Help with Multi-variable Limits

[Euler2718]

Homework Statement

Evaluate or show that the limit does not exist:

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

Homework Equations

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?

 

[Samy_A]

Homework Statement

Evaluate or show that the limit does not exist:

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

Homework Equations

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?

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