Question

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

Short answer

Question: Prove that the given limit does not exist: \(\lim_{(x, y) \rightarrow(0,0)} \frac{a x^{2(p-n)} y^{n}}{b x^{2 p}+c y^{p}}\), where \(a, b,\) and \(c\) are nonzero real numbers, \(n\) and \(p\) are positive integers with \(p \geq n\). Answer: The limit does not exist because along two different paths approaching (0,0), Path 1 with \(x = 0\) and Path 2 with \(x = y^q \) where \(q = \frac{p-n}{2(p-n)}\), the limits are not equal. The limit along Path 1 is 0 and the limit along Path 2 is \(\frac{a}{b+c}\). Since the two limits are not equal, it contradicts the limit's existence, proving that the limit does not exist.

Step-by-step solution

Define two paths

Let's define two paths which approach to point (0,0): Path 1: \(x = 0\) Path 2: \(x = y^q\) where \(q\) is a real number.

Calculate limit along Path 1

For Path 1, we substitute \(x = 0\) into the expression: \(\lim \limits_{y \rightarrow 0} \frac{a * 0^{2(p-n)} * y^{n}}{b * 0^{2 p}+c * y^{p}}\) Which simplifies to: \(\lim \limits_{y \rightarrow 0} \frac{0}{c * y^{p}} = 0\) So, the limit along Path 1 is 0.

Calculate limit along Path 2

For Path 2, we substitute \(x = y^{q}\) into the expression: \(\lim \limits_{y \rightarrow 0} \frac{a * (y^{q})^{2(p-n)} * y^{n}}{b * (y^{q})^{2 p}+c * y^{p}}\) Which simplifies to: \(\lim \limits_{y \rightarrow 0} \frac{a * y^{2q(p-n)+n}}{b * y^{2qp}+c * y^{p}}\) Now, let's choose \(q\) such that \(2q(p-n) + n = p\). The value of \(q\) would be \(q = \frac{p-n}{2(p-n)}\) since \(p \geq n\), this value is positive and valid. Substitute this value of q in our expression: \(\lim \limits_{y \rightarrow 0} \frac{a * y^{p}}{b * y^{2p(\frac{p-n}{2(p-n)})}+c * y^{p}}\) Which simplifies to: \(\lim \limits_{y \rightarrow 0} \frac{a * y^{p}}{b * y^{p}+c * y^{p}}\) Now, we can cancel out the \(y^{p}\) term: \(\lim \limits_{y \rightarrow 0} \frac{a}{b+c}\) Since \(a, b,\) and \(c\) are nonzero real numbers, then \(\frac{a}{b+c}\) is a constant.

Conclusion

We found out that the limit along Path 1 is 0, while the limit along Path 2 is \(\frac{a}{b+c}\). Since these two limits are not equal, it contradicts the limit's existence, and thus, the limit does not exist.

Key concepts

Paths of Approach

When dealing with multivariable limits, the path of approach is crucial. Since multi-variable functions can approach a point from infinite directions, the value of the limit can depend on the path along which we approach it. In simpler terms, imagine arriving at a junction via different streets. The route you choose can affect the scenery you observe as you approach the junction.

In our problem, two paths were chosen:

• Path 1: Setting \(x = 0\), we approach along the y-axis.
• Path 2: Setting \(x = y^q\), we approach along the curve where \(x\) increases as a function of \(y\).

This comparison is essential for checking if the limit exists. If a limit is the same along every path, it is possible that the overall limit might exist. Conversely, different limits along different paths suggest that the overall limit does not exist.

Limit Calculation

Calculating the limit along a path involves substituting the path's equation into the function and examining the resulting expression. This can simplify the problem significantly.

For **Path 1**, substituting \(x = 0\) into the limit expression simplifies the numerator to zero, as we have terms containing \(x\) raised to a power: \(\lim \limits_{y \rightarrow 0} \frac{0}{c \cdot y^{p}} = 0\). This means as \(y\) approaches zero, so does the limit along this path.

For **Path 2**, by substituting \(x = y^{q}\), the expression reflects a more complex relationship: \(\lim \limits_{y \rightarrow 0} \frac{a \cdot y^{p}}{b \cdot y^{p} + c \cdot y^{p}}\). By strategically choosing \(q\) such that \(2q(p-n) + n = p\), the expression simplifies. This approach is a tool used to reveal the behavior of the function over its variable path. Ultimately, deciding carefully on \(q\) allows the path to simplify, yielding \(\frac{a}{b+c}\) as \(y\) approaches zero.

Thus, the limit's value on Path 2 is not zero, indicating a disparity between the computed path limits.

Non-Existence of Limits

The existence of a multivariable limit hinges on the limit being the same along every potential path of approach. If different paths yield different limits, then generally, we conclude that there is no single, unifying limit for the function.

In our exercise, we observed that Path 1 results in a limit of 0, while Path 2 yields \(\frac{a}{b+c}\). Since these two values differ, they contradict each other. This contradiction is pivotal for showing the limit does not exist.

Here are some takeaways:

• Divergent limits indicate the absence of a unique value the function reaches, implying non-existence.
• Such results suggest the function behaves unpredictably close to the point of interest.

Therefore, multivariable limits require careful analysis and highlight the complexities in navigating the infinite approaches within higher-dimensional spaces.