limits - Prove that $lim_{xto 0}frac {sqrt{x^2+x+1}-1}{sin(2x)}= infty$

How do I as precisely as possible prove that the following limit goes to infinity?

$$\lim_{x\to 0}\frac {\sqrt{x^2+x+1}-1}{\sin(2x)}=\infty$$

It seems difficult. I have started the proof by selecting an $M>0$ and attempting to show that the function is $M$ is always greater than the function. My problem seems to be algebraically manipulating the function so that I can extract $|x|$.

Answer

HINT:

The limit is not $\infty$. Note that

$$\begin{align} \frac{\sqrt{x^2+x+1}-1}{\sin(2x)}&=\left(\frac{\sqrt{x^2+x+1}-1}{\sin(2x)}\right)\left(\frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}\right)\\\\ &=\frac{x(x+1)}{\sin(2x)}\,\frac{1}{\sqrt{x^2+x+1}+1} \end{align}$$

Now, use $\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta}=1$.