Question

Evaluate: $\underset{z\to \infty }{lim}\frac{i{z}^3+iz-1}{(2z+3i)(z-i{)}^2}$

Answer 1

$\underset{z\to \infty }{lim}\frac{i{z}^3+iz-1}{(2z+3i)(z-i{)}^2}$

$=\underset{z\to \infty }{lim}\frac{z^3(i+\frac{i}{z^2}-\frac{1}{z^3})}{z.z^2(2+\frac{3i}{z}){(1-\frac{i}{z})}^2}$

$=\underset{z\to \infty }{lim}\frac{(i+\frac{i}{z^2}-\frac{1}{z^3})}{(2+\frac{3i}{z}){(1-\frac{i}{z})}^2}$

As, $z\to \infty ,\frac{1}{z}\to 0$

$=\frac{i+0-0}{(2+0)(1-0{)}^2}$

$=\frac{i}{2}$ (Ans)