Problem 5 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 6 , discuss the continuity of \(f\). \(f(x, y)= \begin{cases}\frac{x y}{|x|+|y|} & \text { if }(x, y) \neq(0,0) \\\ 0 & \text { if }(x, y)=(0,0)\end{cases}\)

Step-by-Step Solution

Answer

The function is continuous at (0,0).

Step 1 - Define Continuity

A function is continuous at a point if the limit as you approach that point equals the function's value at that point. For a function of two variables, we need to check \ \ \( \lim_{(x,y) \to (a,b)} f(x,y) = f(a,b) \).

Step 2 - Identify the Point to Check

The problem requires checking the continuity of \ \( f(x, y) \) at the point \((0,0)\). We are given that \( f(0,0) = 0 \).

Step 3 - Calculate the Limit Along Different Paths

We need to compute \( \lim_{(x,y) \to (0,0)} \frac{xy}{|x|+|y|} \) along different paths to determine if the limit is consistent. \ \ For the path y = 0, \( f(x, 0) = \frac{x \cdot 0}{|x| + |0|} = 0 \), so \( \lim_{{x \to 0}} f(x,0) = 0 \). \ For the path x = 0, \( f(0, y) = \frac{0 \cdot y}{|0| + |y|} = 0 \), so \( \lim_{{y \to 0}} f(0,y) = 0 \).

Step 4 - Check the Limit Along y = x

Choose the path y = x. Then \( f(x,x) = \frac{x \cdot x}{|x| + |x|} = \frac{x^2}{2|x|} \). Since \( |x| = x \) for \( x > 0 \), this simplifies to \( \frac{x^2}{2x} = \frac{x}{2} \). Hence, \( \lim_{{x \to 0}} f(x,x) = 0 \).

Step 5 - Check the Limit Along y = -x

Choose the path y = -x. Then \( f(x, -x) = \frac{x \cdot (-x)}{|x| + |-x|} = \frac{-x^2}{2|x|} \), simplifying to \( \frac{-x^2}{2x} = \frac{-x}{2} \), which also yields \( \lim_{{x \to 0}} f(x,-x) = 0 \).

Step 6 - Conclusion on Continuity

Since the limit \( \lim_{(x,y) \to (0,0)} \frac{xy}{|x|+|y|} = 0 \) is consistent along multiple paths and matches \( f(0,0) = 0 \), \( f \) is continuous at \((0,0)\).

Key Concepts

Limits of Multivariable Functions

When dealing with functions of multiple variables, understanding limits is crucial. The limit of a multivariable function helps us predict the function's behavior as it approaches a specific point. This involves examining the function as the variables approach certain values, often in various directions or paths.
To determine if the limit exists, we check if the function approaches the same value no matter which path is taken. Mathematically, this is expressed as:
\( \ \ \lim_{{(x,y) \to (0,0)}} f(x,y) = L \ \ \ \ \) where L is the value the function is approaching. In our exercise, we look at different paths to see if:
\( \ \ \lim_{{(x,y) \to (0,0)}} \frac{xy}{|x|+|y|} \ \ = \)0 follows:
\

• Path 1: y = 0,
  \ f(x, 0) = \frac{x \cdot 0}{|x| + |0|} = 0 \lim_{{x \to 0}} f(x,0) = 0
  
• Path 2: x = 0:
  \ f(0, y) = \frac{0 \cdot y}{|0| + |y|} =0 \lim_{{y \to 0}} f(0,y) = 0.
  

When considering different paths, such as y = x and y = -x, the limit also approaches 0, suggesting consistency. Thus, we can say the multivariable function's limit exists and equals 0 at the point (0,0).

Continuous Functions

A continuous function is one where there are no breaks, jumps, or holes at any point in its domain. For functions of two variables, continuity at a point \( (a,b) \) means:
\ \ \mathrm{lim}_{{(x,y) \to (a,b)}} f(x,y) = f(a,b) \ \
To check continuity at (0,0), we need to see if:
\

• The value of the function at (0,0), f(0,0) must be defined.
• The limit of the function as (x,y) approaches (0,0) must exist.
  
• The limit must equal the function's value at (0,0).

Using our function, we see:
\ f(0,0) = 0.
From the step-by-step solution, we found that the limit along different paths converges to 0. This matches the function's value at (0,0).
Thus, we verify the function \frac{xy}{|x|+|y|} is continuous at (0,0) because the limit equals the function's value at that point.

Approaching a Point in Multiple Paths

Investigating whether a function is continuous in multivariable calculus involves approaching a specified point from various directions or paths. Doing so helps understand if the limit is the same regardless of the path taken. This significance boils down to the idea that for a function to be continuous at a point, its limit must not depend on the direction from which we approach that point.
Consider the function defined as:

• Along y = x:
  \ f(x, x) = \frac{x \cdot x}{|x| + |x|} =0. \lim_{{x\to 0}} f(x,x)= 0
• Along y = -x:
  \ f(x, -x) = \frac{-x^2}{2|x|}.
• Along y = mx:
  \f(m, xy) =0

Different paths give us comprehensive insight into the function's behavior. If all paths, like y = 0, x = 0, y = x, and y = -x, yield the same limit, the function behaves consistently near that point. This helps confirm continuity in \( R^2 \) for our provided function.