Multivariable Limits- Non Origin Ones?

I need your help in the following limits:

1) How can I prove the following limit is zero? $$\lim_{(x,y)\to(1,1)} \frac{\tan(y-1)\sin^2(x-y)}{(x-1)^2+(y-1)^2} \ ?$$

2) It seems like the following limit does not exist, but I don't know how to prove it rigorously: $$\lim _{(x,y)\to(1,1)} \frac{18(y-1)^4(x-1)^2 + 5(x-y)^2(y-1)^4 } {28(x-1)^6+y(y-1)^6}$$

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I would recommend starting by changing variables so the limit is at $(0,0)$. So for the first one, let $x' = x+1$, $y' = y+1$, find the limit as $(x',y') \to (0,0)$, and if you like, rename $x'$ to be $x$ and $y'$ to be $y$. This makes the problem a lot less confusing. –  Stefan Smith Apr 16 '13 at 16:22
I have already tried this... it gives me the limit of $\frac{tan(y)sin^2(x-y)} {x^2+y^2 }$ , which I don't know how to solve –  czash Apr 16 '13 at 16:23

For $$\lim_{(x,y)\to(1,1)} \frac{\tan(y-1)\sin^2(x-y)}{(x-1)^2+(y-1)^2},$$ for the sake of familiarity let $u=x-1$ and $v=y-1$. We end up wanting $$\lim_{(u,v)\to(0,0)} \frac{\tan u\sin^2(u-v)}{u^2+v^2} .$$ Now multiply top and bottom by $u(u-v)^2$. This is OK, since if $u=v$ our function is $0$, so we can assume that $u\ne v$. We end up wanting $$\lim_{(u,v)\to(0,0)} \frac{\tan u}{u}\frac{\sin^2(u-v)}{(u-v)^2}\frac{u(u-v)^2}{u^2+v^2} .$$ The two front expressions have limit $1$, by results familiar from one variable calculus. So we want to find $$\lim_{(u,v)\to(0,0)} \frac{u(u-v)^2}{u^2+v^2} .$$ It is not hard to show the above limit is $0$. One easy way to do it is to let $u=r\cos\theta$ and $v=r\sin\theta$.

For $$\lim _{(x,y)\to(1,1)} \frac{18(y-1)^4(x-1)^2 + 5(x-y)^2(y-1)^4 } {28(x-1)^6+y(y-1)^6} ,$$ first do the same substitution as above. So we want $$\lim _{(u,v)\to(0,0)} \frac{18v^4u^2 + 5(u-v)^2u^4 } {28u^6+(v+1)v^6}$$

The limit does not exist. Just choose the paths (i) $u=v$ and (ii) $u=2v$ to approach $(0,9)$.

For $$\lim_{(x,y)\to(0,1)} \frac{1-\cos(2xy)}{x^2y\sin(\pi y )},$$ we will use your suggestion and rewrite the top as $2\sin^2(xy)$ (there are other ways).

Now divide top and bottom by $(xy)^2$. That neutralizes the $\sin^2(xy)$ part, and we can focus on the bad guy, $\sin(\pi y)$.

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There is nothing special about limits at the origin or any other point. You could define $z=x-1, w=y-1$ and your first becomes $\lim_{(1,1)} \frac{\tan(y-1)\sin^2(x-y)}{(x-1)^2+(y-1)^2}=\lim_{(0,0)} \frac{\tan(w)\sin^2(z-w)}{z^2+w^2}$ Then you can use the small angle approximation to get $\lim_{(0,0)} \frac{w(z-w)^2}{z^2+w^2}$ Now if $w,z$ are both small you have three small factors in the numerator and only two in the denominator.

For 2, what happens if you approach $(1,1)$ along $x=y$?

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Thanks!!!!!!!!!! –  czash Apr 16 '13 at 16:58