4
$\begingroup$

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

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.

However, WolframAlpha says the limit does not exist.

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

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Git Gud
    Mar 22, 2015 at 21:07
  • 1
    $\begingroup$ 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 $x,y,z$. $\endgroup$
    – GEdgar
    Mar 22, 2015 at 21:07
  • $\begingroup$ @GEdgar: I think what they teach is all about $\mathbb{R^3}$. $\endgroup$
    – science
    Mar 22, 2015 at 22:29

2 Answers 2

Reset to default
3
$\begingroup$

Wolfram does a complex limit, unless you specify otherwise. See the hint in the very link quoted:

WA

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$
    – user223022
    Mar 22, 2015 at 20:37
  • 3
    $\begingroup$ 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. $\endgroup$
    – Git Gud
    Mar 22, 2015 at 20:43
  • $\begingroup$ 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. $\endgroup$
    – user223022
    Mar 22, 2015 at 20:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .