Can the 0.5 constant in the asymptote of sqrt(x*(x-1)) be found?

[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?

 

[Mute]

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?

\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.)

 

[daudaudaudau]

Oh. Right. Thank you :-)