How to calculate this multivariable limit? $$
\lim_{(x,y,z)\to (0,0,0) } \frac{\sin(x^2+y^2+z^2) + \tan(x+y+z) }{|x|+|y|+|z|}
$$
I know the entire limit should not exist. In addition, the limit:
$$
\lim_{(x,y,z)\to (0,0,0) } \frac{\tan(x+y+z) }{|x|+|y|+|z|}
$$
does not exist and it seems like the limit:
$$
\lim_{(x,y,z)\to (0,0,0) } \frac{\sin(x^2+y^2+z^2) }{|x|+|y|+|z|}
$$
is zero. But, how can I calculate this limit (the last one) ? 
Will you please help me?
Thanks a lot in advance

Write $\frac{\sin(x^2+y^2+z^2)}{|x|+|y|+|z|}=\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}\sqrt{x^2+y^2+z^2}$. The first factor tends to $1$, the second is bounded and the last tends to zero.

To answer your question in your comment:
$$x^2+y^2+z^2\leq x^2+y^2+z^2+2|x||y|+2|x||z|+2|y||z|=(|x|+|y|+|z|)^2$$
Therefore $$\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}\leq1$$

A good-to-know fact is that the norms in $\mathbb{R}^n$ are all equivalent. $\|(x,y,z)\|_2=\sqrt{x^2+y^2+z^2}$ and $\|(x,y,z)\|_1=|x|+|y|+|z|$ are to norms of $\mathbb{R}^3$ and equivalence means that there are non-zero constants $c,C$ such that
$$c\|\cdot\|_1\leq \|\cdot\|_2\leq C\|\cdot\|_1$$
In our case $C=1$.

