Problem 1: Find f '(2) by the limit defn if f(x)=(2x-3)/(x-1).
f '(2) =lim_{h->0} (f(2+h)-f(2))/h=
=lim_{h->0} ((2(2+h)-3)/(1+h)-1)/h
lim_{h->0}h/(h*(1+h))=1.
Problem 2: (a) f(x)=(x-2)*e^x, so the product rule gives f '(x)=(x-2)*e^x+1*e^x.
(b) f(x)=\sqrt(x)/(x^2+1), so the quotient rule gives f '(x)= ((x^2+1)*(1/2*x^(-1/2))-\sqrt(x)*2x)/(x^2+1)^2.
Problem 3: (a) lim_{t->3} (t^2-2t-3)/(t-3)=lim_{t->3} (t+1)*(t-3)/(t-3)= lim_{t->3} (t+1)=4.
(b) lim_{x->\infty} (5+3x^7)/(\sqrt(x^14+8)+3x^7). The dominant term is x^7 in both the numberator and denominator, so we divide by x^7:
lim_{x->\infty} (5/x^7+3)/(\sqrt(1+8/x^14)+3) =3/(1+3)=3/4.
Problem 4: y=x^2*g(x) so by the product rule dy/dx= x^2*g '(x)+2x*g(x) which at x=2 is 2^2*3+2*2*5=32. So the slope of the tangent line is 32, and it passes through (2, 2^2*g(2))=(2,20), thus the equation of the tangent line is y-20=32*(x-2).
Problem 5: Sorry cannot draw one in html, but here is a descripton: Take a line of slope -1 for x less than 0, make a cusp at x=0, just like the absolute value function, then make an asymptote when x approaches 3, and finally have a maximum or minimum at x=5 so that the tangent line is horizontal and has slope =0.