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$\lim \limits_{(x,y) \to (2,1)} \frac{\tan(y-1)\sin^2(x-2y)}{(x-2)^2+(y-1)^2}$

I tried this change of variables: $s=x-2, t=y-1$, therefore:

$\lim \limits_{(s,t) \to (0,0)} \frac{\tan(t)\sin^2(s-2t)}{s^2+t^2}$

And I'm pretty much stuck here.

Thanks

Yes
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    In general limit over a plane depends on the path. But here the limit is independent of $x$. To see that just take the limit as $y$ goes to 1 for any $x$. – mhp May 16 '15 at 23:01

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