# How can I evaluate this multivariate limit?

How can I evaluate the following limit? $$\lim_{|(x,y,z)|\to+\infty}(x^4+y^2+z^2-x+3y-z).$$

• Is it my misunderstanding, that the limit is (quite obviously) undefined? Feb 2, 2018 at 7:39
• @ElfHog: It is possible to understand that the limit exists in the sense of the extended real line. Feb 2, 2018 at 7:44
• @hardmath But if $x\rightarrow+\infty$ then the equation reaches $+\infty$, and if $x\rightarrow-\infty$ then the equation reaches $-\infty$.. Does the limit still exist? Feb 2, 2018 at 7:46
• @ElfHog you cannot jump to such conclusions so quickly, see my answer. Feb 2, 2018 at 7:47
• @ElfHog: Look again. Feb 2, 2018 at 7:48

HINT

By completing the squares and squeeze theorem, note that

$$x^4+y^2+z^2-x+3y-z\ge x^2+y^2+z^2-x+3y-z=$$ $$=\left( x-\frac12\right)^2+\left( y+\frac32\right)^2+\left( z-\frac12\right)^2-\frac{11}4\to +\infty$$

• Thanks for appreciation!
– user
Feb 2, 2018 at 8:10
• so what after ? Feb 2, 2018 at 8:18
• You have a sum of squares and at least ine must tend to infinity.
– user
Feb 2, 2018 at 8:21
• ohhh ok thanx!! Feb 2, 2018 at 8:26

Let $x=tu, \ y=tv, \ z=tw\,$ where $\,t = \sqrt{x^2+y^2+z^2}\,$ and $\,u=x/t, \ v=y/t, \ w=z/t$. Assuming $t>1$, we have that $$x^4+y^2+z^2-x+3y-z=t^2(t^2u^4+v^2+w^2)+t(-u+3v-w)\geq t^2(u^4+v^2+w^2)-t(|u|+3|v|+|w|)\geq t^2(t^2u^4+v^2+w^2)-5t.$$ Now, the function defined on the unit ball $\,\mathbb{S^2}\,$ by $(u,v,w)\mapsto u^4+v^2+w^2$ has an absolute minimum $p$ (by Weierstrass's theorem - $\,\mathbb{S^2}\,$ is closed and bounded). Also, $p>0$ because $u^4+v^2+w^2\geq0$, and if $u^4+v^2+w^2=0=p$ then $u=v=w$, but $(0,0,0)\notin \mathbb{S^2}$. Therefore $$x^4+y^2+z^2-x+3y-z>pt^2-5t$$ which clearly approaches $+\infty$ uniformly for $(u,v,w)\in \mathbb{S^2}$.

Some Notes:

I used the following results:

$t^2>1\implies t^2u^2\geq u^2$ and $t(-u+3v-w)\geq-t|-u+3v-w|\geq-t(|u|+3|v|+|w|)$, as well as $|u|,|v|,|w|\leq 1.$