Find the limit (if it exists): $\lim_{(x,y,x)\rightarrow(0,0,0)}\frac{xyz}{x^2+y^2+z^2}$

Question

Find the limit (if it even exists). If not, prove it doesn't exist. $$\lim_{(x,y,x)\rightarrow(0,0,0)}\frac{xyz}{x^2+y^2+z^2}$$

Answer 1

Let's probe the truth of the following statement: $$\forall\epsilon>0,\exists\delta>0,\forall (x,y,z),\left(0<\sqrt{x^2+y^2+z^2}<\delta\Longrightarrow \big|\frac{xyz}{x^2+y^2+z^2}\big|<\epsilon\right)$$

If we set $u=\max\{|x|,|y|,|z|\}$ then from taking $\sqrt{x^2+y^2+z^2}<\delta$ we have $$|x|<\delta,~~|y|<\delta,~~|z|<\delta$$ and then $u<\delta$ and $$\big|\frac{xyz}{x^2+y^2+z^2}\big|<\frac{u^3}{u^2+0+0}=u<\delta^2$$ I think now you are able to do the rest by yourself. :-)

Answer 2

Hint: Use spherical coordinate substitution. You should notice that the radius goes to $0$, and the function $\sin x$ and $\cos x$ are bounded by $1$.

After substitution we get $$|\frac{xyz}{x^2+y^2+z^2}|=|r\cos^2\phi\sin\phi\cos\theta\sin\theta|\leqslant r\rightarrow0$$

Answer 3

Let $f(x,y,z) = \frac{xyz}{x^2 + y^2 + z^2}$. Approaching the origin on the axes and other obvious paths (such as $x=y=z$) yields zero, so we suspect that the limit is zero. Note first that $\left|\frac{f(x,y,z)}{x}\right| < 1$. This is not hard to show.

Given $\varepsilon > 0$ we wish to find some $\delta > 0$ such that $f(x,y,z) < \varepsilon$ whenever $\lVert (x,y,z) \rVert < \delta$. Suppose the latter. This implies that $|x| < \delta$. Note that $|f(x,y,z)| = |x| \left| \frac{f(x,y,z)}{x} \right|$. Combine this with the above observation to conclude.

Answer 4

If you use the facts

$$ |x|=\sqrt{x^2}\leq \sqrt{x^2+y^2+z^2}, \quad |y|=\sqrt{y^2}\leq \sqrt{x^2+y^2+z^2},\quad |z|=\sqrt{z^2}\leq \sqrt{x^2+y^2+z^2}, $$

then the problem falls apart as

$$ \Bigg|\frac{xyz}{x^2+y^2+z^2}\Bigg|= \frac{|xyz|}{x^2+y^2+z^2}= \frac{|x||y||z|}{x^2+y^2+z^2} $$

$$ \leq \frac{(x^2+y^2+z^2)^{3/2}}{x^2+y^2+z^2}=\sqrt{x^2+y^2+z^2}< \epsilon=\delta. $$