# Calculating limits with Wolfram|Alpha

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!

• Be careful wolframalpha is not perfect, sometimes can fail. The best thing you can do is evaluate yourself. – Masacroso Dec 13 '15 at 15:57
• I know. It gives the right result though so I wondered if one can somehow transform $pn^{2/3}\to 0$ to $n\to pn^{2/3}$. Is there a straightforward way of evaluating such limits manually? – akkarin Dec 13 '15 at 15:59

## 2 Answers

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

Wolfram|Alpha

• Where does the $n^{-4}$ in your query come from? – akkarin Dec 14 '15 at 13:56
• $p=\varepsilon.n^{-2/3}$, and so $p^6=\varepsilon ^6.n^{-4}$ – JonMark Perry Dec 14 '15 at 14:21
• Ah, I see. Would you say it is obvious that the limit is $0$ as $\epsilon\to 0$, or does it need more reasoning? – akkarin Dec 14 '15 at 15:03
• $\binom{n}{4}$ is $o(n^4)$ and $n\to\infty$ so $p\to 0$ does it – JonMark Perry Dec 14 '15 at 15:05
• what happens otherwise? – JonMark Perry Dec 14 '15 at 15:19

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