Problem 9 - The Calculus with Analytic Geometry

Question

In Exercises 7 through 12, prove that for the given function \(f, \lim _{(x, y) \rightarrow(0,0)} f(x, y)\) does not exist. \(f(x, y)=\frac{x^{3}+y^{3}}{x^{2}+y}\)

Step-by-Step Solution

Answer

The limit exists and equals 0.

Step 1: Analyze the limit definition

To determine if the limit \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) exists, evaluate the function along different paths approaching the point \((0,0)\). If the limit value differs for different paths, conclude that the limit does not exist.

Step 2: Path along the x-axis (y = 0)

Substitute \(y = 0\) into the function: \[\lim _{x \rightarrow 0} f(x, 0) = \lim _{x \rightarrow 0} \, \frac{x^3 + 0^3}{x^2 + 0} = \lim _{x \rightarrow 0} \, \frac{x^3}{x^2} = \lim _{x \rightarrow 0} x = 0 \] Therefore, along the x-axis, the limit is 0.

Step 3: Path along the y-axis (x = 0)

Substitute \(x = 0\) into the function: \[\lim _{y \rightarrow 0} f(0, y) = \lim _{y \rightarrow 0} \, \frac{0^3 + y^3}{0^2 + y} = \lim _{y \rightarrow 0} \, \frac{y^3}{y} = \lim _{y \rightarrow 0} y^2 = 0 \] Therefore, along the y-axis, the limit is 0.

Step 4: Path along y = x

Substitute \(y = x\) into the function: \[\lim _{x \rightarrow 0} f(x, x) = \lim _{x \rightarrow 0} \, \frac{x^3 + x^3}{x^2 + x} = \lim _{x \rightarrow 0} \, \frac{2x^3}{x^2 + x} = \lim _{x \rightarrow 0} \, \frac{2x^3}{x(x + 1)} = \lim _{x \rightarrow 0} \, \frac{2x^2}{x + 1} = 0 \] Therefore, along the line \(y = x\), the limit is also 0.

Step 5: Path along y = -x

Substitute \(y = -x\) into the function: \[\lim _{x \rightarrow 0} f(x, -x) = \lim _{x \rightarrow 0} \, \frac{x^3 + (-x)^3}{x^2 + (-x)} = \lim _{x \rightarrow 0} \, \frac{x^3 - x^3}{x^2 - x} = \lim _{x \rightarrow 0} 0 = 0 \] Therefore, along the line \(y = -x\), the limit is 0.

Step 6: Consider another path (y = x^2)

Substitute \(y = x^2\) into the function: \[\lim _{x \rightarrow 0} f(x, x^2) = \lim _{x \rightarrow 0} \, \frac{x^3 + (x^2)^3}{x^2 + x^2} = \lim _{x \rightarrow 0} \, \frac{x^3 + x^6}{2x^2} = \lim _{x \rightarrow 0} \, \frac{x(x^2 + x^5)}{2} = 0 \] Therefore, along the parabola \(y = x^2\), the limit is 0.

Step 7: Conclusion

Since \(\forall\) paths tested (x-axis, y-axis, y = x, y = -x, y = x^2) approach the limit 0 for \(f(x, y)\) as \((x,y) \rightarrow (0,0)\), the limit exists and equals 0.

Key Concepts

limit of multivariable functions

In multivariable calculus, understanding the concept of limits is crucial. A limit describes the behavior of a function as its input points approach a certain value. Formally, the limit of the function \(f(x,y)\) as \((x,y)\) approaches \((a,b)\) is written as \[\text{lim}_{(x,y) \rightarrow (a,b)} f(x,y) \]. If the functions tend to a single value regardless of the path taken by \((x,y)\), the limit exists.

For instance, consider the function $$f(x, y)=\frac{x^3+y^3}{x^2+y}$$ and the limit as \((x,y) \rightarrow (0,0)\). Evaluating along different paths helps determine if the limit exists by checking for consistency in the approached value.

path analysis in limits

Path analysis involves checking various paths to see if the function approaches the same limit value. This helps confirm or deny the existence of a limit in multivariable functions. If the limit varies with different paths, it indicates that no single limit exists.

For our given function, \(f(x, y)=\frac{x^3+y^3}{x^2+y}\), checking along different paths like the x-axis (\(y = 0\)), y-axis (\(x = 0\)), and parabolas (\(y = x^2\)) shows that the limit as \((x, y)\rightarrow (0,0)\) always equals \(0\). Thus, path analysis here confirms that our function has a consistent limit of \(0\), meaning the limit exists.

continuity in multivariable functions

For multivariable functions, continuity means the function's value at a point is the same as the limit as the inputs approach that point. Formally, \(f(x,y)\) is continuous at \((a,b)\) if \[\lim_{(x,y) \rightarrow (a,b)} f(x,y) = f(a,b) \].

This indicates there's no sudden jump in value. It's smooth. In our example, if \(f(x, y) = \frac{x^3+y^3}{x^2+y}\) is continuous at \((0,0)\), then \(f(0,0)\) must equate to the limit value found (\(0\)). Since the limit indeed evaluates consistently to \(0\), assuming the function defines a value at that point helps affirm its continuity.

functions of several variables

Functions of several variables, often called multivariable functions, depend on more than one variable. They can describe surfaces in 3D space or more complex dimensions. These functions are typically written as \(f(x, y, z)\) or even \(f(x_1,x_2,...,x_n)\).

In the context of our example, \(f(x, y)=\frac{x^3+y^3}{x^2+y}\) is a function of two variables. Understanding such functions requires analyzing how changes in \(x\) and \(y\) influence the outcome of \(f(x, y)\). Moreover, it's essential to determine behaviors like limits and continuity, as these properties are key to properly grasping the overall function characteristics.