# Limit of a function $\mathbb{R}^3 \to \mathbb{R}$

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.

• 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 $t$. – The Phenotype Dec 20 '17 at 13:25
• I think your plan will work fine. – ziggurism Dec 20 '17 at 13:25
• I suppose the domain was meant to be $(\mathbb R\setminus\{0\})^3$ rather than $\mathbb R^3$? – Henning Makholm Dec 20 '17 at 13:26

Set $u=xyz\to0$
$$\frac{1}{u}\tan\left(\frac{u}{1+u}\right)=\frac{\tan\left(\frac{u}{1+u}\right)}{\frac{u}{1+u}}\frac{1}{1+u}\to1\cdot1=1$$
• if instead of $xyz$ it would have been $x^2+y^2+z^2$ could I have still just equaled it to $t$? – Alexandra Dec 20 '17 at 18:41
• @Alexandra Yes you can for any $f(x,y,z)\to 0$ – 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 $$\frac 1{xyz}\tan\Bigl(\frac{xyz}{1+xyz}\Bigr)=\frac1u\tan\Bigl(\frac{u}{1+u}\Bigr)\sim_0\frac1u\, \frac u{1+u}=\frac 1{1+u}\to 1.$$