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

Question

Does

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

exist?

Answer 1

Given that

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

The value of the limit

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

do not exist as the denominator becomes 0.
Evaluate

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

as shown below

lim_{(x, y) \to (3, - 1)} \frac{(x - 3) (y + 1)}{{(x - 3)}^{2} + {(y + 1)}^{2}} = \frac{(0 - 3) (0 + 1)}{{(0 - 3)}^{2} + {(0 + 1)}^{2}}

= \frac{(- 3) (1)}{9 + 1}

= - \frac{3}{10}

Therefore, the limits when x=0 and y=0

lim_{(x, y) \to (3, - 1)} \frac{(x - 3) (y + 1)}{{(x - 3)}^{2} + {(y + 1)}^{2}} = - \frac{3}{10}

.

Expert Answer