A First Course in Complex Analysis · Problem 11

Integration — Beck, Marchesi, Pixton & Sabalka

Mathematics — Complex Analysis · limits-extended-complex-plane

Problem

Show that our definition of C^\hat{\mathbb{C}} honors the ``finite'' limit rules in Proposition 2.1.6, by proving the following, where a∈Ca \in \mathbb{C}: ((a)) If lim⁡z→z0f(z)=∞\lim_{z\to z_0} f(z) = \infty and lim⁡z→z0g(z)=a\lim_{z\to z_0} g(z) = a then lim⁡z→z0(f(z)+g(z))=∞\lim_{z\to z_0}(f(z) + g(z)) = \infty. ((b)) If lim⁡z→z0f(z)=∞\lim_{z\to z_0} f(z) = \infty and lim⁡z→z0g(z)=a≠0\lim_{z\to z_0} g(z) = a \neq 0 then lim⁡z→z0(f(z)⋅g(z))=∞\lim_{z\to z_0}(f(z) \cdot g(z)) = \infty. ((c)) If lim⁡z→z0f(z)=lim⁡z→z0g(z)=∞\lim_{z\to z_0} f(z) = \lim_{z\to z_0} g(z) = \infty then lim⁡z→z0(f(z)⋅g(z))=∞\lim_{z\to z_0}(f(z) \cdot g(z)) = \infty. ((d)) If lim⁡z→z0f(z)=∞\lim_{z\to z_0} f(z) = \infty and lim⁡z→z0g(z)=a\lim_{z\to z_0} g(z) = a then lim⁡z→z0g(z)f(z)=0\lim_{z\to z_0} \frac{g(z)}{f(z)} = 0. ((e)) If lim⁡z→z0f(z)=0\lim_{z\to z_0} f(z) = 0 and lim⁡z→z0g(z)=a≠0\lim_{z\to z_0} g(z) = a \neq 0 then lim⁡z→z0g(z)f(z)=∞\lim_{z\to z_0} \frac{g(z)}{f(z)} = \infty.

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