Problem 1: Find f '(2) by the limit defn if f(x)=(4-5x)/(x).
f '(2) =lim_{h->0} (f(2+h)-f(2))/h=
=lim_{h->0} ((4-5(2+h))/(2+h)-(-3))/h
=lim_{h->0}(-6-5h+6+3h)/(h*(2+h))= -1.
Problem 2: (a) f(x)=(e^x-x)e^x, so the product rule gives f '(x)=(e^x-1)*e^x+(e^x-x)*e^x.
(b) f(x)=(x^2+2^4)/\sqrt{x}, so the quotient rule gives f '(x)= (\sqrt{x}*(2x)-(x^2+2^4)*1/2*x^(-1/2))/\sqrt(x)^2.
Problem 3: (a) \lim_{h->0} ((1+h)^{11}-1)/h is the derivative of the function f(x)=x^11 at x=1, thus is 11. Or you can see that (1+h)^11=1+11*h+higher order terms in h, so the limit is 11.
(b) \lim_{t->\infty} (-2t+3t^5)(\sqrt{t^{10}+t}+4t^5). The dominant term is t^5 in both the numerator and denominator, so we divide by t^5:
lim_{t->\infty} (-2/t^4+3)/(\sqrt(1+1/t^9)+4) =3/(1+4)=3/5.
Problem 4: Suppose that h(x) is differentiable at x=3, h(3)=4, h'(3)=5. Find the equation of the tangent line to the graph y=h(x)/x at x=3.
Use the quotient rule to find dy/dx=(x*h '(x)-h(x))/x^2, which at x=3 is (3*5-4)/3^2=11/9. if x=3, y=4/3, so the equation of the tangent line is y-4/3=11/9*(x-3).
Problem 5: The curve increases until x=1, then has a cusp at x=1, and increases (or decreases) to the vertical asymptote at x=2. For x>2, there is a horizontal tangent line at x=3.