Math Problem Statement

PLEASE TEACH ME LIMIT AND CONTINUITY FOR FUNCTIONS WITH SEVERAL VARIABLES

Solution

Certainly! Let's dive into limits and continuity for functions of several variables.

1. Limit of a Function with Several Variables

For functions of several variables, say f(x,y)f(x, y), the limit as (x,y)(x, y) approaches a point (a,b)(a, b) is defined as follows:

lim(x,y)(a,b)f(x,y)=L\lim_{{(x, y) \to (a, b)}} f(x, y) = L

This means that as (x,y)(x, y) gets closer and closer to (a,b)(a, b), the function f(x,y)f(x, y) gets closer and closer to a single value LL.

Key Points:

  • Existence: For the limit to exist, the value of f(x,y)f(x, y) as (x,y)(x, y) approaches (a,b)(a, b) must be the same regardless of the path taken.
  • Path Dependence: If the limit depends on the path taken to approach (a,b)(a, b), then the limit does not exist.

Example:

Consider the function f(x,y)=x2y2x2+y2f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}. Let's find the limit as (x,y)(x, y) approaches (0,0)(0, 0).

  • Along the path y=0y = 0: f(x,0)=x202x2+02=1f(x, 0) = \frac{x^2 - 0^2}{x^2 + 0^2} = 1
  • Along the path x=0x = 0: f(0,y)=02y202+y2=1f(0, y) = \frac{0^2 - y^2}{0^2 + y^2} = -1

Since the limit depends on the path taken, the limit lim(x,y)(0,0)x2y2x2+y2\lim_{{(x, y) \to (0, 0)}} \frac{x^2 - y^2}{x^2 + y^2} does not exist.

2. Continuity of a Function with Several Variables

A function f(x,y)f(x, y) is continuous at a point (a,b)(a, b) if:

lim(x,y)(a,b)f(x,y)=f(a,b)\lim_{{(x, y) \to (a, b)}} f(x, y) = f(a, b)

This means that the function’s value at (a,b)(a, b) matches the value approached by f(x,y)f(x, y) as (x,y)(x, y) gets close to (a,b)(a, b).

Example:

Consider the function f(x,y)=xyx2+y2f(x, y) = \frac{xy}{x^2 + y^2} and let's analyze its continuity at (0,0)(0, 0).

  1. Limit as (x,y)(0,0)(x, y) \to (0, 0):

    • Along the path y=xy = x: f(x,x)=xxx2+x2=x22x2=12f(x, x) = \frac{x \cdot x}{x^2 + x^2} = \frac{x^2}{2x^2} = \frac{1}{2}
    • Along the path y=0y = 0: f(x,0)=x0x2+02=0f(x, 0) = \frac{x \cdot 0}{x^2 + 0^2} = 0

    The limit depends on the path taken, so the limit does not exist as (x,y)(0,0)(x, y) \to (0, 0).

  2. Since the limit does not exist, the function is not continuous at (0,0)(0, 0).

3. Summary:

  • Limit: To find if a limit exists, check if the function approaches the same value from all possible paths.
  • Continuity: A function is continuous at a point if the limit at that point equals the function’s value there.

Relative Questions:

  1. How do you determine the limit of a multivariable function when the limit depends on the path taken?
  2. What happens if a function of two variables has different limits along different paths towards a point?
  3. Can a function be continuous if its limit doesn’t exist at a certain point? Why or why not?
  4. How would you approach finding limits for functions with three variables?
  5. What are some common strategies to prove that a limit does not exist?

Tip: Always test multiple paths when checking the existence of a limit in multivariable functions, especially the coordinate axes and curves like y=mxy = mx.

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Math Problem Analysis

Mathematical Concepts

Limits
Continuity
Functions of Several Variables

Formulas

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Theorems

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Suitable Grade Level

College Level