Question

Use the Squeeze Theorem to find the limit

Answer 1

\[\lim_{(x,y) \rightarrow (0,0)} tanxsin(\frac{ 1 }{ \left| x \right| +\left| y \right|})\]

Answer 2

So I know I am supposed to use the fact that sin goes from -1 to 1 but I don't know how to deal with the fact that tan can be negative and positive.

Answer 3

have you tried using that sine ranges from -1 to 1 (inclusive)?

Answer 4

should I use absolute value of tan

Answer 5

so you find the left and right limit you will see they are the same

Answer 6

\[-1 \le \sin(\frac{1}{|x|+|y|}) \le 1 \\ \text{ assume } \tan(x)>0 \\ \text{ then } -\tan(x) \le \tan(x)\sin(\frac{1}{|x|+|y|} ) \le \tan(x) \\ \text{ and if } \tan(x)<0 \text{ then } \\ -\tan(x) \ge \tan(x) \sin(\frac{1}{|x|+|y|}) \ge \tan(x) \\ \text{ we don't care about } \tan(x)=0 \\ \text{ \because that happens where } x=0\]

Answer 7

use the first inequality to evaluate:
\[\lim_{x \rightarrow 0+}\tan(x) \sin(\frac{1}{|x|+|y|}) \\ \]
use the second inequality to evaluate:
\[\lim_{x \rightarrow 0-}\tan(x) \sin(\frac{1}{|x|+|y|}) \\ \]

Answer 8

By first inequality do you mean the case where tanx>0 or do you mean I should use the left side as one and the right side as another one?

Answer 9

for values directly to the right of x=0 we have tan(x)>0 so yes

Answer 10

you could also use

\[-|\tan(x)| \le \tan(x)\sin\left(\frac{1}{|x|+|y|} \right) \le |\tan(x)|\]

Answer 11

So is the answer 0? And I don't need to account for the y direction?

Answer 12

yes the answer is zero.

the value of y does not matter (except when (x,y)=(0,0)...and we don't have to worry about that since (0,0) is the limiting value)