# Calculate $\lim_{(x,y)\to(0,0)}\tan(x)\sin(\frac1{|x|+|y|})$

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.

• Yes, you are correct. Just use $\left| \sin\left(\frac1{|x|+|y|}\right)\right|\le 1$ . The sine function is defined for all $(x,y)\ne (0,0)$. – Mark Viola Sep 26 '17 at 1:13
• What would you say the limit is? – K Split X Sep 26 '17 at 1:14
• The limit is $0$. – Mark Viola Sep 26 '17 at 1:16
• $\sin (\frac{1}{|x|+|y|})$ is defined, except at the origin where it is not. When we talk about $\lim_{(x,y) \to (0,0)} f(x,y)$ we do not care whether or not the function is defined at the origin. All we care about is what happens near the origin. – Ahmed S. Attaalla Sep 26 '17 at 1:19

HINT.-$$0\le\left|\tan(x)\sin\left(\dfrac1{|x|+|y|}\right)\right|\le |\tan(x)|$$
• This is incorrect. The left hand side should be $-\tan(x)$ – K Split X Sep 26 '17 at 22:02