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

Question

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 $|x|$.

score 4 · Accepted Answer · answered Feb 15 '17 at 05:04

4

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

answered Feb 15 '17 at 05:04

Mark Viola

179,405

Maybe Dole didn't know this: $\sin(2x) = 2\sin(x)\cos(x)$. – Hopeless Feb 15 '17 at 05:33
2

@Hopeless That identity isn't required. Just let $\theta =2x$. – Mark Viola Feb 15 '17 at 05:34
Didn't see that way. Thanks! – Hopeless Feb 15 '17 at 05:35
@Hopeless You're welcome. My pleasure. – Mark Viola Feb 15 '17 at 05:38
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 05:41
You're quite welcome. I'm pleased to hear that you have it now! – Mark Viola Feb 15 '17 at 06:07

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

1 Answers1