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) and x0 ∈ (a,b), then f (x0 + h) − f (x0) f ′ (x0) lim = . h→0 h 1! This conclusion follows directly from the defnition of derivative. Similarly, if f is twice differentiable on (a,b) and x0 ∈ (a,b), then ′ (x0) h ′′ (x0) f (x0 + h) − f (x0) − f 1! f lim = . h2 h →0 2! We can prove this by applying the L’Hospital rule to get ′ (x0) h ′′ (x0) f (x0 + h) − f (x0) − f f ′ (x0 + h) − f ′ (x0) f 1! lim = lim = . h2 h →0 h→0 2h 2! It is now clear that we can solve part (a) by using the L’Hospital rule as follows: ′′ (x0) h2 f (x0 + h) − f (x0) − f ′ (x0) h f ′ (x 0 + h) − f ′ (x0) − f ′′ (x0) h f ′′′ (x0) 1! − f 2! 1! lim = lim = . h3 3h2 h →0 h→0 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 n times differentiable on (a,b) and x0 ∈ (a,b), then f (k)(x 0)h k f (x0 + h) − ∑ n−1 k=0 f (n)(x 0) k! lim = . hn+1 h →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 ∪ B is bounded. We have shown that A ∪ B is both closed and bounded. It now follows from Theorem 5.1.5 that A ∪ B is 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) ∈ R or inf f (I) ∈ R you will also need to determine whether they belong to f (I) or not. SECTION 5.3 1 Exercise 5.3.2. (a) Consider the sequence {xn} with xn = π 2 + 2nπ .
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