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

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 $|x|$.

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

• Maybe Dole didn't know this: $\sin(2x) = 2\sin(x)\cos(x)$. – Hopeless Feb 15 '17 at 5:33
• @Hopeless That identity isn't required. Just let $\theta =2x$. – Mark Viola Feb 15 '17 at 5:34
• Didn't see that way. Thanks! – Hopeless Feb 15 '17 at 5:35
• @Hopeless You're welcome. My pleasure. – Mark Viola Feb 15 '17 at 5:38
• The real issue is that I made a sign error somewhere that made it seem like the limit was infinity on my graphic calculator. Thank you very much... – Dole Feb 15 '17 at 5:41