MATH 180 Study Guide - Final Guide: Intermediate Value Theorem, Quotient Rule, Product Rule
Document Summary
Problem 1 solution: let c be a real number. Given f (x) = (cid:26) (x 1)2 x 3. Solution: (a) since x approaches 3 from the left we know that x < 3 so we use f (x) = (x 1)2. The limit is then lim x 3 (x 1)2 = (3 1)2 = 4 . (b) since x approaches 3 from the right we know that x > 3 so we use f (x) = 6 cx. The limit is then lim x 3+ (6 cx) = 6 c(3) = 6 3c . (c) in order for the two-sided limit to exist, the one-sided limits have to be the same. That is, lim x 3+ f (x) = lim x 3 f (x) 3 (d) since x < 3 we use f (x) = (x 1)2. This function is a polynomial and is continuous everywhere in its domain. Therefore, it is continuous at x = 2.
