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

Carol Gates
2021-10-25
Answered

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

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jlo2niT

Answered 2021-10-26
Author has **96** answers

Given that $\underset{(x,y)\to (3,-1)}{lim}\frac{(x-3)(y+1)}{{(x-3)}^{2}+{(y+1)}^{2}}$

The value of the limit$\underset{(x,y)\to (3,-1)}{lim}\frac{(x-3)(y+1)}{{(x-3)}^{2}+{(y+1)}^{2}}$ do not exist as the denominator becomes 0.

Evaluate$\underset{(x,y)\to (3,-1)}{lim}\frac{(x-3)(y+1)}{{(x-3)}^{2}+{(y+1)}^{2}}$ as shown below

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

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

$=-\frac{3}{10}$

Therefore, the limits when x=0 and y=0$\underset{(x,y)\to (3,-1)}{lim}\frac{(x-3)(y+1)}{{(x-3)}^{2}+{(y+1)}^{2}}=-\frac{3}{10}$ .

The value of the limit

Evaluate

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

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