Homework Help: Difference of Squares?

1. Apr 21, 2007

Monochrome

I'm reviewing material for my exams and I came across this:

$$\lim _{x\rightarrow \infty }\sqrt {{x}^{2}+x+1}-\sqrt {{x}^{2}-3\,x}$$

The only explanation it gives is "By the difference of squares" the solution sheet then jumps to:

$$\lim _{x\rightarrow \infty }{\frac {4\,x+1}{\sqrt {{x}^{2}+x+1}+\sqrt {{x}^{2}-3\,x}}}$$

What the hell just happened there? I can solve from then on but I've no idea what's happening on this step. Also an idiot proof link would be appreciated.

2. Apr 21, 2007

cristo

Staff Emeritus
Think of the first line as $$\lim _{x\rightarrow \infty }\frac{\sqrt {{x}^{2}+x+1}-\sqrt {{x}^{2}-3\,x}}{1}$$, then multiply top and bottom of the fraction by $\sqrt {{x}^{2}+x+1}+\sqrt {{x}^{2}-3\,x}$. Does this make the second line any clearer?

3. Apr 21, 2007

Monochrome

Yes, thanks.

4. Apr 21, 2007

symbolipoint

I just tried it myself; how does "2" seem?

5. Apr 21, 2007

Hurkyl

Staff Emeritus
2 sounds good, since the function looks like
$$\frac{4x}{\sqrt{x^2} + \sqrt{x^2}} = 2$$
when x is big.