Answer:
B. 1/2
Step-by-step explanation:
[tex]\lim_{z \to 0} \frac{g(z)e^{-z}-3}{z^{2}-2z}[/tex]
If we plug in 0 for z, we get 0/0. Apply l'Hopital's rule.
[tex]\lim_{z \to 0} \frac{-g(z)e^{-z}+g'(z)e^{-z}}{2z-2}[/tex]
Now when we plug in 0 for z, we get:
[tex]\frac{-g(0)e^{0}+g'(0)e^{0}}{2(0)-2}\\\frac{-g(0)+g'(0)}{-2}\\\frac{-3+2}{-2}\\\frac{1}{2}[/tex]
