Calculating limits with Wolfram|Alpha

Question

in a random graph setting with $n$ vertices and edge-probability $p$ I want to show that $\binom n4p^6\to 0$ as $pn^{2/3}\to 0$. To check this is even true, I submitted this Wolfram|Alpha query.

Now why does Wolfram|Alpha calculate the limit for $n\to pn^{2/3}$ rather than $pn^{2/3}\to 0$? Is it just a misinterpretation, or is there a mathematical connection I fail to see? Surprisingly it yields the desired result ...

I also don't quite understand what Wolfram|Alpha does in the next line, and I am not even sure how to calculate the limit manually, so I'd be grateful for any insight!

Answer 1

Use $pn^{2/3}=\varepsilon$ and then $\varepsilon\to0$.

Wolfram|Alpha

Answer 2

Simple answer, in wolfram alpha you can't have the product of two variables approach a value. In fact you only can have one variable approach a value. You would have to change it so one variable is approaching a certain value. If you want to keep it like you have it, then this is a job for mathematica, not wolfram alpha. See:https://mathematica.stackexchange.com/questions/21544/finding-limits-in-several-variables