Two reasonably hard limit questions

a) If lim (x->infinity) g(x) = 6 and lim (x->infinity) g(x)/f(x)-1 = 3, find lim (x->infinity) f(x)

b) If a function y = f(x) is differentiable at x = 3 and f'(3) = 5, find the limit
lim (x->3) (x^2 - 3x) / ( f(3) - f(x) )

Cannot fathom how to do these limit questions, would be great if someone could show the steps on how it would be done, thank you!

a)
as x approaches ∞, g(x)/f(x)-1 approaches 3.
Since at ∞, g(x) goes to 6, you can say as x→∞, 6/f(x)-1 approaches 3.
This means that at ∞, f(x) must approach (6/3)+1 = 3

So
lim f(x) = 3
x→∞

b)

By definition:

lim (f(x) - f(3))/(x-3) = f ' (3) = 5
x→3

Multiply this by -1

lim (f(3) - f(x))/(x-3) = -5
x→3

This means that

lim (x-3)/(f(3)-f(x)) = -1/5 ....(1)
x→3

We also know that

lim x = 3 ....(2)
x→3


Multiply (1) by (2) to get

lim x(x-3)/(f(3)-f(x)) = -3/5
x→3

Or equivalently

lim (x²-3x)/(f(3)-f(x)) = -3/5
x→3