1. (a) Find the slope of the tangent line to the curve y=x-x^3 at point (1,0) (i) using the 1st definition of a limit: lim(x->a)- (f(x)-f(a))/(x-a) (ii) using the 2nd equation of a limit: lim(h->a)- (f(a+h)-f(a))/h 3. The attempt at a solution In my attempt I got two different values (the same integer value, except one's a negative). i) (0-(x-x^3))/(x-1)= (-x+x^3)/(x-1) Then I factored an x out of the numerator which gives me a difference of squares. (x(x^2-1))/(x-1) = (x(x-1)(x+1))/(x-1) Then (x-1) cancels out =x(x+1) Plug in lim(x->1) =1(1+1)=2 Then I tried the other equatin: lim(h->0)- (f(a+h)-f(a))/h since 1 is the limit, we put 1 in as the a and it equals: ((1+h-(h+3)^3)-(1-1^3))/h and 1 minus 1 cubed=0 so we drop the minus sign at the end along with the zero and get: =(1+h-(h^3+3h^2+3h+1))/h =(1+h-h^3-3h^2-3h-1)/h =(-h^3-3h^2-2h)/h =(-h^2-3h-2) Then we plug in the lim(h->0) =(-0^2-3(0)-2 lim(h->0)=-2 So as you can see, by doing both equations I get 2 and -2, but only one is tangent to the curve and the other one crosses the graph but never touches it again. I re-did these problems three times to check my calculations and nothing seems off.