# Regarding $\lim \limits_{(x,y,z)\to (0,0,0)}\left(\frac{x^2z}{x^2+y^2+16z^2}\right)$--is WolframAlpha incorrect?

$$\lim_{x,y,z\to 0} {zx^2\over x^2+y^2+16z^2}$$

So I am trying to evaluate this limit..

To me, by using the squeeze theorem, it seems that the answer must be zero.

I trying using the spherical coordinates, which also gives in the same result.

Could I know whether I am missing something or WolframAlpha is incorrect?(as it happens occasionally)

• By the way, I can't prove this. But I'm convinced that when computing multivariate limits, WA finds the iterated limits. So even has a confirmation of existence it is very shallow. – Git Gud Mar 22 '15 at 21:07
• Are you sure Wolfram is wrong? It is doing the limit in complex space $\mathbb C^3$. You have to tell it if you want to restrict to real $x,y,z$. – GEdgar Mar 22 '15 at 21:07
• @GEdgar: I think what they teach is all about $\mathbb{R^3}$. – science Mar 22 '15 at 22:29

## 2 Answers

Wolfram does a complex limit, unless you specify otherwise. See the hint in the very link quoted: But look: substitute $$(x,y,z) = (4it,4t^{3/2},t)$$ into $${zx^2\over x^2+y^2+16z^2}$$ to get $$\frac{t(-16)t^2}{-16t^2+16t^3+16t^2} = -1$$ along the whole curve. So anyone claiming the limit is zero is wrong.

hint: $0 \leq \dfrac{|zx^2|}{x^2+y^2+16z^2} \leq |z|$

• Thanks for your input. But that's exactly what I did, which gives me the answer of L=0. I just wanted to clarify that I have the right answer and WolframAlpha is giving out an incorrect answer? – user223022 Mar 22 '15 at 20:37
• WA is wrong plenty. But when WA says a limit doesn't exist, that is usually the case. This is the first time I see WA saying a limit doesn't exist when it exists. – Git Gud Mar 22 '15 at 20:43
• Thanks for that! Guess I learnt a lesson not to always trust WA.. This is interesting as this was my first time trying to evaluate a limit on WA. – user223022 Mar 22 '15 at 20:50