Question

How do you do this question?

Answer 1

B. 1/2

Explanation:

`\lim_(z \to 0) (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) (-g(z)e^(-z)+g'(z)e^(-z))/(2z-2)`

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

`(-g(0)e^(0)+g'(0)e^(0))/(2(0)-2)\\(-g(0)+g'(0))/(-2)\\(-3+2)/(-2)\\(1)/(2)`