Show limit of a function

Question

Show that $$\lim_{(x,y)\to (0,0)} (x+y) \sin\frac1x \sin\frac1y=0$$

I obviously can't approach from the positive/negative x or y axis as function would be undefined. I can't think of where to approach from to show the limit.

score 3 · Accepted Answer · answered Mar 23 '15 at 05:37

3

Can you see this

$$ |(x+y)\sin(1/x)\sin(1/y) | \leq |x+y| ? $$

answered Mar 23 '15 at 05:37

science

2,900

does that imply $-(x+y)\le |(x+y)sin(1/x)sin(1/y)|≤(x+y)$ Therefore as (0,0) is approached by pinching theorem limit is 0. – user117498 Mar 23 '15 at 05:47
1

Do not forget The absolute value for $(x+y)$ on both sides. – science Mar 23 '15 at 05:48

Show limit of a function

1 Answers1