How can I evaluate the following limit? $$\lim_{|(x,y,z)|\to+\infty}(x^4+y^2+z^2-x+3y-z).$$
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$\begingroup$ Is it my misunderstanding, that the limit is (quite obviously) undefined? $\endgroup$– ElfHogFeb 2, 2018 at 7:39
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2$\begingroup$ @ElfHog: It is possible to understand that the limit exists in the sense of the extended real line. $\endgroup$– hardmathFeb 2, 2018 at 7:44
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$\begingroup$ @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? $\endgroup$– ElfHogFeb 2, 2018 at 7:46
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2$\begingroup$ @ElfHog you cannot jump to such conclusions so quickly, see my answer. $\endgroup$– Denis28Feb 2, 2018 at 7:47
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7$\begingroup$ @ElfHog: Look again. $\endgroup$– hardmathFeb 2, 2018 at 7:48
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2 Answers
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HINT
By completing the squares and squeeze theorem, note that
$$x^4+y^2+z^2-x+3y-z\ge x^2+y^2+z^2-x+3y-z=$$ $$=\left( x-\frac12\right)^2+\left( y+\frac32\right)^2+\left( z-\frac12\right)^2-\frac{11}4\to +\infty$$
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$\begingroup$ You have a sum of squares and at least ine must tend to infinity. $\endgroup$– userFeb 2, 2018 at 8:21
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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 $|u|,|v|,|w|\leq 1.$