How to show $\frac{|x|^{\delta}-(1+|x|^2)^{\delta/2}}{(1+|x|^2)^{\delta/2}} \sim \frac{1}{|x|^2}.$

Question

Suppose $x\ \in \mathbb R^{n}$, then for $1<\delta<2$ how to show $$\frac{|x|^{\delta}-(1+|x|^2)^{\delta/2}}{(1+|x|^2)^{\delta/2}} \sim \frac{1}{|x|^2}.$$

It seems to me the numerator and the denominator have the same order. I also tried to foil by fractional binomial but still couldn't get it.

Answer 1

For large $|x|$,$$(1+|x|^2)^{\delta/2}=|x|^\delta(1+|x|^{-2})^{\delta/2}\approx|x|^\delta\left(1+\frac{\delta}{2}|x|^{-2}\right)\implies\frac{|x|^\delta}{(1+|x|^2)^{\delta/2}}-1\approx-\frac{\delta}{2}|x|^{-2}.$$