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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?

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    $\begingroup$ 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. $\endgroup$ – RiRi Aug 2 at 13:57
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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 $0$.

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  • $\begingroup$ thank you so much!!! $\endgroup$ – RiRi Aug 2 at 13:58
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$|f(x,y) ≤ |x| + |y|$ 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.

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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 $f\le g\le h$.

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$x,y \not =0;$

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

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| $.

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