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

Question

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

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)

Answer 1

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.

Answer 2

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