# Square root asymptote

1. Jun 29, 2008

### daudaudaudau

Hi.

I'd like to show that sqrt(x*(x-1)) has the asymptote x-0.5. The coefficient on "x" is found by saying

$$\lim_{x\rightarrow\infty}\frac{\sqrt{x(x-1)}}{x}=1$$

but how does one find the 0.5 constant?

2. Jun 29, 2008

### Mute

$$\sqrt{x(x-1)} = \sqrt{x^2\left(1-\frac{1}{x}\right)} \rightarrow_{x \rightarrow \infty} x\left(1-\frac{1}{2x}\right) = x - \frac{1}{2}$$

At the arrow I've used a binomial expansion ($\sqrt{1+a} \simeq 1 + a/2$, to first order, when $a \ll 1$) and kept the first order term, which gives the result you're looking for. (any further terms in the expansion will be $\mathcal{O}(1/x)$, which is why they get neglected but the 1/2 is kept, I guess.)

Last edited: Jun 29, 2008
3. Jun 29, 2008

### daudaudaudau

Oh. Right. Thank you :-)