Problem 20 - Calculus for Biology and Medicine

Question

Compute $$\lim _{(x, y) \rightarrow(0.0)} \frac{3 x^{2} y}{\left(2 x^{4}+y^{2}\right)}$$ along lines of the form \(y=m x\), for \(m \neq 0\), and along the parabola \(y=x^{2} .\) What can you conclude?

Step-by-Step Solution

Answer

The limit does not exist as it is path-dependent.

Step 1: Consider the Line

Let's evaluate the limit as \((x, y) \to (0, 0)\) along lines of the form \(y = mx\), where \(m eq 0\) is a constant. Substitute \(y = mx\) into the function:\[\frac{3x^2(mx)}{2x^4 + (mx)^2} = \frac{3mx^3}{2x^4 + m^2x^2}\]

Step 2: Simplify the Expression

Factor out \(x^2\) from the numerator and the denominator:\[\frac{3mx^3}{2x^4 + m^2x^2} = \frac{3mx^3}{x^2(2x^2 + m^2)} = \frac{3mx}{2x^2 + m^2}\]

Step 3: Evaluate the Limit Along the Line

As \(x \to 0\), the expression \(\frac{3mx}{2x^2 + m^2}\) simplifies. Notice that the numerator tends to zero faster than the denominator:\[\lim_{x \to 0} \frac{3mx}{2x^2 + m^2} = \frac{3m \cdot 0}{2 \cdot 0^2 + m^2} = 0\] So, along any line \(y = mx\), the limit results in 0.

Step 4: Consider the Parabola

Now, let's evaluate the limit as \((x, y) \to (0, 0)\) along the parabola \(y = x^2\). Substitute \(y = x^2\) into the function:\[\frac{3x^2(x^2)}{2x^4 + (x^2)^2} = \frac{3x^4}{2x^4 + x^4}\]

Step 5: Simplify the Parabolic Expression

Combine like terms in the denominator:\[\frac{3x^4}{3x^4} = 1\]

Step 6: Evaluate the Limit Along the Parabola

The expression simplifies to 1 for all \(x eq 0\). Therefore, as \(x \to 0\), the function value remains 1:\[\lim_{x \to 0} \frac{3x^4}{3x^4} = 1\] So, along the parabola \(y = x^2\), the limit is 1.

Step 7: Conclusion from Different Paths

Since the limit depends on the path taken towards \((0, 0)\)—0 along lines \(y=mx\) and 1 along the parabola \(y=x^2\)—the overall limit does not exist. The limit of the function is path-dependent.

Key Concepts

Path-Dependent Limits

When discussing multivariable limits, understanding how the limit can change based on the path taken is crucial. A path-dependent limit implies that as you approach a specific point from different directions (or paths), the limit values can vary, which indicates that a single overall limit does not exist.
In the context of our exercise, we evaluated the limit of a function as the point \((x, y) \to (0, 0)\) by considering different paths—lines and a parabola—to reach the origin. As we observed:

• Along lines of the form \(y = mx\), where \(m eq 0\), the limit turned out to be 0.
• Along the parabola \(y = x^2\), the limit was 1.

Since these limits differ, the conclusion is that the overall limit at \((0, 0)\) does not exist. Simply put, when a limit is path-dependent, there isn't a universal approach to calculate it, hence it highlights the importance of examining different paths in multivariable calculus.

Limit Along a Line

One common method to explore multivariable limits is by examining a function along a straight line. This means substituting a form such as \(y = mx\) into the function. The slope \(m\) gives us the direction of the line.
In the exercise provided, we calculated the limit as \((x, y) \to (0, 0)\) along lines defined by \(y = mx\). By making this substitution, we transformed our original expression into \(\frac{3mx^3}{2x^4 + m^2x^2}\). Further simplification revealed that when \(x\) approaches 0, the limit becomes 0 because the numerator \(3mx\) approaches 0 faster than the denominator \(2x^2 + m^2\).
This approach effectively demonstrates how a limit behaves along linear paths and is a powerful technique for testing path-dependence. Understanding this concept helps reinforce the observation that if different lines yield the same limit result, it suggests—but doesn't confirm—the existence of a limit at the point.

Limit Along a Parabola

To further analyze multivariable functions, it's insightful to test limits along non-linear paths, like parabolas. In our exercise, we examined the path where \(y = x^2\), a simple parabolic curve.
By substituting \(y = x^2\) into the function, we simplified the expression from \(\frac{3x^2y}{2x^4 + y^2}\) to \(\frac{3x^4}{3x^4}\). This simplification showed that the limit is 1 for all \(x eq 0\). As \(x \to 0\), the function's value stabilizes at 1, indicating this specific limit along the parabola.
By evaluating a limit along a parabola, we introduce new dimensions to our analysis, especially important because while linear paths showed a limit of 0, the parabolic path yielded a different limit. Such differences further emphasize the path-dependent nature of some limits in multivariable calculus, showcasing the diverse behaviors that can emerge based on the chosen approach.