# Show that $\lim\limits_{(x,y)\to(0,0)}[x\sin(1/y) + y\sin(1/x)]$ exist

I don't really know how to solve this, and I have seen a duplicate of this question else where which is this: Does $\lim_{(x,y)\to(0,0)}[x\sin (1/y)+y\sin (1/x)]$ exist?

There is only 1 answer and it doesn't really clear my doubt. Basically, what I know about squeeze theorem is this: If we have 3 functions let's say $$f(x)$$, $$g(x)$$ and $$h(x)$$, and the limit of each function is laid out like this:

$$\lim_{(x,y)\to(0,0)} f(x,y)\le\lim_{(x,y)\to(0,0)} g(x,y)\le\lim_{(x,y)\to(0,0)}h(x,y)$$

But in his answer, he said $$|f(x,y)|\le |x| +|y|$$ This doesn't make sense how can squeeze theorem apply here?

• I don't think there was any need of marking this as duplicate, I have already mentioned it clearly in my question about the reason why i asked this question again. – RiRi Aug 2 at 13:57

The squeeze theorem applies here because $$0 \le |f(x,y)| \le |x| + |y|$$ (i.e., the missing lower bound function is just $$0$$), so as $$(x,y) \to (0,0)$$, the lower bound is already $$0$$ and the upper bound goes to $$0$$.

• thank you so much!!! – RiRi Aug 2 at 13:58

$$|f(x,y) ≤ |x| + |y|$$ because $$|\sin(a)| ≤ 1$$ for all $$a \in R$$. The case for the reciprocal also holds: $$|\sin(\frac{1}{b})| ≤ 1$$ except for $$b=0$$ which doesn't apply here.

Take $$g(x,\,y):=x\sin\tfrac{1}{y}+y\sin\tfrac{1}{x},\,h(x,\,y):=|x|+|y|,\,f(x,\,y):=-h(x,\,y)$$ so $$|g|\le|x|\left|\sin\tfrac{1}{y}\right|+|y|\left|\sin\tfrac{1}{x}\right|\le h,$$where the first $$\le$$ follows from the triangle inequality while the second follows from $$|\sin t|\le1$$. Hence $$f\le g\le h$$.

$$x,y \not =0;$$

$$|x\sin (1/y) +y \sin (1/x)| \le |x| +|y| \lt 2\sqrt{x^2+y^2}$$.

Choose $$\delta =\epsilon/2$$

Then

$$|x^2+y^2| \lt \delta$$ implies

$$|x \sin (1/y)+y \sin (1/x)| \lt$$

$$2\sqrt{ x^2+y^2} \lt 2\delta = \epsilon.$$

Note: $$|x| =\sqrt{x^2} \lt \sqrt{x^2+y^2}$$,

similarly for $$|y|$$.