Answers to the review problems given Oct. 6 in class(^ indicates exponent):

1.(a) x^2-7x+10 = (x-5)(x-2). Cancelling, the limit is 1/3. (b) Divide top and bottom by x^3. The
numerator is then 3+2/x^2+1/x^3 which tens to 3in the limit. The denominator is |x^3|/x|3| +1/x^2
which tens to -1 in the limit. Thus the limit is -3. (c) Dividing numerator and denominator by theta,
the numerator is 1 and denominator is 1+tan(theta)/theta. Since tan is sin over cos, and 
sin(theta)/theta tends to 1, as does cos(theta), as theta tends to zero, the limit of the function is 1/2. 

2. 3x^2 -2x-5 tends to 3 as x tends to 2, as does x+1. So if we define c=3, the
function will be continuous at x=2.

3. The instantaneous velocity is f'(t)= 24-32t, and so it will be zero at time t=3/4. When t=2 
the instantaneous velocity is 24-64= -40.

4. 3 (sqrt(x+h)-sqrt(x) )= 3 (x+h-x)/(sqrt(x+h)+sqrt(x)). Dividing by h and taking the limit,
we get 3/2sqrt(x) for the derivative.

5. (a) f'(x)= -csc^2(x). (b) Use quotient rule. (c) f'(x)= e^x (cos x - sin x).