By E. Hohne

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**Extra resources for Crystal Structure Analysis using Superposition, Complimentary Structures [short article]**

**Example text**

61. Take n = 17. 9 that there exists a δ-neighborhood Vδ (w) such that f (x) < 0 for all x ∈ Vδ (w). But since w < b, Chapter 5 — Continuous Functions 12. 13. 14. 15. 16. 17. 18. 19. 37 this contradicts the fact that w = sup W . There is a similar contradiction if we assume that f (w) > 0. Therefore f (w) = 0. Since f (π/4) < 1 while f (0) = 1 and f (π/2) > 1, it follows that x0 ∈ (0, π/2). If cos x0 > x20 , then there exists a δ-neighborhood Vδ (x0 ) ⊆ I on which f (x) = cos x, so that x0 is not an absolute minimum point for f .

Thus, if f is continuous at x0 , then it is continuous at 0, and hence everywhere. 13. First show that f (0) = 0 and (by Induction) that f (x) = cx for x ∈ N, and hence also for x ∈ Z. Next show that f (x) = cx for x ∈ Q. Finally, if x ∈ / Q, let x = lim(rn ) for some sequence in Q. 14. First show that either g(0) = 0 or g(0) = 1. Next, if g(α) = 0 for some α ∈ R and if x ∈ R, let y := x − α so that x = α + y; hence g(x) = g(α + y) = g(α)g(y) = 0. Thus, if g(α) = 0 for some α, then it follows that g(x) = 0 for all x ∈ R.

If we substitute y := 1/n, we get f (1/n) = f (1)/n, whence L = lim f (x) = lim(f (1/n)) = 0. Since f (x) − f (c) = x→0 31 Chapter 4 — Limits f (x − c), we infer that lim (f (x) − f (c)) = lim f (x − c) = lim f (z) = 0, so that x→c x→c z→0 lim f (x) = f (c). x→c 13. (a) g(f (x)) = g(x + 1) = 2 if x = 0, so that lim g(f (x)) = 2, but g( lim f (x)) = x→0 x→0 g(f (0)) = g(1) = 0. Not equal. (b) f (g(x)) = g(x) + 1 = 3 if x = 1, so that lim f (g(x)) = 3, and f ( lim g(x)) = x→1 x→1 f (2) = 3. Equal.