# Does \lim_{(x,y)\to(3,-1)}\frac{(x-3)(y+1)}{(x-3)^2+(y+1)^2} exist?

Does $\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}$ exist?
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Given that $\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}$
The value of the limit $\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}$ do not exist as the denominator becomes 0.
Evaluate $\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}$ as shown below
$\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}=\frac{\left(0-3\right)\left(0+1\right)}{{\left(0-3\right)}^{2}+{\left(0+1\right)}^{2}}$
$=\frac{\left(-3\right)\left(1\right)}{9+1}$
$=-\frac{3}{10}$
Therefore, the limits when x=0 and y=0 $\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}=-\frac{3}{10}$.