Answer
By Squeeze Theorem we obtain the limit:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x\sin \left( {\frac{1}{{|x| + |y|}}} \right) = 0$
Work Step by Step
We have $ - 1 \le \sin \left( {\frac{1}{{|x| + |y|}}} \right) \le 1$.
Suppose $x > 0$, so
$ - \tan x \le \tan x\sin \left( {\frac{1}{{|x| + |y|}}} \right) \le \tan x$
Taking limits of the inequalities gives
$\mathop { - \lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x\sin \left( {\frac{1}{{|x| + |y|}}} \right) \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x$
The two limits on both ends are equal to $0$.
Now, if $x<0$, then
$\tan x \le \tan x\sin \left( {\frac{1}{{|x| + |y|}}} \right) \le - \tan x$
Taking limits of the inequalities gives
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x \le \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x\sin \left( {\frac{1}{{|x| + |y|}}} \right) \le \mathop { - \lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x$
In this case too, the two limits on both ends are equal to $0$.
Therefore, by Squeeze Theorem we obtain the limit:
$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \tan x\sin \left( {\frac{1}{{|x| + |y|}}} \right) = 0$
