Problem 76 - Calculus Early Transcendentals

Question

Show that $$\lim _{(x, y) \rightarrow(0,0)} \frac{a x^{2(p-n)} y^{n}}{b x^{2 p}+c y^{p}} \text { does }$$ not exist when \(a, b,\) and \(c\) are nonzero real numbers and \(n\) and \(p\) are positive integers with \(p \geq n\)

Step-by-Step Solution

Answer

Question: Prove that the limit does not exist for the given function: $$ \lim_{(x,y) \to (0,0)} \frac{a x^{2(p-n)} y^{n}}{b x^{2p}+c y^{p}} $$ where \(a, b, c\) are nonzero real numbers, \(n\) and \(p\) are positive integers, and \(p \geq n\). Answer: The limit does not exist because we found two different paths, with different values of \(k\), that give different limit values. For \(k=1\), the limit is \(\frac{a}{b}\), and for \(k=2\), the limit is \(\frac{a}{b+c}\). Since these limits are not equal, the limit does not exist when approaching \((0,0)\).

Step 1: Rewrite the given expression in a more convenient form

Let's rewrite the given expression to easily perform further operations: $$ L = \frac{a x^{2(p-n)} y^{n}}{b x^{2p}+c y^{p}} $$

Step 2: Analyze the given constraints

We are given that \(a\), \(b\), and \(c\) are nonzero real numbers, \(n\) and \(p\) are positive integers, and \(p \geq n\).

Step 3: Find a suitable path and limits along the path

Let's consider the path \(y = x^k\), where \(k\) is a positive integer. We can later try different values of \(k\) to find two paths that yield different limit values: $$ L = \frac{a x^{2(p-n)} (x^k)^{n}}{b x^{2p}+c (x^k)^{p}} $$ Now, we'll simplify the expression: $$ L = \frac{a x^{2p}}{b x^{2p}+c x^{kp}} $$ As \((x, y) \rightarrow (0,0)\), \(x \rightarrow 0\). Hence, by applying the limit, we get: $$ \lim_{x \rightarrow 0} L = \lim_{x \rightarrow 0} \frac{a x^{2p}}{b x^{2p}+c x^{kp}} = \lim_{x \rightarrow 0} \frac{ax^{2p}}{bx^{2p}+\frac{cx^{kp}}{ax^{2p}}} \cdot \frac{1}{x^{2p}} $$ Simplifying further, we get: $$ \lim_{x \rightarrow 0} L = \frac{a}{b+\frac{c}{x^{(k-2)p}}} $$

Step 4: Find two different limits using different values of \(k\)

Let's evaluate the limit using different values of \(k\): 1. When \(k=1\), the limit becomes: $$ \lim_{x \rightarrow 0} L = \frac{a}{b+\frac{c}{x^{(1-2)p}}} = \frac{a}{b} \quad(\text{since } p \geq n \Rightarrow (1-2)p \leq 0) $$ 2. When \(k=2\), the limit becomes: $$ \lim_{x \rightarrow 0} L = \frac{a}{b+\frac{c}{x^{(2-2)p}}} = \frac{a}{b+c} \quad(\text{since } p \geq n \Rightarrow (2-2)p=0) $$ Since we have found two different paths (values of \(k\)) that give different limit values (which are not equal), the limit does not exist.

Key Concepts

Limits in Multiple Variables

When dealing with limits in multiple variables, we are often interested in understanding how a function behaves as it approaches a particular point in a two-dimensional or higher-dimensional space. For instance, in our exercise, we approach the origin (0, 0) in a two-dimensional plane.
In single variable calculus, you can only approach a point in two directions: from the left and from the right. However, in multivariable calculus, there are infinitely many directions you can approach a point. This makes analyzing limits in multiple variables more complex.

To study these limits, it's crucial to substitute different paths and observe if the function's limit remains constant as you approach the target point. If the limit is the same regardless of the path taken, then the limit at that point exists. If not, it does not exist.

Path Dependence in Limits

Path dependence in limits is a concept that arises specifically in the context of multivariable limits. It indicates that the limit of a function can vary depending on the path taken to approach a particular point.

For example, in our exercise, we used the paths where \(y = x^k\), allowing us to substitute different values for \(k\). We observed different limit values along these paths.

• When \(k=1\), the limit simplifies to \(\frac{a}{b}\).
• When \(k=2\), the limit changes to \(\frac{a}{b+c}\).

Since these two paths yield different limit values, the limit does not exist as a unique value.
This showcases the importance of checking multiple paths to determine whether multivariable limits exist or differ based on the path taken. By revealing different values, path dependence proves that a function’s behavior is not uniform across the approaching paths.

Nonexistence of Limits in Calculus

In calculus, demonstrating the nonexistence of a limit usually involves showing that the function does not approach a single value as it nears a certain point.

For multivariable functions, one common method is to find two or more paths that result in distinct limits, as was done in our given problem.
This evidence of path-dependent limits proves that the limit cannot settle onto a single value.

Nonexistence is critical to recognize because it indicates that the behavior of the function near a point is unpredictable or discontinuous.

• The different values along selected paths signify the absence of a unique limit.
• It helps in understanding and categorizing functions based on continuity and differentiability.

The concept of nonexistence plays a significant role in higher-level calculus, as it guides us in understanding the intricacies and limitations of behavior within the multiple dimensions of functions.