Problem 3 - Active Calculus Multivariable

Question

Show that the function $$f(x, y)=\frac{x^{3} y}{x^{6}+y^{3}}$$ does not have a limit at (0,0) by examining the following limits. (a) Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) along the line \(y=x\). \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\) ________. (b) Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) along the curve \(y=x^{3}\). \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\) ___________.

Step-by-Step Solution

Answer

(a) \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)=0\) (b) \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)=1\)

Step 1: Calculate the limit along y=x

First, let's calculate the limit of the function as (x, y) → (0,0) along the line y=x. Replace y with x, to obtain: \(f(x, x)=\frac{x^{3} x}{x^{6}+x^{3}}=\frac{x^{4}}{x^{6}+x^{3}}\). Now, evaluate the limit: \(\lim_{(x, x) \rightarrow (0,0)} f(x, x)=\lim_{x\rightarrow 0} \frac{x^{4}}{x^{6}+x^{3}}\).

Step 2: Simplify the expression

Factor out \(x^{3}\) from the denominator: \(\lim_{x\rightarrow 0} \frac{x^{4}}{x^{3}(x^{3}+1)}\). Now cancel out the common term \(x^{3}\) from the numerator and denominator: \(\lim_{x\rightarrow 0} \frac{x}{x^{3}+1}\).

Step 3: Evaluate the limit along y=x

Evaluate the limit as x approaches 0: \(\lim_{x\rightarrow 0} \frac{x}{x^{3}+1}=\frac{0}{1}=0\). So, the limit of the function as (x, y) → (0,0) along the line y=x is 0.

Step 4: Calculate the limit along y=x^3

Now, let's calculate the limit of the function as (x, y) → (0,0) along the curve y=x^3. Replace y with \(x^3\), to obtain: \(f(x, x^3)=\frac{x^{3} (x^3)}{x^{6}+(x^3)^{3}}=\frac{x^{6}}{x^{6}+x^{9}}\). Now, evaluate the limit: \(\lim_{(x, x^3) \rightarrow (0,0)} f(x, x^3)=\lim_{x\rightarrow 0} \frac{x^{6}}{x^{6}+x^{9}}\).

Step 5: Simplify the expression

Factor out \(x^{6}\) from the denominator: \(\lim_{x\rightarrow 0} \frac{x^{6}}{x^{6}(1+x^{3})}\). Now cancel out the common term \(x^{6}\) from the numerator and denominator: \(\lim_{x\rightarrow 0} \frac{1}{1+x^{3}}\).

Step 6: Evaluate the limit along y=x^3

Evaluate the limit as x approaches 0: \(\lim_{x\rightarrow 0} \frac{1}{1+x^{3}}=\frac{1}{1+0}=1\). So, the limit of the function as (x, y) → (0,0) along the curve y=x^3 is 1. Since the limits along the line y=x and the curve y=x^3 are different (0 and 1, respectively), we can conclude that the function \(f(x, y)=\frac{x^{3} y}{x^{6}+y^{3}}\) does not have a limit at (0,0).

Key Concepts

Limits at a Point

Understanding the concept of limits at a point is essential in multivariable calculus. It is analogous to the concept in single-variable calculus but with its unique complexities due to the additional dimensions. A limit at a point for a multivariable function expresses the value that the function approaches as the input variables converge to a specific point.
In our example, we're looking at the limit of a two-variable function as both variables approach zero. It's important to note that in multivariable functions, how you approach the point can affect the limit. If the limit exists, the value should be the same regardless of the path taken to approach the point. However, if different paths lead to different limits, as with our function along the line y=x versus the curve y=x^3, this indicates that the overall limit at the point does not exist.
This is critical because it illustrates a principle in multivariable calculus: for a function to have a limit at a point, it must approach the same value from all possible directions. Our example showcases how analyzing different paths can give insight into the behavior of the function at a specific point.

Multivariable Calculus

Moving into the realm of multivariable calculus, we delve deeper into studying functions with more than one independent variable. When dealing with such functions, the behavior becomes more intricate than in single-variable scenarios. In multivariable calculus, we have to consider not just a single line of approach to a particular point, but an entire plane (in two dimensions) or even space (in three or more dimensions).
Our function, expressed as \(f(x, y)=\frac{x^{3} y}{x^{6}+y^{3}}\), is a perfect example of a two-variable function. Evaluating limits and understanding the behavior around a point like (0,0) requires careful path-specific analysis. The essence of multivariable calculus lies in these enriched procedures, where each variable contributes to the overall outcome. The interplay of these variables dictates the outcomes we observe, such as continuity, differentiability, and optimization within this multidimensional world.

Limit Evaluation Methods

To fully unravel the complexities of multiple variable functions, various limit evaluation methods are employed. These methods can include algebraic simplification, substitution, and looking at the behavior along specific paths.
In our textbook solution, the algebraic simplification method is used extensively. This involves factoring and canceling out common terms to simplify the expression before evaluating the limit. For instance, we simplified \(\lim_{x\rightarrow 0} \frac{x^{4}}{x^{6}+x^{3}}\) by factoring out \(x^{3}\) to cancel common factors and straightforwardly evaluate the limit.
Analyzing along different paths—like y=x and y=x^3—is another approach to gain insights into the function's behavior as it approaches the point of interest. When these methods lead to different results, it vividly demonstrates that the function's limit at that point does not exist. Educating on these methods encourages a deeper understanding of their application and the logic behind interpreting results from different approaches.