TITLE: Calculate $\lim_{(x,y)\to(0,0)}\tan(x)\sin(\frac1{|x|+|y|})$
QUESTION [8 upvotes]: Calculate $\lim_\limits{(x,y)\to(0,0)}\tan(x)\sin\left(\dfrac1{|x|+|y|}\right)$
We know that
$$-1\leq \sin\left(\dfrac{1}{|x|+|y|}\right)\leq 1$$
$$-\tan(x)\leq \tan(x)\sin\left(\dfrac{1}{|x|+|y|}\right)\leq \tan(x)$$
Taking the limit on both sides gives us $0$, therefore the limit is $0$.
Does this work? I am doubtful of the fact that $\sin\left(\dfrac{1}{|x|+|y|}\right)$ is undefined, so I am not sure.

REPLY [3 votes]: HINT.-$$0\le\left|\tan(x)\sin\left(\dfrac1{|x|+|y|}\right)\right|\le |\tan(x)|$$