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A Collection of Limits Contents 1 Short theoretical introduction 1 2 Problems 12 3 Solutions 23 2 Chapter 1 Short theoretical introduction Consider a sequence of real numbers (a n ) n≥1 , and l ∈ R. We’ll say that l represents the limit of (a n ) n≥1 if any neighborhood of l contains all the terms of the sequence, starting from a certain index. We write this fact as lim n→∞ a n = l, or a n → l. We can rewrite the above definition into the following equivalence: lim n→∞ a n = l ⇔ (∀)V ∈ V(l), (∃)n V ∈ N ∗ such that (∀)n ≥ n V ⇒ a n ∈ V . One can easily observe from this definition that if a sequence is constant then it’s limit is equal with the constant term. We’ll say that a sequence of real numbers (a n ) n≥1 is convergent if it has limit and lim n→∞ a n ∈ R, or divergent if it doesn’t have a limit or if it has the limit equal to ±∞. Theorem: If a sequence has limit, then this limit is unique. Proof: Consider a sequence (a n ) n≥1 ⊆ R which has two different limits l , l ∈ R. It follows that there exist two neighborhoods V ∈ V(l ) and V ∈ V(l ) such that V ∩ V = ∅. As a n → l ⇒ (∃)n ∈ N ∗ such that (∀)n ≥ n ⇒ a n ∈ V . Also, since a n → l ⇒ (∃)n ∈ N ∗ such that (∀)n ≥ n ⇒ a n ∈ V . Hence (∀)n ≥ max{n , n } we have a n ∈ V ∩ V = ∅. Theorem: Consider a sequence of real numbers (a n ) n≥1 . Then we have: (i) lim n→∞ a n = l ∈ R ⇔ (∀)ε > 0, (∃)n ε ∈ N ∗ such that (∀)n ≥ n ε ⇒ |a n −l| < ε. 1 2 A Collection of Limits (ii) lim n→∞ a n = ∞ ⇔ (∀)ε > 0, (∃)n ε ∈ N ∗ such that (∀)n ≥ n ε ⇒ a n > ε. (iii) lim n→∞ a n = −∞ ⇔ (∀)ε > 0, (∃)n ε ∈ N ∗ such that (∀)n ≥ n ε ⇒ a n < −ε Theorem: Let (a n ) n≥1 a sequence of real numbers. 1. If lim n→∞ a n = l, then any subsequence of (a n ) n≥1 has the limit equal to l. 2. If there exist two subsequences of (a n ) n≥1 with different limits, then the sequence (a n ) n≥1 is divergent. 3. If there exist two subsequences of (a n ) n≥1 which cover it and have a common limit, then lim n→∞ a n = l. Definition: A sequence (x n ) n≥1 is a Cauchy sequence if (∀)ε > 0, (∃)n ε ∈ N such that |x n+p − x n | < ε, (∀)n ≥ n ε , (∀)p ∈ N. Theorem: A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Theorem: Any increasing and unbounded sequence has the limit ∞. Theorem: Any increasing and bounded sequence converge to the upper bound of the sequence. Theorem: Any convergent sequence is bounded. Theorem(Cesaro lemma): Any bounded sequence of real numbers contains at least one convergent subsequence. Theorem(Weierstrass theorem): Any monotonic and bounded sequence is convergent. Theorem: Any monotonic sequence of real numbers has limit. Theorem: Consider two convergent sequences (a n ) n≥1 and (b n ) n≥1 such that a n ≤ b n , (∀)n ∈ N ∗ . Then we have lim n→∞ a n ≤ lim n→∞ b n . Theorem: Consider a convergent sequence (a n ) n≥1 and a real number a such that a n ≤ a, (∀)n ∈ N ∗ . Then lim n→∞ a n ≤ a. Theorem: Consider a convergent sequence (a n ) n≥1 such that lim n→∞ a n = a. Them lim n→∞ |a n | = |a|. Short teoretical introduction 3 Theorem: Consider two sequences of real numbers (a n ) n≥1 and (b n ) n≥1 such that a n ≤ b n , (∀)n ∈ N ∗ . Then: 1. If lim n→∞ a n = ∞ it follows that lim n→∞ b n = ∞. 2. If lim n→∞ b n = −∞ it follows that lim n→∞ a n = −∞. Limit operations: Consider two sequences a n and b n which have limit. Then we have: 1. lim n→∞ (a n + b n ) = lim n→∞ a n + lim n→∞ b n (except the case (∞, −∞)). 2. lim n→∞ (a n · b n ) = lim n→∞ a n · lim n→∞ b n (except the cases (0, ±∞)). 3. lim n→∞ a n b n = lim n→∞ a n lim n→∞ b n (except the cases (0, 0), (±∞, ±∞)). 4. lim n→∞ a b n n = ( lim n→∞ a n ) lim n→∞ b n (except the cases (1, ±∞), (∞, 0), (0, 0)). 5. lim n→∞ (log a n b n ) = log lim n→∞ a n ( lim n→∞ b n ). Trivial consequences: 1. lim n→∞ (a n − b n ) = lim n→∞ a n − lim n→∞ b n ; 2. lim n→∞ (λa n ) = λ lim n→∞ a n (λ ∈ R); 3. lim n→∞ k √ a n = k lim n→∞ a n (k ∈ N); Theorem (Squeeze theorem): Let (a n ) n≥1 , (b n ) n≥1 , (c n ) n≥1 be three se- quences of real numbers such that a n ≤ b n ≤ c n , (∀)n ∈ N ∗ and lim n→∞ a n = lim n→∞ c n = l ∈ R. Then lim n→∞ b n = l. Theorem: Let (x n ) n≥1 a sequence of real numbers such that lim n→∞ (x n+1 −x n ) = α ∈ R. 1. If α > 0, then lim n→∞ x n = ∞. 2. If α < 0, then lim n→∞ x n = −∞. 4 A Collection of Limits Theorem (Ratio test): Consider a sequence of real positive numbers (a n ) n≥1 , for which l = lim n→∞ a n+1 a n ∈ R. 1. If l < 1 then lim n→∞ a n = 0. 2. If l > 1 then lim n→∞ a n = ∞. Proof: 1. Let V = (α, β) ∈ V(l) with l < β < 1. Because l = lim n→∞ a n+1 a n , there is some n 0 ∈ N ∗ such that (∀)n ≥ n 0 ⇒ a n+1 a n ∈ V , hence (∀)n ≥ n 0 ⇒ a n+1 a n < 1. That means starting from the index n 0 the sequence (a n ) n≥1 is strictly decreasing. Since the sequence is strictly decreasing and it contains only positive terms, the sequence is bounded. Using Weierstrass Theorem, it follows that the sequence is convergent. We have: a n+1 = a n+1 a n · a n ⇒ lim n→∞ a n+1 = lim n→∞ a n+1 a n · lim n→∞ a n which is equivalent with: lim n→∞ a n (1 −l) = 0 which implies that lim n→∞ a n = 0. 2. Denoting b n = 1 a n we have lim n→∞ b n+1 b n = 1 l < 1, hence lim n→∞ b n = 0 which implies that lim n→∞ a n = ∞. Theorem: Consider a convergent sequence of real non-zero numbers (x n ) n≥1 such that lim n→∞ n x n x n−1 − 1 ∈ R ∗ . Then lim n→∞ x n = 0. Theorem(Cesaro-Stolz lemma): 1. Consider two sequences (a n ) n≥1 and (b n ) n≥1 such that: (i) the sequence (b n ) n≥1 is strictly increasing and unbounded; (ii) the limit lim n→∞ a n+1 − a n b n+1 − b n = l exists. Then the sequence a n b n n≥1 is convergent and lim n→∞ a n b n = l. Proof: Let’s consider the case l ∈ R and assume (b n ) n≥1 is a strictly increasing sequence, hence lim n→∞ b n = ∞. Now let V ∈ V(l), then there exists α > 0 such Short teoretical introduction 5 that (l − α, l + α) ⊆ V . Let β ∈ R such that 0 < β < α. As lim n→∞ a n b n = l, there exists k ∈ N ∗ such that (∀)n ≥ k ⇒ a n+1 − a n b n+1 − b n ∈ (l − β, l + β), which implies that: (l −β)(b n+1 − b n ) < a n+1 − a n < (l + β)(b n+1 − b n ), (∀)n ≥ k Now writing this inequality from k to n − 1 we have: (l −β)(b k+1 − b k ) < a k+1 − a k < (l + β)(b k+1 − b k ) (l −β)(b k+2 − b k+1 ) < a k+2 − a k+1 < (l + β)(b k+2 − b k+1 ) . . . (l −β)(b n − b n−1 ) < a n − a n−1 < (l + β)(b n − b n−1 ) Summing all these inequalities we find that: (l −β)(b n − b k ) < a n − a k < (l + β)(b n − b k ) As lim n→∞ b n = ∞, starting from an index we have b n > 0. The last inequality rewrites as: (l −β) 1 − b k b n < a n b n − a k b n < (l + β) 1 − b k b n ⇔ ⇔ (l − β) + a k + (β − l)b k b n < a n b n < l + β + a k − (β + l)b k b n As lim n→∞ a k + (β − l)b k b n = lim n→∞ a k − (β + l)b k b n = 0 there exists an index p ∈ N ∗ such that (∀)n ≥ p we have: a k + (β − l)b k b n , a k − (β + l)b k b n ∈ (β −α, α −β) We shall look for the inequalities: a k + (β − l)b k b n > β −α and a k − (β + l)b k b n < α −β 6 A Collection of Limits Choosing m = max{k, p}, then (∀)n ≥ m we have: l −α < a n b n < l + α which means that a n b n ∈ V ⇒ lim n→∞ a n b n = l. It remains to prove the theorem when l = ±∞, but these cases can be proven analogous choosing V = (α, ∞) and V = (−∞, α), respectively. 2. Let (x n ) n≥1 and (y n ) n≥1 such that: (i) lim n→∞ x n = lim n→∞ y n = 0, y n = 0, (∀)n ∈ N ∗ ; (ii) the sequence (y n ) n≥1 is strictly decreasing; (iii) the limit lim n→∞ x n+1 − x n y n+1 − y n = l ∈ R. Then the sequence x n y n n≥1 has a limit and lim n→∞ x n y n = l. Remark: In problem’s solutions we’ll write directly lim n→∞ x n y n = lim n→∞ x n+1 − x n y n+1 − y n , and if the limit we arrive to belongs to R, then the application of Cesaro-Stolz lemma is valid. Trivial consequences: 1. Consider a sequence (a n ) n≥1 of strictly positive real numbers for which exists lim n→∞ a n+1 a n = l. Then we have: lim n→∞ n √ a n = lim n→∞ a n+1 a n Proof: Using Cesaro-Stolz theorem we have: lim n→∞ (ln n √ a n ) = lim n→∞ ln a n n = lim n→∞ ln a n+1 − ln a n (n + 1) −n = lim n→∞ ln a n+1 a n = ln l Then: lim n→∞ n √ a n = lim n→∞ e ln n √ a n = e lim n→∞ (ln n √ a n ) = e ln l = l 2. Let (x n ) n≥1 a sequence of real numbers which has limit. Then: lim n→∞ x 1 + x 2 + . . . + x n n = lim n→∞ x n Short teoretical introduction 7 3. Let (x n ) n≥1 a sequence of real positive numbers which has limit. Then: lim n→∞ n √ x 1 x 2 . . . x n = lim n→∞ x n Theorem (Reciprocal Cesaro-Stolz): Let (x n ) n≥1 and (y n ) n≥1 two se- quences of real numbers such that: (i) (y n ) n≥1 is strictly increasing and unbounded; (ii) the limit lim n→∞ x n y n = l ∈ R; (iii) the limit lim n→∞ y n y n+1 ∈ R + \{1}. Then the limit lim n→∞ x n+1 − x n y n+1 − y n exists and it is equal to l. Theorem (exponential sequence): Let a ∈ R. Consider the sequence x n = a n , n ∈ N ∗ . 1. If a ≤ −1, the sequence is divergent. 2. If a ∈ (−1, 1), then lim n→∞ x n = 0. 3. If a = 1, then lim n→∞ x n = 1. 4. If a > 1, then lim n→∞ x n = ∞. Theorem (power sequence): Let a ∈ R. Consider the sequence x n = n a , n ∈ N ∗ . 1. If a < 0, then lim n→∞ x n = 0. 2. If a = 0, then lim n→∞ x n = 1. 3. If a > 0, then lim n→∞ x n = ∞. Theorem (polynomial sequence): Let a n = a k n k + a k−1 n k−1 + . . . + a 1 n + a 0 , (a k = 0). 1. If a k > 0, then lim n→∞ a n = ∞. 2. If a k < 0, then lim n→∞ a n = −∞. 8 A Collection of Limits Theorem: Let b n = a k n k + a k−1 n k−1 + . . . + a 1 n + a 0 b p n p + b p−1 n p−1 + . . . + b 1 n + b 0 , (a k = 0 = b p ). 1. If k < p, then lim n→∞ b n = 0. 2. If k = p, then lim n→∞ b n = a k b p . 3. If k > p, then lim n→∞ b n = a k b p · ∞. Theorem: The sequence a n = 1 + 1 n n , n ∈ N ∗ is a strictly increasing and bounded sequence and lim n→∞ a n = e. Theorem: Consider a sequence (a n ) n≥1 of real non-zero numbers such that lim n→∞ a n = 0. Then lim n→∞ (1 + a n ) 1 a n = e. Proof: If (b n ) n≥1 is a sequence of non-zero positive integers such that lim n→∞ b n = ∞, we have lim n→∞ 1 + 1 b n b n = e. Let ε > 0. From lim n→∞ 1 + 1 n n = e, it follows that there exists n ε ∈ N ∗ such that (∀)n ≥ n ε ⇒ 1 + 1 n n − e < ε. Also, since lim n→∞ b n = ∞, there exists n ε ∈ N ∗ such that (∀)n ≥ n ε ⇒ b n > n ε . Therefore there exists n ε = max{n ε , n ε } ∈ N ∗ such that (∀)n ≥ n ε ⇒ 1 + 1 b n b n − e < ε. This means that: lim n→∞ 1 + 1 b n b n = e. The same property is fulfilled if lim n→∞ b n = −∞. If (c n ) n≥1 is a sequence of real numbers such that lim n→∞ c n = ∞, then lim n→∞ 1 + 1 c n c n = e. We can assume that c n > 1, (∀)n ∈ N ∗ . Let’s denote d n = c n ∈ N ∗ . In this way (d n ) n≥1 is sequence of positive integers with lim n→∞ d n = ∞. We have: d n ≤ c n < d n + 1 ⇒ 1 d n + 1 < 1 c n ≤ 1 d n Hence it follows that: 1 + 1 d n + 1 d n < 1 + 1 c n d n ≤ 1 + 1 c n c n < 1 + 1 c n d n +1 ≤ 1 + 1 d n d n +1 Observe that: [...]... convergent and evaluate xn+1 (a − xn ) > 4 lim xn n→∞ 73 Evaluate: lim cos nπ n→∞ √ 2n e 74 Evaluate: lim n→∞ n+1 n tan (n−1)π 2n 20 A Collection of Limits 75 Evaluate: n lim n n→∞ k=1 n k 76 If a > 0, evaluate: a+ lim √ √ 3 a+ n→∞ a + + ln n √ n a n 77 Evaluate: π π + 4 n 78 Let k ∈ N and a0 , a1 , a2 , , ak ∈ R such that a0 + a1 + a2 + + ak = 0 Evaluate: lim n ln tan n→∞ √ √ √ 3 a0 3 n + a1 3 n... be a sequence of positive real numbers such that x1 > 0 and a 3xn = 2xn−1 + 2 , where a is a real positive number Prove that xn is xn−1 convergent and evaluate lim xn n→∞ 54 Consider a sequence of real numbers (an )n≥1 such that a1 = 12 and an+1 = 3 an 1 + Evaluate: n+1 n lim n→∞ k=1 1 ak 18 A Collection of Limits 55 Evaluate: √ lim n→∞ n n n2 + 1 56 If a ∈ R, evaluate: n lim n→∞ k=1 k2 a n3 57 Evaluate:... lim n→∞ nan n ak = lim k=1 n→∞ 1 n+3 = 2n + 2 2 1 (an + 3 2 an−1 + b), where 0 ≤ b < 1 Prove that the sequence is convergent and evaluate lim an 4 Consider the sequence (an )n≥1 such that a1 = a2 = 0 and an+1 = n→∞ b Solution: We have a2 − a1 = 0 and a3 − a2 = ≥ 0, so assuming an−1 ≥ an−2 3 and an ≥ an−1 , we need to show that an+1 ≥ an The recurrence equation gives us: 1 (an − an−1 + a2 − a2 ) n−1... x2 < 1 and xn+1 = xn + n , (∀)n ≥ 1 1 n2 1 1 Prove that the sequences (xn )n≥1 and (yn )n≥2 , yn = − are convergent xn n − 1 97 Evaluate: n lim n→∞ sin i=1 2i n2 98 If a > 0, a = 1, evaluate: xx − ax x a ax − aa lim 99 Consider a sequence of positive real numbers (an )n≥1 such that an+1 − 1 1 = an + , (∀)n ≥ 1 Evaluate: an+1 an 1 1 1 1 lim √ + + + n→∞ a2 an n a1 100 Evaluate: 2arctan x − 2arcsin x... Let an , bn ∈ Q such that (1 + an lim n→∞ bn √ √ 2)n = an + bn 2, (∀)n ∈ N∗ Evaluate 35 If a > 0, evaluate: (a + x)x − 1 x→0 x lim 36 Consider a sequence of real numbers (an )n≥1 such that a1 = 3 and an+1 = 2 a2 − an + 1 n Prove that (an )n≥1 is convergent and find it’s limit an 37 Consider a sequence of real numbers (xn )n≥1 such that x0 ∈ (0, 1) and xn+1 = xn − x2 + x3 − x4 , (∀)n ≥ 0 Prove that... of the sequence: xn+1 = xn + a , n ≥ 1, x1 ≥ 0, a > 0 xn + 1 69 Consider two sequences of real numbers (xn )n≥0 and (yn )n≥0 such that x0 = y0 = 3, xn = 2xn−1 + yn−1 and yn = 2xn−1 + 3yn−1 , (∀)n ≥ 1 Evaluate xn lim n→∞ yn 70 Evaluate: lim x→0 tan x − x x2 71 Evaluate: lim x→0 tan x − arctan x x2 72 Let a > 0 and a sequence of real numbers (xn )n≥0 such that xn ∈ (0, a) and a2 , (∀)n ∈ N Prove that... Prove that this sequence is an arithmetical progression and evaluate: 3 Consider the sequence (an )n≥1 , such that 23 ak = (n3 − 1)2 2 n 2 Evaluate: √ 3 5x + 2 + 2 lim √ = lim x→−2 3x + 10 − 2 x→−2 3 1+ 2 n + 1 n3 + 3 1− 1 2 n3 24 A Collection of Limits 1 n→∞ nan n lim ak k=1 Solution: For n = 1 we get a1 = 6 Then a1 + a2 = 15, so a2 = 9 and the ratio is r = 3 Therefore the general term is an = 6 +... , evaluate: lim x→0 1 − cos x · cos 2x · · cos nx x2 93 Consider a sequence of real numbers (xn )n≥1 such that xn is the real root of the equation x3 + nx − n = 0, n ∈ N∗ Prove that this sequence is convergent and find it’s limit 94 Evaluate: arctan x − arctan 2 x→2 tan x − tan 2 lim 95 Evaluate: 22 A Collection of Limits lim 1+ √ 22 2! + √ 32 n→∞ 3! + + n √ n2 n! x2 96 Let (xn )n≥1 such that... cos a x2 − a2 Solution: −2 sin x +a · sin x a cos x − cos a 2 2 = lim x→ a x→ a x2 − a2 (x − a) (x + a) sin x +a sin x a 2 2 = lim · lim x +a x→ a x→ a a − x 2 lim sin x a 2 x→ a a − x sin a =− 2a = lim 14 If n ∈ N∗ , evaluate: ln(1 + x + x2 + + xn ) x→0 nx lim Solution: Using lim x→0 ln(1 + x) = 1, we have: x αj−1 αj n +1+ αj+1 αj n + + αp αj n 28 A Collection of Limits x + x2 + + xn ln(1 + x +... 27 Evaluate: lim x→0 1 + sin2 x − cos x √ 1 − 1 + tan2 x Problems 15 28 Evaluate: √ x+ x √ x− x lim x→∞ x 29 Evaluate: 1 lim (cos x) sin x x→0 x>0 30 Evaluate: 1 lim (ex + sin x) x x→0 31 If a, b ∈ R∗ , evaluate: + lim n→∞ a 1+ a √ n n b 32 Consider a sequence of real numbers (an )n≥1 defined by: 1 if n ≤ k, k ∈ N∗ (n + 1)k − nk an = if n > k n k−1 i)Evaluate lim an n→∞ n k · lim an , evaluate: . n→∞ a n b n . 35. If a > 0, evaluate: lim x→0 (a + x) x − 1 x 36. Consider a sequence of real numbers (a n ) n≥1 such that a 1 = 3 2 and a n+1 = a 2. convergent and evaluate lim n→∞ x n . 54. Consider a sequence of real numbers (a n ) n≥1 such that a 1 = 12 and a n+1 = a n 1 + 3 n + 1 . Evaluate: lim