sole9 - ANALYSIS EXERCISE 9–SOLUTIONS 1. (a) Let lim f...

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Unformatted text preview: ANALYSIS EXERCISE 9–SOLUTIONS 1. (a) Let lim f (x) = A. By the definition x→ a (∀ε > 0)(∃δ > 0)(∀x ∈ D(f )) (0 < |x − a| < δ ) ⇒ (|f (x) − A| < ε) . Denote h = x − a, then x = a + h. We obtain (∀ε > 0)(∃δ > 0)(∀a + h ∈ D(f )) (0 < |h| < δ ) ⇒ (|f (a + h) − A| < ε) . (b) Write down the definition again to see this. 2. 0.1 Let us prove, for instance,(ii) using the Heine definition of the limit. Let xn → a and (∀n ∈ N)(xn = a). Then lim f (xn ) = A n→∞ and lim g (xn ) = B . By the corresponding theorem for limits n→∞ of sequences we have that n→∞ lim f (xn )g (xn ) = lim f (xn ) lim g (xn ) = AB. n→∞ n→∞ 0.2 Let ε > 0. Choose δ1 such that (∀x ∈ D(f )) (0 < |x − a| < δ1 ) ⇒ (|f (x) − A| < ε/2) . Choose δ2 such that (∀x ∈ D(g )) (0 < |x − a| < δ2 ) ⇒ (|g (x) − B | < ε/2) . Let γ = min{δ, δ1 , δ2 }. Then (∀x ∈ (D(f )∩D(g ))) (0 < |x−a| < γ ) implies A − B = A − f (x) + f (x) − g (x) + g (x) − B ≤ A − f (x) + g (x) − B < ε. Therefore we have proved that (∀ε > 0) (A − B < ε), which implies A ≤ B . 3. From the definition (∀ε > 0)(∃δ > 0)(∀x ∈ D(f )) (0 < |x| < δ ) ⇒ (|f (x)| < ε) . (a) Fix ε > 0. Take δ1 = δ/2 (δ as above). Then (∀x ∈ D(f )) (0 < |x| < δ1 ) ⇒ (|f (2x)| < ε) . √ (b) Fix ε > 0. Take δ2 = δ (δ as above). Then (∀x ∈ D(f )) (0 < |x| < δ2 ) ⇒ (|f (x2 )| < ε) . 4. (a) 2. (b) x→1 lim x−1 = 0. x(x + 1) 1 2 ANALYSIS EXERCISE 9–SOLUTIONS (c) x→1 √ lim ( x + 1) = 2. √1 . M 5. (a) Fix M > 0. Take δ = (∀x ∈ D(f )) Then 1 >M (x − 3)2 . 0 < |x − 3| < δ ⇒ x→a+ (b) We say that lim f (x) = +∞ if (∀M ∈ R)(∃δ > 0)(∀x ∈ D(f )) Fix M > 0.Take δ = (∀x ∈ D(f )) 1 M. 0<x−a<δ ⇒ Then 1 >M x f (x) > M . 0<x<δ ⇒ . ...
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sole9 - ANALYSIS EXERCISE 9–SOLUTIONS 1. (a) Let lim f...

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