f(3(-2)-1)/(-2)^2 - 4 = f(-7)/4 - 4 = 0 - 4 = -4
Now, if you meant
f(3x-1)/(x^2-4) then that would be 0/0 which is a problem.
But, l'Hôpital's Rule says that the limit is the same as
d/dx f(3x-1) / d/dx (x^2-4)
= (f'(3x-1) * 3) / (2x)
= (3*3) / -6
= -3/2
see what you can do with the other one
The table below gives selected values of a twice differentiable function f(x)
x|. -7. -6. -4. -2.
f(x)|. 0. -1. -2. 0
f'(x)|. 3. 2. -1. 7
What is the limit of f(3x-1)/(x^2)-4 as x approaches -2?
What is the limit of f(f(x))/5x + 20 as x approaches -4?
1 answer
1 answer