Limit of the function $$\lim\limits_{x,y,z\to 0,0,0} \frac{1}{xyz}\tan\bigg(\frac{xyz}{1+xyz}\bigg)$$

If this was a one-dimensional function, this would look like an oportunity to apply the limit $\frac{\sin(x)}{x} = 1$. Is there a way to substitute $(x,y,z)$ with $t$ or something and apply the limit? I would be interested in specifically this transformation and not another way to compute the limit.

  • $\begingroup$ Yes, there is. If you take $t(x,y,z)=xyz$, then $\lim_{(x,y,z)\to (0,0,0)}t(x,y,z)=\lim_{(x,y,z)\to (0,0,0)}xyz=0$, so $(x,y,z)\to (0,0,0)$ implies $t\to 0$, so you can change $\lim_{(x,y,z)\to (0,0,0)}$ to $\lim_{t\to 0}$ and replace $xyz$ by $t$. $\endgroup$ – The Phenotype Dec 20 '17 at 13:25
  • $\begingroup$ I think your plan will work fine. $\endgroup$ – ziggurism Dec 20 '17 at 13:25
  • $\begingroup$ I suppose the domain was meant to be $(\mathbb R\setminus\{0\})^3$ rather than $\mathbb R^3$? $\endgroup$ – Henning Makholm Dec 20 '17 at 13:26

Set $u=xyz\to0$


  • $\begingroup$ if instead of $xyz$ it would have been $x^2+y^2+z^2$ could I have still just equaled it to $t$? $\endgroup$ – Alexandra Dec 20 '17 at 18:41
  • $\begingroup$ @Alexandra Yes you can for any $f(x,y,z)\to 0$ $\endgroup$ – gimusi Dec 20 '17 at 18:43

Yes, you just have to set $u=xyz$ and use the fact that $\tan u\sim_0 u$, so $$\frac 1{xyz}\tan\Bigl(\frac{xyz}{1+xyz}\Bigr)=\frac1u\tan\Bigl(\frac{u}{1+u}\Bigr)\sim_0\frac1u\, \frac u{1+u}=\frac 1{1+u}\to 1.$$


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