I was wondering if someone could help me with the problems within the picture:
I am in dire need of a step by step explanation as I answered this as -1 and was incorrect
couldn't f be a radical like this (-x+1)/(x-1) so that the horizontal asymptote would be y =-1 and so the derivative of that function would be 0 so the limit of the derivative would also be zero. I think I got. Well, I hope so. Can someone confirm this please
You are right - more or less:
1. The example you gave is a constant function with a hole at x = 1. Better use $\displaystyle f(x)=\dfrac{1-x}x$ (as an example of course!)
2. If an asymptote y = -1 exists then the function can be replaced by f(x) = -1 for very large values of x.
3. If f(x) = -1 then f'(x) = 0