&= \lim_{x\to0}{\frac{x^2(x+2)^2}{x^2(x+1)^2}}\\
\\
&=4
\end{align}
