4. Evaluate each of the following littles. a. Jim - 2x 8x - Him x( X - 2X -8) _ Mm x(X-4) (x + 2) x -4 x - 4 ( x - 4) As x approaches 4 from elther side, x is positive, x + 2 is positive, did the two factors of x - 4 will have the same sign, so their quotient will be positive. So all the values inside the square root will be positive, so we may bring the limit inside the square root. Him * ( X 4) (x+2) flim (x(*+2))=4(4+2)-V24=206 e2x - 1 b. lim ( hint: try factoring the numerator!) 82x - 1 (e")-1 2-40 e " - 1 lim (e 1) (e+1) ex -1 = lim =lim(e* +1)=e' +1=2 e1 Section 3.1/3.2 Group Work Solutions 1. (Problem 1 from the 2.1/2.2 handout) Suppose that a ball is dropped from the top of the Seattle Space Needle, 605 ft above the ground. According to Galileo's law, the distance s(t) feet the ball has fallen after t seconds, neglecting air resistance, is s(t) = 16t a. Find s'(2) if s(t) = 16t using the limit definition of the derivative. To find s'(2) , we need to find s'(t) and then evaluate it at t=2. (16(t +h)?) -(16t2 ) s'(t) = lim s(t +h) - s(t)_1 - lim 161 +32th + 16h - 1812 = lim h - lim- h (32t+16h) - lim(32t+16h) =32t+16(0) = 32t so, s'(2)=32(2)=64 (Which is what we thought it might be..) b. Interpret your answer in the context of the problem. s'(2)=64 means that two seconds into the fall, the instantaneous velocity of the ball is 64 ft/sec. 2. If f (x)=4-3x3, find f'(x) using the limit definition of the derivative. f'(x) =lim /(xth)-f(x)_ (4-3(xth)-(4-3x3 lim 4 - 3x