How do I expand $\frac{\left(x^{2}-2x-3\right)}{\sqrt{x-1}}$ when $x\to \infty$?

Question

What I have so far is:

$$\frac{\left(x^{2}(1-\frac{2}{x}-\frac{3}{x^{2}})\right)}{\sqrt{x\left(1-\frac{1}{x}\right)}}$$

Can someone just give me a tip so I could continue? Thank you.

Answer 1

$$ \frac{x^2-2x-3}{\sqrt{x-1\,}} = \frac{(x-1)^2-4}{\sqrt{x-1\,}} = (x-1)^{3/2} - \frac{4}{\sqrt{x-1}}\;. $$ For $x\longrightarrow\infty$, the first term diverges, while the second term vanishes. Hence, $$ \frac{x^2-2x-3}{\sqrt{x-1}} \longrightarrow\infty\;. $$

A different method implies factorising out the maximal powers of $x$, both in the numerator and denominator: $$ \frac{x^2-2x-3}{\sqrt{x-1\,}} = \frac{x^2 \left(1-2x^{-1}-3 x^{-2}\right)}{x^{1/2}\sqrt{1-x^{-1}\,}} = x^{3/2}\; \frac{1-2x^{-2}-3 x^{-2}}{\sqrt{1-x^{-1}}}\;. $$ For $x\longrightarrow\infty$, the fraction on the right-hand side approaches unity. Therefore, the overall expression behaves, for large $x$, as $O(x^{3/2})$ -- and therefore diverges..

Answer 2

We have that by binomial series

$$\frac1{\sqrt{x-1}}=\frac1{\sqrt{x}}\left(1-\frac1x\right)^{-\frac 12} =\frac1{\sqrt{x}}\left(1+\frac1{2x}+O\left(\frac1{x^2}\right)\right)$$

therefore

$$\frac{\left(x^{2}-2x-3\right)}{\sqrt{x-1}}=\left(x^{2}-2x-3\right)\frac1{\sqrt{x}}\left(1+\frac1{2x}+O\left(\frac1{x^2}\right)\right)=$$

$$=x^{\frac32}-\frac32x^{\frac12}+O\left(\frac1{\sqrt x}\right)$$