Question

Use polar coordinates to find the limit. If \((r,\theta )\) are polar coordinates of the point \((x, y)\) with \(r \ge 0\) note that \(r \to {0^ + }\) as \((x,y) \to (0,0)\)

\(\mathop {lim}\limits_{(x,y) \to (0,0)} \left( {{x^2} + {y^2}} \right)\ln \left( {{x^2} + {y^2}} \right)\)

Short answer

Using polar coordinates: \(\mathop {\lim }\limits_{(x,y) \to (0,0)} \left( {{x^2} + {y^2}} \right)\ln \left( {{x^2} + {y^2}} \right) = 0\)

Step-by-step solution

use polar coordinates:

\(x = r\cos \theta \), \(y = r\sin \theta \) where \(r \ge 0\)

When \((x,y) \to (0,0)\), \(r \to {0^ + }\)

calculating the limit:

\(\mathop {\lim }\limits_{r \to {0^ + }} \left( {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } \right)\ln \left( {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } \right)\)

\( = \mathop {\lim }\limits_{r \to {0^ + }} {r^2}\ln {r^2}\)

\( = \mathop {\lim }\limits_{r \to {0^ + }} \frac{{\ln {r^2}}}{{\frac{1}{{{r^2}}}}} = \mathop {\lim }\limits_{r \to {0^ + }} \frac{{\left( {\frac{1}{{{r^2}}}} \right)2r}}{{ - \frac{2}{{{r^3}}}}}\)(using L’ Hospital’s rule)

\( = \mathop {\lim }\limits_{r \to {0^ + }} ( - {r^2}) = 0\)

Hence, \(\mathop {\lim }\limits_{(x,y) \to (0,0)} \left( {{x^2} + {y^2}} \right)\ln \left( {{x^2} + {y^2}} \right) = 0\)