# Help with Multi-variable Limits

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1. Mar 20, 2016

### Euler2718

1. The problem statement, all variables and given/known data

Evaluate or show that the limit does not exist:

$$\lim_{(x,y) \to (0,0)}\frac{ 2x^{4} + 5y^{3} }{8x^{2}-9y^{3}}$$
$$\lim_{(x,y) \to (0,-2)}\frac{ xy+2x }{3x^{2}+(y+2)^{2}}$$

2. Relevant equations

3. The attempt at a solution

So the first one is indeterminate and cannot be factored to make a direct substitution. So I try to show then that the limit does not exist by taking the limit along y=0 and x=0

$$y=0 \implies \lim_{(x,0) \to (0,0)} \frac{2x^{4}+5(0)^{3} }{8x^{2} -9(0)^{3}} \implies \lim_{(x,0)\to (0,0)} \frac{2x^4}{8x^{2}} = 0$$

$$x=0 \implies \lim_{(0,y) \to (0,0)} \frac{2(0)^{4}+5y^{3} }{8(0)^{2} -9y^{3}} \implies \lim_{(0,y)\to (0,0)} \frac{5y^3}{-9y^{3}} = -\frac{5}{9}$$

So the limit doesn't exist, since as x and y approach zero from two different paths they don't equal the same value? Wolfram alpha however tells me the limit is zero (assuming x and y are both real). What have I missed?

The second problem results in a indeterminate form. Could this be the correct method to show it?

First approach along x-2:

$$\lim_{(x,x-2) \to (0,-2)}\frac{ x(x-2)+2x }{3x^{2} +((x-2)+2)^{2}} = \frac{1}{4}$$

Than along x^2-2:

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

So since both paths do not equal the limit does not exist? Wolfram confirms this but is the reasoning sound?

2. Mar 20, 2016

### Samy_A

If you find different limits along different paths, then the limit doesn't exist. That is indeed sound reasoning.