128 - 2c the limit is 2, after expanding the suare and algebraic simplification including cancelling the x's in the numerator and denominator (which is allowed since the limit ignores what happens at the limit point), one must use only the limit of the sum is the sum of the limit. - 5 Since f is bounded in a neighborhood of c, there exists a K and a delta_k such that |f| LT K if |x-c| LT delta_k. Choose epsilon GT 0. Since lim_(x goto c) g = 0, there exists a delta_(epsilon/K) such that |g| LT epsilon/K if 0 LT |x-c| LT delta_(epsilon/K). Hence if we let delta + min(delta_K, delta_(epsilon/K), |fg| LT epsilon if 0 LT |x-c| LT delta. - 11d sin is bounded and x^.5 goes to zero, hence the limit is 0. 139 - 3 Showing the limit is +_ infinity will shoo that both the right hand and left hand limits are equal to that. Choose K. Let delta_k = 1/K^2 If 0 LT |x| LT 1/K^2, 1/|x|^.5 GT K. 5h The limit is -1, since x grows more rapidly than x^.5 9 lim_(x goto inf) xf(x) = L entails that xf(x) LT (L + epsilon) for x sufficiently large (what is sufficiently large depends on epsilon). If we choose epsilon = L/10, xf(x) LT (1.1)L, hence f(x) LT (1.1)L/x. This is arbitrarily small for x arbitrarily large.

Sun Oct 19 1997