# 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.