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$$\lim_{(x,y,z)->(0,0,0)}{\frac{xyz}{\sqrt{x} + \sqrt{y} + \sqrt{z}}}$$

I tried thinking about a solution using the epsilon-delta definition: I assumed that the limit is 0 but I couldn't figure out a inequality that would lead me to the solution.

oren revenge
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    The function is restricted to non-negative values of $;x,yz;$ and thus that limit, as given, cannot exist... – DonAntonio Dec 17 '16 at 12:08
  • @DonAntonio Come on, it is obvious that we are computing the limit into its natural domain. – Crostul Dec 17 '16 at 12:17
  • @DonAntonio That would depend on the definition, but isn't it common to take the intersection of $(x,y,z)$ with the (maximal, if unspecified) domain of the function? Then there is no problem, as long as $(0,0,0)$ is a limit point of that domain - which it is... The same would apply to the very common $\lim_{x \to 0} \sqrt{x}$ for example. – StackTD Dec 17 '16 at 12:17
  • @Crostul Well, "come on": I'd as something like this to my studets and I'd expect an answer as the above I wrote. This, of course, depends on how formal you've been in your classes: in engineery I wouldn't be so strict, but in mathematics or physics I would. First thing one must knowpay attention to, imo, is what's the domain of definition of things. Interestingly enough, two out of the three answers below remark that "for $;x,y,z>0;$ ..." I think these things matter. – DonAntonio Dec 17 '16 at 12:54

3 Answers3

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If $x,y,z\in (0,1)$ then $$0<xyz<x=\sqrt{x}\sqrt{x}<\sqrt{x}(\sqrt{x}+\sqrt{y}+\sqrt{z}).$$ That is,

$$0<\frac{xyz}{\sqrt{x}+\sqrt{y}+\sqrt{z}}<\sqrt{x}.$$

So, if $x\to 0$ then $\frac{xyz}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\to 0.$

mfl
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Hint. Note that by AM-GM inequality, $$(xyz)^{1/6}\leq \frac{\sqrt{x} + \sqrt{y} + \sqrt{z}}{3}.$$ Therefore for $x> 0$, $y> 0$, $z> 0$, $$0\leq {\frac{xyz}{\sqrt{x} + \sqrt{y} + \sqrt{z}}} \leq \frac{xyz}{3(xyz)^{1/6}}=\frac{(xyz)^{5/6}}{3}.$$

Robert Z
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Using the AM-GM inequality $$0 \le \frac{xyz}{\sqrt{x}+\sqrt{y}+\sqrt{z}} = \frac{\left( \sqrt[3]{\sqrt{x}\sqrt{y}\sqrt{z}} \right)^{3/2}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \le \frac{\left( \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \right)^{3/2}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} = \frac{1}{3\sqrt{3}}\sqrt{\sqrt{x}+\sqrt{y}+\sqrt{z}}$$ so by the squeeze theorem the limit is $0$.

Crostul
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