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 \((x,y) \to (0,0)\)

\(\mathop {lim}\limits_{(x,y) \to (0,0)} \left( {\frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}}} \right)\)

Short answer

Using polar coordinates

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \left( {\frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}}} \right)\)

Step-by-step solution

 Step 1: 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 limit:

\(\mathop {\lim }\limits_{r \to {0^ + }} \frac{{{r^2}{{\cos }^3}\theta + {r^3}{{\sin }^3}\theta }}{{{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta }}\)

\( = \mathop {\lim }\limits_{r \to {0^ + }} \frac{{{r^2}({{\cos }^3}\theta + {{\sin }^3}\theta )}}{{{r^2}({{\cos }^2}\theta + {{\sin }^2}\theta )}}\)

\( = \mathop {\lim }\limits_{r \to {0^ + }} {r^2}({\cos ^3}\theta + {\sin ^3}\theta )\)

\( = 0\)

Hence, \(\mathop {\lim }\limits_{(x,y) \to (0,0)} \left( {\frac{{{x^3} + {y^3}}}{{{x^2} + {y^2}}}} \right) = 0\)