214 - 1 If A does not equal 0, f is bounded away from 0 in a neighorhood of c by continuity, hence since B = 0, lim(x goes to c) [f/g] does not exist (blows up). There hint suggests the perhaps simpler argument that the limit of the product is the product of the limits when the limits exist, and anything times 0 is 0. 6a Taking the derivatives of the numerator and denominator produces [1/(x+1)]/[cos(x)] which is 1/1 = 1 8b Taking the derivatives: [1/x]/[.5 x^(-.5)] = 2/x^(.5) which goes to 0 as x goes to infinity. 226 - 3 We shall use C(n,k) to denote combinations. Since C(1,0) = C(1,1) = 1, the case n=1 is just the product rule. assuming the equation is true for n, taking the derivative of the left hand side entails applying the product rule to each summand on the right hand side. Terms involving f^(n+1-k)g^(k) will result only from taking the derivative of f^(n-k)g^(k) and f^(n+1-k)g^(k-1). Since C(n,k) + C(n,k-1) = C(n+1,k) the induction step has been demonstrated. 9 The remainder term in Taylor's theorem is [f^(n+1)(c)][(x-x0)^(n+1)]/[(n+1)!]. Since the first factor (the derivative) is bounded by 1, the fact that a^n/n! goes to zero for a fixed provides the desired result. 14c Heuristically, you are looking at x + .17x^3 which is an odd function, so there is no extremum at 0. Or we can cite Thm. 6.4.4, since the the first non-zero derivative is the first derivative.

Sun Nov 16 1997