172 Solutions and Hints for Selected Exercises Exercise 4.5.5. (a) Observe that a simpler version of this problem can be stated as follows: If f is differentiable on (a,b) andx0 ∈(a,b), then lim h→0 f (x0 +h)−f (x0) h = f ′(x0) 1! . This conclusion follows directly from the definition of derivative. Similarly, if f is twice differentiable on(a,b) andx0 ∈(a,b), then lim h→0 f (x0 +h)−f (x0)−f ′(x0) h 1! h2 = f ′′(x0) 2! . We can prove this by applying the L’Hospital rule to get lim h→0 f (x0 +h)−f (x0)−f ′(x0) h 1! h2 =lim h→0 f ′(x0 +h)−f ′(x0) 2h = f ′′(x0) 2! . It is now clear that we can solve part (a) by using the L’Hospital rule as follows: lim h→0 f (x0 +h)−f (x0)−f ′(x0) h 1! −f ′′(x0) h2 2! h3 =lim h→0 f ′(x0 +h)−f ′(x0)−f ′′(x0) h 1! 3h2 = f ′′′(x0) 3! . Note that the last equality follows from the previous proof applied to the function f ′. (b) With the analysis from part (a), we see that if f is ntimes differentiable on (a,b) and x0 ∈(a,b), then lim h→0 f (x0 +h)−∑ n−1 k=0 f (k)(x 0)h k k! hn+1 = f (n)(x 0) n! . This conclusion can be proved by induction. This general result can be applied to obtain the Taylor expansion with Peano’s remainder in Exercise 4.5.6. SECTION 5.1 Exercise 5.1.3. Suppose A and B are compact subsets of R. Then, by Theorem 5.1.5, A and B are closed and bounded. From Theorem 5.1.2(iii) we get that A∪B is closed. Moreover, let MA,mA, MB,mB be upper and lower bounds for A and B, respectively. Then M=max{MA,MB} and m=min{mA,mB} are upper and lower bounds for A∪B. In particular, A∪Bis bounded. We have shown that A∪Bis both closed and bounded. It now follows from Theorem 5.1.5 that A∪Bis compact. SECTION 5.2 Exercise 1.5.1. Consider both sup f (I) and inf f (I). These may be real numbers of either ∞or −∞. When sup f (I) ∈Ror inf f (I) ∈Ryou will also need to determine whether they belong to f (I) or not. SECTION 5.3 Exercise 5.3.2. (a) Consider the sequence {xn}withxn = 1 π 2 +2nπ .
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