Equation with prime numbers

Question

I am reading a paper which, in order to prove a result about graphs, states that if $p$ is a prime, then $$\frac{1}{2}(p^2-1)(p-1) \approx \frac{1}{2} (p^2-1)^{3/2}.$$ In other words, $$\frac{1}{2}(p^2-1)(p-1) = \bigg(\frac{1}{2}-o(1)\bigg)(p^2-1)^{3/2}.$$ This is stated without further explanation, but I can't see why it is true. Can anyone help?

Thank you in advance.

Answer 1

$$\lim_{p \rightarrow \infty}\frac{p-1}{\sqrt{p^2-1}}=\lim_{p \rightarrow \infty}\frac{1}{\frac{p}{\sqrt{p^2-1}}}=\lim_{p \rightarrow \infty}\sqrt{1-\frac{1}{p^2}}=1$$

Use this to see where the $o(1)$ comes from. Also, note that I've never used that is $p$ prime.

Answer 2

Dividing by $(p^2-1)$ yields $(p-1) \approx ((p-1)(p+1))^{1/2}$, i.e. $ (p-1)^{1/2} \approx (p+1)^{1/2}$.