Problem 86 - Calculus 3

Question

For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not. $$ \lim _{(x, y) \rightarrow(0,0)} \frac{x y+y^{3}}{x^{2}+y^{2}} $$ a. Along the \(x\) -axis \((y=0)\) b. Along the \(y\) -axis \((x=0)\) c. Along the path \(y=2 x\)

Step-by-Step Solution

Answer

The limit does not exist as it varies with the path taken.

Step 1: Evaluate the limit along the x-axis

To evaluate the limit along the x-axis, substitute \( y = 0 \) into the function: \[\lim _{(x, y) \rightarrow(0,0)} \frac{x y+y^{3}}{x^{2}+y^{2}} = \lim _{x \to 0} \frac{x \cdot 0 + 0^{3}}{x^{2}+0^{2}} = \lim _{x \to 0} \frac{0}{x^{2}}\]Since the numerator is 0, the limit along the x-axis is 0.

Step 2: Evaluate the limit along the y-axis

Now evaluate the limit along the y-axis by setting \( x = 0 \): \[\lim _{(x, y) \rightarrow(0,0)} \frac{x y+y^{3}}{x^{2}+y^{2}} = \lim _{y \to 0} \frac{0 \cdot y + y^{3}}{0^{2}+y^{2}} = \lim _{y \to 0} \frac{y^{3}}{y^{2}}\]This simplifies to \( \lim _{y \to 0} y = 0 \). Hence, the limit along the y-axis is 0.

Step 3: Evaluate the limit along the path y = 2x

Substituting \( y = 2x \) into the function gives:\[\lim _{(x, y) \rightarrow(0,0)} \frac{x (2x)+(2x)^{3}}{x^{2}+(2x)^{2}} = \lim _{x \to 0} \frac{2x^2+8x^3}{x^2+4x^2} = \lim _{x \to 0} \frac{2x^2+8x^3}{5x^2}\]Simplifying the expression results in:\[\lim _{x \to 0} \left(\frac{2x^2}{5x^2} + \frac{8x^3}{5x^2}\right) = \lim _{x \to 0} \left( \frac{2}{5} + \frac{8x}{5} \right)\]As \( x \to 0 \), the second term \( \frac{8x}{5} \to 0 \), thus the limit becomes \( \frac{2}{5} \).

Step 4: Analyze the results for the existence of the limit

The limits along the x-axis and y-axis both yield 0, but along the path \( y = 2x \), the limit is \( \frac{2}{5} \). Since these values are not the same, the overall limit does not exist.

Key Concepts

Limit evaluation

Evaluating limits in multivariable calculus can be a bit tricky compared to single-variable calculus. In single-variable calculus, we only deal with one path as we approach a specific point. However, in multivariable calculus, we must consider different paths as a point is approached in a multi-dimensional space.
To evaluate limits, we check if the function approaches a single value as the variables near the point of interest.
Often, we substitute specific paths to simplify the problem, such as fixing variables to constants or expressing one variable in terms of the other. This allows us to observe the behavior of the function more easily, as seen in the approach along the x-axis, y-axis, and specific paths like line equations or curves.

Paths of approach

In the exercise, the function's limit is explored through different paths: along the x-axis, the y-axis, and the line described by the equation \(y=2x\).
Each path gives different insights into the behavior of the function as it approaches the origin. When analyzing paths of approach, it is crucial to note:

• The x-axis path sets \(y=0\), reducing the function to a simpler form to evaluate the limit.
• Similarly, the y-axis path involves \(x=0\), again simplifying the expression.
• For the path \(y=2x\), we substitute in this relationship to see how the function behaves.

These paths are chosen specifically to understand how the function reacts when variables change in one way or another, demonstrating that while some limits converge to the same point, others might not.

Two-variable functions

Two-variable functions or bivariate functions depend on two inputs to produce an output.
This adds an extra dimension of complexity compared to single-variable functions. In the context of limits, this necessitates considering how both variables simultaneously approach certain values.
The function in the exercise involves the variables \(x\) and \(y\), examining them together in a fraction that combines both in the numerator and denominator. Understanding how these interact as both approach zero is key.
Such functions often model real-world scenarios, where several factors contribute to outcomes at once, further emphasizing the importance of thorough limit analysis.

Limit existence

Existence of a limit means that no matter the path chosen, the function approaches the same value when nearing a specific point.
If limits differ based on the path, it indicates the overall limit does not exist. In this exercise, we discover that:

• Along the x-axis and y-axis, the limit evaluates to 0.
• However, along the path \(y=2x\), the evaluation yields \(\frac{2}{5}\).

Since these result in different values, the limit does not exist.
This lack of uniformity across paths shows the importance of checking various approaches to confirm or refute a potential limit for multivariable functions.

Limit computation steps

Computing limits in multivariable calculus involves specific steps to ensure clarity and correctness.
First, substitute specific paths into the function and simplify until you can take the limit as a single variable approaches zero.

• In the example, \(y=0\) for the x-axis simplifies calculations radically, revealing the behavior as \(x\) tends to 0.
• Similarly, \(x=0\) fixes a constant in the expression when analyzing the y-axis path.
• To explore the path \(y=2x\), substitution helps in transforming the function into a simplified form just involving \(x\).

After substitution and simplification, observe if the limits are consistent across all paths or not. If not, the overall limit at that point does not exist.
These steps ensure a thorough understanding of how the function behaves as it approaches the desired point from various directions.