Question

Evaluate :
$\underset{z\to 0}{lim}\frac{\sin \left(2z\right)+7z^2-2z}{z^2{\left(z+1\right)}^2}$

Answer 1

$\underset{z\to 0}{lim}\frac{\sin \left(2z\right)+7z^2-2z}{z^2{\left(z+1\right)}^2}$

$=\underset{z\to 0}{lim}\frac{\sin \left(2z\right)+7z^2-2z}{z^2\left(z^2+1+2z\right)}$

$=\underset{z\to 0}{lim}\frac{\sin \left(2z\right)+7z^2-2z}{z^4+z^2+2z^3}$

$L.H.Rule$

$=\underset{z\to 0}{lim}\frac{2\cos \left(2z\right)+14z-2}{4z^3+2z+6z^2}$

$L.H.Rule$

$=\underset{z\to 0}{lim}\frac{-2.2\sin \left(2z\right)+14}{12z+2+12z}$

$=\frac{14}{2}$

$=7$