Question

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

Short answer

The function \( \ln(x^{2}+y^{2}) \) does not have a limit as \((x, y) \rightarrow (0,0)\) and thus, it is not continuous at the point \((0,0)\).

Step-by-step solution

Locate the point

We need to evaluate the limit as \((x, y) \rightarrow (0,0)\). This is the point of interest.

Substitute the point

To find the limit as \((x, y) \rightarrow (0,0)\), substitute \(x = 0\) and \(y = 0\) into the equation to see what we find: \(f(x, y) = \ln (0 + 0) = \ln 0 \). We have a problem here as the natural logarithm of zero is undefined.

Evaluate limits from different paths

Since there is a problem at \((0,0)\), we need to test the limit from different paths to confirm if the limit exists. Two usual paths are when \(y = 0\) and when \(y = x\). For \(y = 0\), the function becomes \(f(x, 0) = \ln(x^2)\), and the limit as \(x \rightarrow 0\) is also undefined. Similarly, for \(y = x\), the function becomes \(f(x, x) = \ln(2x^2)\), and again the limit as \(x \rightarrow 0\) is also undefined.

Conclude continuity and limits

It is clear that the function is not defined at \((0,0)\), and any approach from different paths leads to an undefined limit. Thus, the limit \(\lim_{{(x, y) \rightarrow (0,0)}} \ln(x^{2}+y^{2}) \) does not exist. And therefore, the function isn't continuous at \((0,0)\) as the function is not defined at that point.

Key concepts

Continuity in Calculus

Continuity is a fundamental concept in calculus that describes a situation where a function has no breaks, jumps, or holes at a certain point. In formal terms, a function f(x, y) is said to be continuous at a point (a, b) if the following three conditions are met:

• The function is defined at (a, b).
• The limit of f(x, y) as (x, y) approaches (a, b) exists.
• The limit of the function as (x, y) approaches (a, b) is equal to f(a, b).

Applying this to the problem, we can see that the function f(x, y) = \(ln\left(x^{2} + y^{2}\right)\) is not continuous at the point (0,0) because it does not satisfy all of these conditions. Specifically, the function is not defined at (0,0) since the natural logarithm of zero is undefined. Hence, the function f(x, y) breaks the first condition for continuity at that point.

Evaluating Limits

Evaluating limits in multi-variable calculus can sometimes be more challenging than in single-variable calculus due to the possibility of approaching a point from infinitely many paths. When determining the limit of a function f(x, y) as (x, y) approaches a point, if the limit values are the same from all possible directions, the limit exists. Otherwise, it is undefined.

In the given exercise, we are trying to evaluate \(\lim_{{(x, y) \rightarrow (0,0)}} \ln(x^{2}+y^{2}) \) . Approaching this limit directly by substituting x = 0 and y = 0 into the function presents an immediate problem as the natural logarithm of zero does not exist. In such cases, investigating the limit along different paths can reveal whether the limit exists or if it's undefined. The solution process indicates that the limit remains undefined regardless of the path taken—whether along y = 0 or y = x.

Undefined Limits

Limits can be undefined for a number of reasons, such as the function heading towards infinity or the value at a certain point not existing. The expression 'undefined limit' often refers to the non-existence of a single, unique limit at a certain point.

In the context of the provided problem, \(\lim_{{(x, y) \rightarrow (0,0)}} \ln(x^{2}+y^{2})\) is undefined because \(\ln(0)\)—logarithm of zero—is not a real number. Logarithmic functions are defined only for positive numbers, and any attempt to evaluate the natural logarithm at zero is bound to yield an undefined result. Moreover, since the limit yields different forms of undefined expressions as you approach the point (0,0) from different paths (yet all undefined), we confirm that the limit indeed does not exist.