MA2108S Tutorial 5 Solution

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1 MA08S Tutorial 5 Solutio Prepared by: LuJigyi LuoYusheg March 0 Sectio 3. Questio 7. Let x := / l( + ) for N. (a). Use the difiitio of limit to show that lim(x ) = 0. Proof. Give ay ɛ > 0, sice ɛ > 0, e ɛ >. By the Archimedea Property, K N such that K > e ɛ. i.e. l(k + ) > ɛ. i.e. l(k+) < ɛ. Hece l(+) 0 = l(+) l(k+) < ɛ K. Hece lim(x ) = 0. 7(b). Fid a specific value of K(ɛ) as required i the defiitio of limit for each of (i) ɛ = /, ad (ii) ɛ = /0. (i) For ɛ =, take K(ɛ) = 7. The 0 l(+) < < ɛ K(ɛ). l(k(ɛ)+) l(k(ɛ)+) < ɛ. Hece by part(a) (ii) For ɛ =, take K(ɛ) = 06. The 0 0 l(+) < < ɛ K(ɛ). l(k(ɛ)+) l(k(ɛ)+) < ɛ. Hece by part(a) Questio 8. Prove that lim(x ) = 0 if ad oly if lim( x ) = 0. Give a example to show that the covergece of ( x ) eed ot imply the comvergece of (x ).

2 Proof. We ca see that lim(x ) = 0 ɛ > 0, K N such that x 0 = x = x 0 < ɛ K lim( x ) = 0. Hece lim(x ) = 0 if ad oly if lim( x ) = 0 The covergece of ( x ) eed ot imply the covergece of (x ). Example: x := ( ). The lim( x ) =. But (x ) does ot coverge. Questio 9. Show that if x 0 for all N ad lim(x ) = 0, the lim( x ) = 0. Proof. For all ɛ > 0, sice ɛ > 0 ad lim(x ) = 0, K N such that x < ɛ K. Sice x > 0 for all N, x = x < ɛ K. This implies that x < ɛ K. Hece by defiitio, lim( x ) = 0. Questio 0. Prove that if lim(x ) = x ad if x > 0, the there exists a atural umber M such that x > 0 for all M. Proof. Sice lim(x ) = x > 0, take ɛ = x > 0, the M N such that x x < x M. Hece x x < x < x + x M. Hece x > 0 M. Questio. Show that lim( + ) = 0. Proof. N, observe that + 0 = (+) = + < Archimedea Property, K N such that K < ɛ. Hece + 0 < Hece by defiitio, lim( + ) = 0. Questio. Show that lim( 3 ) = 0.. ɛ > 0, by the < ɛ K. Proof. Observe that for all, 0 3 = = (by the Beroulli s Iequality)< 3 (+) +. Give ay ɛ > 0, sice ɛ > 0, by the Archimedea Property, K N such that < ɛ, K which implies that K < ɛ. Hece 3 0 < K < ɛ lim( 3 ) = 0. Questio 5. Show that lim(! ) = 0. K. Hece by defiitio,

3 Proof. Observe that 0! = < = < = > 3.! ( )( ) ɛ > 0, by the Archimedea Property, K N such that < ɛ. The 0 K! < 3 < ɛ K K + 3. Hece by defiitio, lim( ) = 0.! Questio 6. Show that lim(! ) = 0. Proof. For all 3, observe that! 0 = ( ) 3 3 = ( 3). Sice <, let = where h > 0. Hece by the Beroulli s iequality, ( 3 3 +h 3) = ( ) ( ) +h < if 3. +( )h ( )h Give ay ɛ > 0, sice hɛ > 0, by the Archimedea Property, K N such that K < hɛ, which meas Kh < ɛ. Hece K +,! 0 < ( )h kh 0. < ɛ. Hece lim(! ) = Questio 7. If lim(x ) = x > 0, show that there exists a atural umber K such that if K, the x < x < x. Proof. Sice lim(x ) = x > 0, take ɛ = x > 0. The by defiitio, K N such that x x < ɛ = x K. Hece x < x < 3 x < x if K. Sectio 3. Questio (b). Establish either the covergece or the divergece of the sequece X = (x ). x := ( ) +. Aswer. X is diverget. Proof. Suppose x is coverget ad lim(x ) = l. Take ɛ =. By defiitio, K N such that ( ) l + < K. I particular, for all K, ( ) l + < ad ( ) + (+) l (+)+ <. These imply that l + < ad + + l + <. Hece by the triagle iequality, + = > l l > + = if K. This implies that + + >, which is a cotradictio. Hece the sequece is diverget. 3

4 Questio. Give a example of two diverget sequeces X ad Y such that: (a). their sum X + Y coverges. Aswer. Let X := (( ) ) ad Y := (( ) + ). The X ad Y diverge but X + Y = 0 coverges. (b). their product XY coverges. Aswer. Let X := (0,, 0, ) ad Y := (, 0,, 0 ). The X ad Y diverge but XY = 0 coverges. Questio 3. Show that if X ad Y are sequeces such that X ad X + Y are coverget, the Y is coverget. Proof. Sice X ad X + Y are coverget sequeces. By limit theorem, Y = (X + Y ) X also coverges. Questio 4. Show that if X ad Y are sequeces such that X coverges to x 0 ad XY coverges, the Y coverges. Proof. Claim: there exists a K such that for all K, x 0. Proof of Claim: Case : x > 0. The take ɛ = x > 0. Sice lim(x) = x, by defiitio, K N such that x x < x K. Hece x > x x = 0 K. Case : x < 0. The take ɛ = x > 0. Sice lim(x) = x, by defiitio, K N such that x x < x K. Hece x < x x = 0 K. Hece K N such that x 0 K. Sice XY coverges, let lim(xy ) = z. Cosider the K-tail of X, Y, XY, the K-tail of X coverges to x ad the K-tail of XY coverges to z. Sice Y = XY X z. Hece Y coverges. x ifx 0, by limit theorem, the K-tail of Y coverges to Questio 5. Show that the followig sequeces are ot coverget. 4

5 (a). ( ) Proof. Accordig to Ex.3, > N. Suppose to the cotrary, lim( ) = l exists. The take ɛ =, by defiitio, K N such that l < K. Hece < < l + K. Sice l + R, by the Archimedea Property, H N such that H > l +. Hece max{h, K}, > > l +. This is a cotradictio. Hece ( ) is ot coverget. (b). (( ) ) Proof. Suppose to the coutrary, lim(( ) ) = l exists. Take ɛ =, by defiitio, K N such that ( ) l < K. I particular, ( ) 4 l < ad ( )+ ( + ) l < K. These imply that 4 l < ad 4k + 4k + + l < K. Hece by the triagle iequality, > 4 l + 4k + 4k + + l 8k + 4k + > K. This implies that >, which is a cotradictio. Hece (( ) ) is diverget. Questio 6. Fid the limits of the followig sequeces. (b). lim ( ( ) + ) Aswer. Sice ( ). Also, lim ( ( ) Hece by the Squeeze Theorem, lim ( ) = ) = lim ( +) = 0. ( ) + (d). lim ( Aswer. lim + ) ( ) = lim + = lim ( ) + lim ( ) ( ) lim = 0. Questio 7. If (b ) is a bouded sequece ad lim(a ) = 0, show that lim(a b ) = 0. Explai why Theorem 3..3 caot be used. 5

6 Proof. Sice (b ) is bouded, there exists M > 0 such that b M N. Give ay ɛ > 0, sice lim(a ) = 0, ad ɛ M > 0, by defiitio, K N such that a < ɛ M K. Hece a b < a M < ɛ K. Hece lim(a b ) = 0. Alteratively, it also follows from the squeeze theorem sice M a b M a for all. Theorem 3..3 caot be used because (b ) may ot be coverget. Questio 9. Let y := + for N. Show that (y ) ad ( y ) coverge. Fid their limits. Aswer. Observe that y = + = ++. Hece 0 ++ ( ) Sice lim(0) = lim = 0, by the Squeeze Theorem, lim(y ) = 0.. Observe that y = ( + ) = = (+)+ (+) lim( y ) = +lim( =. )+ + = + +. Hece Questio 0. Determie the followig limits. (a). lim((3 ) / ). Aswer. Note that by Example 3..(c), if c > 0, the lim(c / ) =. By Example 3..(d), lim( / ) =. Hece, lim((3 ) / ) = lim(3 / ) lim( /4 ) = lim(3 / ) lim ( 4) /4 lim((4) /4 ) = =. (b). lim(( + ) / l(+) ). Aswer. Let ( + ) / l(+) = a, the + = a l(+). Hece a = e. Hece lim(( + ) / l(+) ) = lim(e) = e. Questio If a > 0, b > 0, show that lim ( ( + a)( + b) ) = a+b. Proof. 6

7 lim ( ( + a)( + b) ) = lim ( (+a)(+b) (a+b)+ab ) = lim ( ) (+a)(+b)+ (+a)(+b)+ = lim ( a+b+ ab (+ a )(+ b )+) = lim (a+b)+lim ab lim (+ a ) lim (+ b )+ = a+b. Questio 3 Proof. Use the Squeeze Theorem 3..7 to determie the limits of the followig. (a) ( ) (b) ((!) ) (a) Sice for all, the () ( ) =. But lim = lim ( ) =, hece by the Squeeze Theorem, we have lim (() )) =. (b) Sice! for all, the (!) ( ) =. But lim = lim ( ) =, hece by the Squeeze Theorem, we have lim ((!) )) =. Questio 4 Show that if z := (a + b ) where 0 < a < b, the lim (z ) = b. Proof. Sice 0 < a < b, the b a + b b, the (b ) (a + b ) ( b ). But lim (b ) = b = lim lim (b ) = lim ( b ), hece by the Squeeze Theorem, we have lim (a + b ) = b, i.e. lim (z ) = b. Questio 5 Apply theorem 3.. to the followig sequeces, where a, b satisfy 0 < a <, b >. (a) (a ) (c) ( b ) (b) ( b ) (d) ( 3 3 ) 7

8 Proof. (a) Sice a > 0, ad lim a+ a = a <, hece by Theorem 3.., we have lim (a ) = b (b) Sice b > 0, ad lim b = b, but b ca be either greater or smaller tha, hece Theorem 3.. does ot apply here. (c) Sice > 0, ad lim + b b + = <, b b hece by Theorem 3.., we have lim ( b ) = 0. (a) Sice 3 3 > 0, ad lim 3(+) 3 (+) 3 3 = 8 9 <, hece by Theorem 3.., we have lim ( 3 3 ) = 0. Questio 6 (a) Give a example of a coverget sequece (x ) of positive umbers with lim x + x =. (b) Give a example of a diverget sequece with this property. Aswer. (a) Cosider the coverget sequece (x ) = ( ). We have lim (x ) = 0 ad lim ( x + x ) = lim =. + (b) Cosider the diverget sequece (x ) = (). We have (x ) is ot bouded ad hece diverge, ad lim ( x + x ) = lim + =. Questio 7 Let X = (x ) be a sequece of positive real umbers such that lim ( x + x ) = L >. Show that X is ot a bouded sequece ad hece is ot coverget. P roof. 8

9 Sice lim ( x + x ) >, the ρ > ad N N, such that x + x > ρ N. Hece x N +k ρ k x N, k N. Let α be ay give positive real umber. Sice (ρ k ) is ubouded, hece N N, such that ρ k > α x N k N. The N + N, x ρ N x N > α x N x N = α. Therefore, X is ot a bouded sequece ad hece ot coverget. Questio 8 Discuss the covergece of the followig sequeces, where a, b satisfy 0 < a <, b >. P roof. (a) ( a ) (b) ( b ) (c) ( b! ) (d) ( )! (a) Sice a > 0, ad lim (+) a + a = (lim ( + )) a = a <, hece by Theorem 3.., we have lim (a ) = 0. b + (b) Sice b (+) > 0, ad lim b = (lim ( + )) b = b >, hece by Questio 7, we have ( b ) diverges. (c) Sice b > 0, ad lim b! + (+)! b! = lim b + = 0 <, hece by Theorem 3.., we have lim ( b ) = 0.! (d) Sice! > 0, ad lim (+)! (+) +! hece by Theorem 3.., we have lim (! ) = 0. = lim ( + ) = lim (+ ) = e <, Questio 9 Let (x ) be a sequece of positive real umbers such that lim (x ) = L <. Show that there exists a umber r with 0 < r < such that 0 < x < r for all sufficietly large N. Use this to show that lim (x ) = 0. 9

10 P roof. Sice lim (x ) = L <, the r < ad N N, such that x < r N. Hece 0 x < r N, but lim 0 = lim (r : N) = 0. Hece, by squeeze theorem, we have lim (x : N) = 0. Hece lim (x ) = lim (x : N) = 0. Questio 0 (a) Give a example of a coverget sequece (x ) of positive umbers with lim x =. (b) Give a example of a diverget sequece with this property. Aswer. (a) Cosider the coverget sequece (x ) = ( ). We have lim (x ) = 0 ad lim x = lim ( ) =. (b) Cosider the diverget sequece (x ) = (). We have (x ) is ot bouded ad hece diverge, ad lim x = lim =. Questio Suppose that (x ) is a coverget sequece ad (y ) is such that for ay ε > 0 there exists M such that x y < ε for all M. Does it follow that (y ) is coverget? P roof. The aswer is yes. Let x := lim (x ), ε > 0, the N N, such that x x < ε, N. By the assumptio, N N, such that y x < ε, N. 0

11 Hece, y x y x + x x < ε + ε = ε, max{n, N }, i.e. (y ) coverges (to x). Questio Show that if (x ) ad (y ) are coverget sequeces, the the sequeces (u ) ad (v ) defied by u := max{x, y } ad v := mi{x, y } are also coverget. P roof. Notice thatu = (x + y + x y ) ad y = (x + y x y ), N by Exercise..6. Sice (x ) ad (y ) are coverget sequeces, lim (u ) = lim ( (x + y + x y )) = (lim (x ) + lim (y ) + lim (x ) lim (y ) ), ad lim(v ) = lim( (x + y x y )) = (lim(x ) + lim(y ) lim(x ) lim(y ) ). Hece, (u ) ad (v ) are also coverget.

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