# How can I prove using the definition of limits that $\lim _{(x,y)\rightarrow (2,3)} y\sin(xy-6)= 0$?

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?

• HINT: What is making the limit come out $0$? Apr 17 '17 at 23:43
• It will simplify the analysis a good bit if you appeal to a result that the limit of the product $y\sin (xy-6)$ is the product of the limits of $y$ and $\sin (xy-6)$. Apr 18 '17 at 0:03
• the fact that xy-6 -> 0 when (x,y)->(2,3) but I still cant see how this could help in the proof of the limit using the definition of limit, sorry. Apr 18 '17 at 0:03
• yeah i could use the product of two limits, but i thought it wouldnt be necesary, i will try it that way then if there is not another path, haha Apr 18 '17 at 0:05
• I ended up using that $sin x \leq x$ then i got that $|ysin (xy - 6)| \leq |y||(xy - 6)| = |y-3 + 3||(x-2)(y-3 +3) + 2(y-3)|$ and could show that given a $\delta$ we can arrive to $|ysin (xy - 6)|< \epsilon$ and that would prove the limit by definition. I used that sin x <= x and i thought one could arrive to the same by some algebraic manipulations and that would be fantastic because i suppose dont have the level to show that sin x <= x because i have only reached to limits in multivariables. However I found the proof on the internet. Is there a proof without knowing that sin x<= x? Apr 18 '17 at 0:38

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.

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 $|y|$