Problem 71 - Advanced Calculus

Question

Let \(\mathrm{f}(\mathrm{x}, \mathrm{y})\) be given by $$ \begin{array}{cl} \mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{xy}^{2} /\left(\mathrm{x}^{2}+\mathrm{y}^{4}\right) & (\mathrm{x}, \mathrm{y}) \neq(0,0) \\ \text { and }=0 & \mathrm{x}=\mathrm{y}=0 \end{array} $$ Is f continuous at the origin?

Step-by-Step Solution

Answer

The function \(f(x, y) = \frac{xy^2}{x^2+y^4}\) is continuous at the origin. It was determined by converting Cartesian coordinates to polar coordinates, evaluating the limit as (x, y) approaches (0, 0), and using the squeeze theorem to show that the limit exists and equals the function value at the origin.

Step 1: Write down the function and the limit

We are given the function \[\mathrm{f}(\mathrm{x}, \mathrm{y})=\frac{\mathrm{xy}^{2}}{\left(\mathrm{x}^{2}+\mathrm{y}^{4}\right)}\] for \((\mathrm{x}, \mathrm{y}) \neq (0,0)\] and \[\mathrm{f}(0,0) = 0.\] We need to evaluate the limit \[\lim_{(x,y) \to (0, 0)} \frac{xy^2}{x^2 + y^4}.\]

Step 2: Convert to polar coordinates

To make this limit easier to evaluate, convert the Cartesian coordinates (x, y) to polar coordinates (r, θ) using the relationships \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). We get \[\mathrm{f}(r, \theta) = \frac{r\cos{\theta}(r\sin{\theta})^2}{(r\cos{\theta})^2 + (r\sin{\theta})^4}.\]

Step 3: Simplify the function in polar coordinates

Simplify the equation above: \[\mathrm{f}(r, \theta) = \frac{r^3\cos{\theta}\sin^2{\theta}}{r^2\cos^2{\theta} + r^4\sin^4{\theta}}.\] Now, divide both the numerator and denominator by \(r^2\): \[\mathrm{f}(r, \theta) = \frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta} + r^2\sin^4{\theta}}.\]

Step 4: Evaluate the limit in polar coordinates

Now, we need to evaluate the limit: \[\lim_{(r,\theta) \to (0, \theta)} \frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta} + r^2\sin^4{\theta}}.\] Since 0 ≤ \(r\cos{\theta}\sin^2{\theta}\) ≤ r, we can apply the squeeze theorem: \[\lim_{r \to 0} 0 \le \lim_{r \to 0} \frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta} + r^2\sin^4{\theta}} \le \lim_{r \to 0} r.\] The limits on both sides are equal to 0. Thus, by the squeeze theorem, the limit of the function \(\mathrm{f}(r, \theta)\) as \(r \to 0\) is equal to 0.

Step 5: Conclude that the function is continuous at the origin

Since both the limit as \((x,y) \to (0,0)\) exists and equals the function value \(\mathrm{f}(0,0) = 0\), we can conclude that the function is continuous at the origin.

Key Concepts

Polar Coordinates

When dealing with limits in multivariable calculus, transforming coordinates can simplify the problem. Polar coordinates are particularly helpful when analyzing limits involving \((x, y)\) approaching \((0,0)\). They utilize \((r, \theta)\) to represent points, where \(r\) is the distance from the origin and \(\theta\) is the angle from the positive x-axis.
For any point \((x, y)\):

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)

By changing variables from Cartesian to polar coordinates, simplifying functions becomes easier as you can often reduce \((x^2 + y^2)\) to \(r^2\). This is because both components are described simply in terms of \(r\) and \(\theta\), allowing for an easier evaluation of limits.

Squeeze Theorem

The Squeeze Theorem is a fundamental tool for evaluating limits, especially when direct computation is difficult. It states that if a function \(g(r)\) is squeezed between two functions \(f(r)\) and \(h(r)\) such that \(f(r) \le g(r) \le h(r)\) for values around a point, and the limits of \(f(r)\) and \(h(r)\) are the same at this point, then the limit of \(g(r)\) must also equal this common limit.
In our context, with polar coordinates, if \(0 \le \frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta} + r^2\sin^4{\theta}} \le r\), as \(r \to 0\), the limit can be evaluated as 0. This theorem is particularly useful when functions are not easy to solve directly and helps confirm continuity by finding common limits around critical points.

Continuous Function

A function is continuous at a point if its limit as it approaches that point exists and equals the function's value at that point. For multivariable functions, this concept extends to limits as they approach a point in two-dimensional or higher spaces.
In this example, we checked the limit of \(f(x, y)\) as \((x, y)\) approaches \((0,0)\) using polar coordinates:

• The function value at \((0,0)\) was given as 0.
• By applying the Squeeze Theorem, the limit as \(r \to 0\) was found to be 0.

Since the limit equals the function's value at the origin, the function is continuous at that point. Continuous functions do not have any jumps, gaps, or points of discontinuity around, making them smooth and predictable.

Multivariable Calculus

Multivariable calculus extends concepts of single-variable calculus to functions of multiple variables, allowing analysis of more complex systems. It involves several new considerations, such as partial derivatives and multiple limits.
In multivariable calculus, limits examine behavior as all variables approach critical points in space. Unlike single-variable functions, paths can significantly impact the result in multivariable functions. Therefore, evaluating limits might require testing multiple paths or converting to alternative coordinate systems, like polar.
This exploration can confirm whether a function is well-behaved (continuous, differentiable) near key points or if special techniques like the Squeeze Theorem are needed for limit evaluation.