Question

In Exercises 25-28, discuss the continuity of the function and evaluate the limit of \(f(x, y)\) (if it exists) as \((x, y) \rightarrow(0,0)\). \(f(x, y)=\frac{x^{2}}{\left(x^{2}+1\right)\left(y^{2}+1\right)}\)

Short answer

The function \(f(x, y)=\frac{x^{2}}{\left(x^{2}+1\right)\left(y^{2}+1\right)}\) is continuous everywhere and the limit of this function as \((x, y) \rightarrow(0,0)\) equals 0.

Step-by-step solution

Determine the Continuity of the Function

In order for the function to be continuous at a point, the limit of the function as it approaches that point should exist and equals the value of the function at that point. For \(f(x, y)=\frac{x^{2}}{\left(x^{2}+1\right)\left(y^{2}+1\right)}\), the function is defined and finite for all real values of x and y. As there are no discontinuities in the function, it can be concluded that the function \(f(x, y)\) is continuous everywhere.

Evaluate the Limit as \((x, y) \rightarrow(0,0)\)

To evaluate the limit of \(f(x, y)\) as \((x, y) \rightarrow(0,0)\), we substitute the values of x and y with 0. This gives: \[lim_{(x, y)\rightarrow (0, 0)} f(x, y) = \frac{0^{2}}{\left(0^{2}+1\right)\left(0^{2}+1\right)} = 0\] Thus, the limit of \(f(x, y)\) as \((x, y) \rightarrow(0,0)\) equals 0.

Make The Final Conclusion

From the steps above, it is concluded that the function \(f(x, y)=\frac{x^{2}}{\left(x^{2}+1\right)\left(y^{2}+1\right)}\) is continuous everywhere and its limit as \((x, y) \rightarrow(0,0)\) equals 0.

Key concepts

Continuity of Functions

Understanding the continuity of functions is pivotal in multivariable calculus. A function is continuous at a point if three criteria are met: the function exists at that point; the limit of the function as it approaches that point exists; and both values are equal.

When analyzing the given function, \(f(x, y)=\frac{x^{2}}{(x^{2}+1)(y^{2}+1)}\), we observe that it is defined for all real numbers since there are no values for \(x\) or \(y\) that would make the denominator zero. Hence, the function exists everywhere. As the function is composed of polynomials and the division by non-zero polynomials, it inherits their continuous properties, leading us to conclude that \(f(x, y)\) is continuous at every point in the \(xy\)-plane, including \( (0,0) \).

This concept underlines the seamless behavior of the function across its entire domain where it does not have abrupt changes or gaps, and its graph would depict a smooth surface.

Limits of Multivariable Functions

When dealing with the limits of multivariable functions, the approach is to determine the value that the function approaches as the variables independently approach their own limits. Unlike single variable functions where the limiting behavior is checked from both sides along one path, in multivariable calculus, we need to consider all possible paths to the point of interest.

For the given function, we aim to find \[\lim_{(x, y)\rightarrow (0, 0)} f(x, y)\]. Since the function is a quotient of polynomials and the limit point \( (0,0) \) is not a point of discontinuity, we may simply substitute \(x=0\) and \(y=0\) directly into the function to find that \[\lim_{(x, y)\rightarrow (0, 0)} \frac{x^{2}}{(x^{2}+1)(y^{2}+1)} = 0\], which suggests that from any direction towards \( (0,0) \) the function value tends to zero.

This exercise exemplifies the need to explore possible paths towards the limit point in multivariable functions to ensure that the function does not behave unexpectedly as it approaches from different trajectories.

Evaluate Limits in Calculus

Evaluating limits is a fundamental task in calculus that involves finding the value that a function is approaching as the input approaches some value. This concept extends into multivariable calculus with greater complexity due to the involvement of multiple variables.

To effectively evaluate limits, we often substitute the approaching values into the function if continuous at that point. Alternatively, techniques such as factoring, rationalizing, or using L'Hôpital's Rule are employed, especially when dealing with indeterminate forms like \(0/0\).

In the context of our function, direct substitution was sufficient since the function \(f(x, y)\) was continuous at the point of interest. However, if the function had presented more complex behavior, the path to evaluation could include analyzing multiple paths, applying squeeze theorem or transforming the function to simplify the limit evaluation.

It is important to note that finding the limit does not necessarily indicate the value of the function at that point but rather where it trends as it approaches the point. Understanding this relationship between limits and function values is crucial for grasping the broader concepts in calculus.