Question

How do you do this question?

Answer 1

Answer:

B. 1/2

Step-by-step explanation:

$\lim_{z \to 0} \frac{g(z)e^{-z}-3}{z^{2}-2z}$

If we plug in 0 for z, we get 0/0.  Apply l'Hopital's rule.

$\lim_{z \to 0} \frac{-g(z)e^{-z}+g'(z)e^{-z}}{2z-2}$

Now when we plug in 0 for z, we get:

$\frac{-g(0)e^{0}+g'(0)e^{0}}{2(0)-2}\\\frac{-g(0)+g'(0)}{-2}\\\frac{-3+2}{-2}\\\frac{1}{2}$