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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.

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    $\begingroup$ Yes, you are correct. Just use $\left| \sin\left(\frac1{|x|+|y|}\right)\right|\le 1$ . The sine function is defined for all $(x,y)\ne (0,0)$. $\endgroup$ – Mark Viola Sep 26 '17 at 1:13
  • $\begingroup$ What would you say the limit is? $\endgroup$ – K Split X Sep 26 '17 at 1:14
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    $\begingroup$ The limit is $0$. $\endgroup$ – Mark Viola Sep 26 '17 at 1:16
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    $\begingroup$ $\sin (\frac{1}{|x|+|y|})$ 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. $\endgroup$ – Ahmed S. Attaalla Sep 26 '17 at 1:19
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HINT.-$$0\le\left|\tan(x)\sin\left(\dfrac1{|x|+|y|}\right)\right|\le |\tan(x)|$$

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  • $\begingroup$ I did exactly this. its in the comments. Thanks for answer though $\endgroup$ – K Split X Sep 26 '17 at 14:33
  • $\begingroup$ This is incorrect. The left hand side should be $-\tan(x)$ $\endgroup$ – K Split X Sep 26 '17 at 22:02
  • $\begingroup$ When using absolute values this is unnecessary. $\endgroup$ – Piquito Sep 26 '17 at 23:47

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