Evaluating $\lim_{x \to 0} \frac{1-\cos(x^2+2x)}{1-\cos(x^2+x)}$ without L'Hopital

Question

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.

Answer 1

\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}

Answer 2

Well since $$\forall x\in \mathbf{R}: \quad \frac{1-\cos(x^{2}+2x)}{1-\cos(x^{2}+x)}=\csc^{2}\left(\frac{x}{2}+\frac{x^{2}}{2}\right)\sin^{2}\left(x+\frac{x^{2}}{2}\right)$$ and using that $$\lim_{x\to 0}\csc^{2}\left(\frac{x}{2}+\frac{x^{2}}{2}\right)\sin^{2}\left(x+\frac{x^{2}}{2}\right)=4$$ so the answer is $$\lim_{x\to 0}\frac{1-\cos(x^{2}+2x)}{1-\cos(x^{2}+x)}=4.$$