| On that problem, you have to substitute in the initial conditions right after using implicit differentiation. So, I remember this from the test, I had:
-1/x + c= ln|y-1| Plugging in (2,0):
-.5 + c = ln|-1|
c=.5 (ln(1) = 0)
And then, you can solve for y.
That gave you... umm.. (1/2)e^(-1/x) + 1 i THINK, I am doing this from memory so I don't remember exactly).
And, so the next limit question, the answer was 3/2. Limit as x->infinity e^0 is 1, so 1/2 + 1 = 3/2. |