Problem 5 - Active Calculus Multivariable

Question

Find the limit, if it exists, or type \(\mathrm{N}\) if it does not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{(x+13 y)^{2}}{x^{2}+13^{2} y^{2}}=$$ _____________.

Step-by-Step Solution

Answer

The limit of the given function as (x, y) approaches (0, 0) is 0.

Step 1: Check for a direct substitution

If the function is defined at the point (0,0), we can directly substitute x and y with 0 and find the limit. However, substituting x and y with 0, the function becomes \(\frac{(0+13\cdot0)^2}{0^2 + (13\cdot0)^2} = \frac{0}{0}\), which is an indeterminate form. Thus, we need to use other methods to find the limit.

Step 2: Polar coordinate transformation

Since we have a 2-variable limit, we can try transforming the function into polar coordinates and then find the limit by using the transformed form. To do this, we will substitute x with \(r\cos\theta\) and y with \(r\sin\theta\), where r is the radial distance from the origin, and \(\theta\) is the angle. Our given function becomes: \(\frac{((r\cos\theta)+13(r\sin\theta))^2}{(r\cos\theta)^2+ (13r\sin\theta)^2}\) After substituting, we can simplify the function:

Step 3: Simplify the function

Simplifying the given function after the polar coordinate transformation: \(\lim_{r \to 0} \frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\cdot r^2\) Now, notice that the function does not depend on \(\theta\) anymore. Instead, we can remove the limits depending on r: \(\frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\)

Step 4: Evaluate the limit

Now, let us see if this limit exists: \(\lim_{(x, y) \to (0,0)} \frac{(x+13 y)^{2}}{x^{2}+13^{2} y^{2}} = \lim_{r \to 0} \frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\cdot r^2 = 0\) Since the part \(\frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\) approaches a constant as r approaches 0, the limit converges to 0 as the whole expression is multiplied by \(r^2\), which approaches 0 as r approaches 0. Thus, the limit of the function as (x, y) approaches (0, 0) is 0.

Key Concepts

Polar Coordinates Transformation

When evaluating limits in two variables, polar coordinates can be a powerful tool. In this system, each point in the plane is represented by

• an angle \(\theta\)
• a radial distance \(r\) from the origin

We substitute \(x = r\cos\theta\) and \(y = r\sin\theta\) into the given expression. This changes the problem of a 2-variable limit into a simpler form. By doing so, the distance to the origin becomes the main variable, which can help in determining the limit more easily.
In the problem, this substitution transformed our expression into:\[ \frac{((r\cos\theta)+13(r\sin\theta))^2}{(r\cos\theta)^2+ (13r\sin\theta)^2} \]After simplifying, we were able to isolate and evaluate the dependence on \(r\) and \(\theta\). This transformation helps us see if the limit uniformly approaches the origin from all directions.

Indeterminate Form

An indeterminate form occurs when direct substitution into a function leads to expressions like \(\frac{0}{0}\). These forms don't provide clear information about a limit's behavior. In the problem at hand, substituting \((x, y) = (0, 0)\) directly yielded the indeterminate form \(\frac{0}{0}\). This suggested that we couldn't evaluate the limit directly.
Indeterminate forms often require us to apply other methods, such as:

• L'Hôpital's Rule (for certain 1-variable functions)
• Algebraic manipulation
• Coordinate transformations (as seen in this exercise)

These approaches help us analyze the behavior of the function around the point of indeterminacy and guide us towards finding the actual limit.

Multivariable Calculus

Multivariable calculus expands on principles from single-variable calculus to functions with more than one variable. It deals with topics like:

• Partial derivatives
• Multiple integrals
• Vector fields
• Limits involving multiple variables

Evaluating limits in two variables, like in this problem, is a classic challenge in multivariable calculus. When approaching the origin from different paths, we need to ensure the limit is consistent no matter the path. Converting to polar coordinates simplifies the problem by focusing on the distance to the origin and angle, making limit evaluation clearer.
Understanding these concepts helps when navigating through complex geometries and varying directions within multivariable functions.

Limit Evaluation Techniques

To evaluate limits in two variables, various techniques are employed to provide accuracy and insights. The common strategies include:

• Direct substitution (if applicable)
• Path approach, where different paths towards the limit point yield consistent results
• Using a polar coordinates transformation, as seen in the problem
• Simplifying expressions to remove indeterminate forms

In our exercise, we faced \(\frac{0}{0}\), steering us toward a conversion to polar coordinates. This revealed that the expression's dependence on \(r\) diminished the complexity of evaluating the limit. As \(r \to 0\), multiplying by \(r^2\) leads to a limit that clearly tends towards zero.
Mastering these techniques provides a robust foundation for handling limits in multivariable calculus efficiently.