MATH116 Lecture 19: M116F15_P4sol
9 Dec 2015
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Solution: (cid:16) (cid:17) x2 + 2x x + (b) lim x (a) divide the top and bottom by y2, which is the highest power of y appearing: = lim y (cid:16) 1 (cid:16) 1 y2 (cid:17) (cid:17) = lim y (cid:17) (cid:16) 2 3y2 (cid:17) = lim (cid:16) 5y2+4y y y2 y2 y2. 5 (b) since this limit seems to be of the type + , we must do something to turn it (x into another type of limit. We will proceed by multiplying (and dividing) by the conjugate, (cid:16) x2 + 2x): x2 + 2x x + (cid:17) lim x (cid:16) x + x2 + 2x. = 1: sketch the graph of a function g for which g(0) = g(cid:48)(0) = 0, g(cid:48)( 1) = 1, g(cid:48)(1) = 3, and g(cid:48)(2) = 1. Solution: find the derivative of the function using the de nition of derivative.
