$$ \lim_{(x,y,z)\to (0,0,0) } \frac{\sin(x^2+y^2+z^2) + \tan(x+y+z) }{|x|+|y|+|z|} $$
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 $C=1$.
Another way to proceed is to transform variables to spherical coordinates.
To that end, let $x=r \sin \theta \cos \phi$, $y=r\sin \theta \sin \phi$, and $z=r\cos \theta$, where $0\le \theta \le \pi$ and $0\le \phi \le 2\pi$. Then, the limit that is in question is
$$\lim_{r\to 0}\left(\frac{\sin r^2}{r(|\sin \theta \cos \phi|+|\sin \theta \sin \phi|+|\cos \theta|)}\right)=0$$
To see this, note that
$$\lim_{r\to 0} \frac{\sin r^2}{r} =0,$$
which can be shown using L'Hospital's Rule. It is important to note that for all $0\le \theta \le \pi$ and $0\le \phi \le 2\pi$, $1\le|\sin \theta \cos \phi|+|\sin \theta \sin \phi|+|\cos \theta| \le \sqrt{3}$ is bounded.
Now, for the entire problem, we need to examine the limit
$$\lim_{r\to 0} \left(\frac{\sin r^2+\tan (r(\sin \theta \cos \phi +\sin \theta \sin \phi + \cos \theta))}{r(|\sin \theta \cos \phi|+|\sin \theta \sin \phi|+|\cos \theta|)}\right)$$
Clearly, the limit as $r \to 0$ does not exist inasmuch as its value depends on the values of $\theta$ and $\phi$. For example, if $\theta =0$, then the limit is $1$ since $\sin \theta \cos \phi +\sin \theta \sin \phi + \cos \theta = 1$, $|\sin \theta \cos \phi|+|\sin \theta \cos \phi|+|\cos \theta| =1$ and
$$\lim_{r\to 0} \frac{\sin r^2}{r} =0$$
$$\lim_{r\to 0} \frac{\tan r}{r} = 1$$
which can be easily shown by using L'Hospital's Rule. This first limit is, in fact, the limit that was in question.
If $\theta = \pi$, then the limit is $-1$ since $\sin \theta \cos \phi +\sin \theta \sin \phi + \cos \theta = -1$, $|\sin \theta \cos \phi|+|\sin \theta \cos \phi|+|\cos \theta| =1$ and
$$\lim_{r\to 0} \frac{\sin r^2}{r} =0$$
$$\lim_{r\to 0} \frac{\tan (-r)}{r} = -1$$
Inasmuch as the limit is non-unique, it does not exist.
