I have a strong feeling that the following limit is zero, can anybody help me prove it.

  1. $ \lim\limits_{(x,y,z) \to (0,0,0)} \frac{{xyz}}{{x+y+z}}$


  • $\begingroup$ Hint: For any $k$, $xyz=k(x+y+z)$ is a surface on which you can take a path to approach the limit. $\endgroup$ – Macavity Sep 16 '14 at 1:23

Hint: what happens when, say, $y = x$ and $z = -2x$?

  • 2
    $\begingroup$ But there is at least one convention for evaluating limits of functions of several variables in which you restrict the neighborhood of the limit point to be an open set on which the function is defined. $\endgroup$ – Barry Cipra Sep 16 '14 at 0:27
  • $\begingroup$ OK, $y=x$ and $z$ very close to $-2x$. $\endgroup$ – Robert Israel Sep 16 '14 at 1:30
  • $\begingroup$ Robert, I had a question for you over here if you have a moment: math.stackexchange.com/questions/920896/… $\endgroup$ – Eric Auld Sep 16 '14 at 1:45
  • $\begingroup$ Moved it to its own question: math.stackexchange.com/questions/933124/… $\endgroup$ – Eric Auld Sep 16 '14 at 2:12

Write $z=u-(x+y)$ so that

$${xyz\over x+y+z}={xy(u-(x+y))\over u}=xy-{xy(x+y)\over u}$$

Now let $x=au^{1/3}$ and $y=bu^{1/3}$, giving

$${xyz\over x+y+z}=abu^{2/3}-ab(a+b)$$

Note that $(x,y,z)\to0$ as $u\to0$ regardless of what $a$ and $b$ are, but this gives $${xyz\over x+y+z}\to -ab(a+b)$$ which can be anything. So the limit is undefined.

  • $\begingroup$ I'm not sure I understand; isn't $a=\frac{x}{(x+y+z)^{1/3}}$ and $b=\frac{y}{(x+y+z)^{1/3}}$, and don't you mean $u\rightarrow 0$ as $(x,y,z)\rightarrow0$? $\endgroup$ – user84413 Sep 16 '14 at 1:04
  • $\begingroup$ @user84413, things might be more clear if we write $u=t^3$. Then we have $(x,y,z)=(at,bt,t^3-(a+b)t)$, which gives a parameterized path that goes to $(0,0,0)$ as $t$ goes to $0$, along which the function $xyz/(x+y+z)$ tends to $-ab(a+b)$. So different paths give different limits, which means the function has no well defined limit. $\endgroup$ – Barry Cipra Sep 16 '14 at 2:03
  • $\begingroup$ Thanks for your reply, but I guess I still don't understand completely. I can see that you have $\frac{xyz}{x+y+z}=abu^{\frac{2}{3}}-ab(a+b)=xy-\frac{xy(x+y)}{x+y+z}$, and that $xy\rightarrow0$, but it's not any clearer to me that $\lim_{(x,y,z)\to0}\frac{xy(x+y)}{x+y+z}$ fails to exist than the original limit. [I would understand this argument if a and b were constants, so forgive me if I'm just being dense.] $\endgroup$ – user84413 Sep 16 '14 at 17:21
  • $\begingroup$ @user84413, $a$ and $b$ are constants! The idea is, different choices for the constants give different limits when $(x,y,z)$ approaches $(0,0,0)$ along the chosen path. But a multivariate function is said to have a limit if and only if it has the same limit along all paths. $\endgroup$ – Barry Cipra Sep 16 '14 at 17:36
  • $\begingroup$ Thanks! - it took me quite a while, but I understand now. (I didn't understand why a and b were constants.) $\endgroup$ – user84413 Sep 16 '14 at 18:16

1) If $(x,y,z)\rightarrow(0,0,0)$ along the x-axis, we get $\displaystyle\lim_{x\to 0}\frac{0}{x}=0$.

2) If $(x,y,z)\rightarrow(0,0,0)$ along the path $x=\frac{t}{1+2t^2}, y=t, z=-2t$,

$\;\;\;\;$we get $\displaystyle\lim_{t\to0}\frac{\frac{-2t^3}{1+2t^2}}{\frac{t}{1+2t^2}-t}=\lim_{t\to0}\frac{-2t^3}{-2t^3}=\lim_{t\to0}1=1.$

Therefore the limit does not exist.


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