Question

L'Hospital's Rule 3.78. Evaluate (a) lim 2z2+(3-40z-6i, 3.79. Evaluate (a) lim 22 +4 (b) zlim3 (z-em/3)(Z3 1), (c) lim za+2+1' z sin z (b).li.u--.n(2) xm Evaluate sin 2

Solve by using complex variables.

Answer 1

Lin z^2 + 4/2 z rightarrow 2i 2z^2 + (3-4i)z -6i (2i)^2 + 4/2(xi)^2 + (3o-4i) (2i) -6i = -4 + 4/-8 + 6i + 8-6 (0/0) form we can use L. Hospital rule lim_zrightarrow 2i (2z^2 + (3-4i) x-6i) = Lt_z rightarrow 2i 2z + O/4z + (3-4i) = 2(2i)/4(2i) + 3-4i = 4i/8i + 3 -4i = 4i/4i + 3 times 4i-3/4i-3 = 4i(4i-3)/(4i)^2-3^2 = 16i^2 -3^2 16;^2 -12i/-16-9 = 1/25 [12i + 16] lim_z rightarrow e^pi i/3 (z-e^pi i /3) (z/z^3 + 1) = e^pi i/3 -e^pi i/3 (e^pi i/3)/ (e^pi i/3)^3 + 1 (0/0) therefore e^i pi = -1\r\ suing L Hospital rule lim_z rightarrow e^pi i/3 (z-e^pi i/3)z/z^3 + 1 = lim_z rightarrow e^pi i/3 z- e^pi i/3 + z .(1)/3z^2 = lim_z rightarrow e^pi i/3 2z - e^pi i/2 2z -e^pi i/3/3z^2 = 2e^pi i/3 - e6pi i/3 i /3 (e^pi i.3)^2 = = e^pi i/3/3(e^pi i/3)^2 = 1/3e^pi i/3 e^pi i/3 = cos pi/3 + i sin pi/3 = 1/2 + i squareroot 3/2