Help with Multi-variable Limits

• Euler2718
In summary, the first problem results in an indeterminate form which cannot be solved by factoring, so the limit is evaluated along two different paths (y=0 and x=0), resulting in different limits and showing that the limit does not exist. The second problem also results in an indeterminate form, and by evaluating the limit along two different paths (x-2 and x^2-2), we get different limits and can conclude that the limit does not exist.
Euler2718

Homework Statement

Evaluate or show that the limit does not exist:

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

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?

Morgan Chafe said:

Homework Statement

Evaluate or show that the limit does not exist:

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

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.

Euler2718

1. What is a multi-variable limit?

A multi-variable limit is a mathematical concept that describes the behavior of a function as it approaches a specific point in a multi-dimensional space. It is used to determine the value of a function at a particular point by analyzing its behavior near that point.

2. How is a multi-variable limit calculated?

A multi-variable limit is calculated by approaching the point in question from various directions, such as along different lines or paths. The limit is then determined by evaluating the function at these different points and seeing if they approach a common value.

3. What is the significance of multi-variable limits?

Multi-variable limits are important in many areas of science and engineering, as they help us understand the behavior of functions in complex systems. They are used in fields such as physics, economics, and biology to model and predict the behavior of systems with multiple variables.

4. What are some common techniques for solving multi-variable limits?

There are several techniques for solving multi-variable limits, including substitution, algebraic manipulation, and using trigonometric identities. In some cases, it may also be helpful to graph the function or use a calculator to evaluate the limit numerically.

5. Can multi-variable limits have different values depending on the approach?

Yes, multi-variable limits can have different values depending on the approach. This is because the behavior of a function may differ along different paths or directions. It is important to consider all possible approaches when evaluating a multi-variable limit.

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