Problem 47 - Calculus Early Transcendentals

Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\ln \left(x^{2}+y^{2}\right)$$

Step-by-Step Solution

Answer

Answer: The function is continuous at all points in \(\mathbb{R}^2\) except for the point \((0, 0)\).

Step 1: Identify possible points of discontinuity

Recall that the natural logarithm function, \(\ln\), is undefined for non-positive values. That is, \(\ln(t)\) is defined only for \(t>0\). In this case, our argument inside the logarithm is \(x^2+y^2\). So, we must identify when \(x^2+y^2\) becomes non-positive.

Step 2: Investigate the argument of the logarithm

We know that both \(x^2\) and \(y^2\) are non-negative, because any real number squared will result in a non-negative value. Given any point \((x, y) \neq (0, 0)\), we have that \(x^2+y^2>0\). However, if \(x=0\) and \(y=0\), we have \(x^2+y^2=0\). Therefore, the only issue we have to worry about is when both \(x\) and \(y\) are equal to \(0\).

Step 3: Check continuity at \((0, 0)\)

For all points \((x, y) \neq (0, 0)\), we have already determined that the function \(f(x, y)\) is defined, and the function appears to be continuous. Therefore, we only need to check the continuity at the point \((0, 0)\). As \(x^2+y^2\) approaches \(0\), the logarithm function, \(\ln(t)\), approaches \(-\infty\). This implies a discontinuity at the point \((0, 0)\) because the logarithm function is undefined at this point.

Step 4: State the conclusion

The function \(f(x, y) = \ln \left(x^{2}+y^{2}\right)\) is continuous at all points in \(\mathbb{R}^{2}\) except for the point \((0, 0)\).