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

Question

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.

Answer 1

Set $u=xyz\to0$

$$\frac{1}{u}\tan\left(\frac{u}{1+u}\right)=\frac{\tan\left(\frac{u}{1+u}\right)}{\frac{u}{1+u}}\frac{1}{1+u}\to1\cdot1=1$$

Answer 2

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