How can I prove using the definition of limits that $\lim _{(x,y)\rightarrow (2,3)} y\sin(xy-6)= 0$?
By Sophia Vance
I want to prove that $\lim_{(x,y)\rightarrow (2,3)} y\sin(xy-6)= 0$ using the definition of limit. However I am unable to make the step of transforming $| y\sin(xy-6)|$ into some inequality that could let me use the fact that $|x-2|, |y-3|\leq \left \| (x-2,y-3)) \right \|< \delta $. I have tried arguing that $| y\sin(xy-6)| \leq |y|\leq |y-3| + 3$. That didn't work so I also tried using some simple trigonometric identities so I could separate some terms inside $sin(xy-6)$, even saying that $\sin(xy-6)= \sin((x-2)y + 2(y-3))$, but couldn't reach anything. I see the problem is the $|y|$ next to the sin function and I should transform it into a $|y-3|$ but i don't see how, may be I should do something with the sin?
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$\begingroup$It suffices to show that $sin(xy-6)$ goes to zero.
Indeed $sin(xy-6)=sin(xy)cos(6)-cos(xy)sin(6)\hspace{10mm}(1)$
So the problem is reduced to showing that
$\lim_{(x,y)\to (2,3)}sin(xy)=sin(6)$
and
$\lim_{(x,y)\to (2,3)}cos(xy)=cos(6)$
I'll show the first one.
Indeed, by mean value theorem in one variable:
$|sin(xy)-sin(6)|=cos(c)|xy-6|$ (for some $c\in \mathbb{R}$that may depend on x and y)
But $cos(c)|xy-6|\leq |xy-6|\to 0$ as $(x,y)\to (2,3)$
So that $\lim_{(x,y)\to (2,3)}sin(xy)=sin(6)$
By an essentially identical arugment:
$\lim_{(x,y)\to (2,3)}cos(xy)=cos(6)$
This proves our claim, in light of $(1)$
As a remark, this follows immediately from continuity of composition of continuous functions. But as you said, you wanted a proof by definition. The above is what I could come up with. Hope this helps.
$\endgroup$ $\begingroup$How about adding and substracting $2y$? You would be left with something like this:
$|y \cdot \sin(xy - 6)| \leq |y \cdot (xy - 6)| \leq $
$|y \cdot (xy - 6 + 2y - 2y)| \leq |y \cdot (y (x - 2) + 2 (y - 3))|$
You can then ask $\delta < 1$ in order to bound $|y|$
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