[PDF] Jiwen He 11 Properties of Limits - UH

Consider lim xln x) This is an indeterminate form of the

Consider lim x0+ (xlnx) This is an indeterminate form of the type 0 1 To apply l’H^opital’s rule we must rewrite it as a quotient First try: lim x0+ x (lnx)−1 is an indeterminate form of type 0

Limits involving ln(

lim x1 lnx = 1; lim x0 lnx = 1 : I We saw the last day that ln2 > 1=2 I Using the rules of logarithms, we see that ln2m = mln2 > m=2, for any integer m I Because lnx is an increasing function, we can make ln x as big as we choose, by choosing x large enough, and thus we have lim x1 lnx = 1: I Similarly ln 1 2n = nln2 < n=2 and as x

Jiwen He 11 Properties of Limits - UH

lnx → 0, as n → ∞ Since f(u) = eu is continuous at 0, we have lim n→∞ x1 n = lim n→∞ e1 n lnx = lim u→0 eu = e0 = 1 2 Some Important Limits: 3 lim

Fiche technique sur les limites

= 0 En 0 lim x0 x>0 xln(x) = 0 ; lim x0 x>0 xn ln(x) = 0 5 2 Fonction exponentielle Comparaison de la fonction exponentielle avec la fonction puissance en +1et en 1

Chapitre 5 Fonction Logarithme Népérien

lnx lim → x = −; le développement limité d’ordre 1 en 1 de ln est : ln (1 + h) = h + h ℇ(h) avec 0 ( ) 0 h lim h → ε = , que l’on écrit aussi : ln (x) = x – 1 + (x – 1) ℇ(x -1) avec 0 ( ) 0 X lim X → ε = Remarque : En physique, on dit que pour h « voisin » de 0, ln (1 + h) ≈ h ou que pour x « voisin » de 1, ln(x

EXERCICES : FONCTION LOGARITHME

1 lim x ∞ ln x2 x =0 2 lim x 0 1 x lnx =∞ 3 lim x 0 5x 1 lnx=0 4 lim x ∞ x2 3x lnx =∞ Exercice 25: Déterminer la limite en ∞ de la fonction f définie par : a f x= x – lnx ; b f x= x – ln x2; c f x= ln2 x – x2 11; d f x= ln 2 1 x2 Exercice 26 : Déterminer la limite en 0 de la fonction f définie par : a f x

Indeterminate Forms

0 0, ∞ ∞, 0 ·∞, 00, ∞0, 1∞, ∞−∞ These are the so called indeterminate forms One can apply L’Hopital’s rule directly to the forms 0 0 and ∞ ∞ It is simple to translate 0 ·∞into 0 1/∞ or into ∞ 1/0, for example one can write lim x→∞xe −x as lim x→∞x/e xor as lim x→∞e −x/(1/x) To see that the

Fonction logarithme népérien

b) Déterminer alors lim x +∞ lnx x 3) En déduire lim x 0+ x lnx Retenons : lim x +∞ lnx x =0 ; lim x 0+ x lnx=0 Activité4 : Soit f : x lnx On désigne par (C) sa courbe représentative selon un orthonormé (O, ⃗ , ⃗) du plan

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Lecture 18Section 10.3 Limit of Sequence Section 10.4 Some

Important Limits

Jiwen He

1 Limit of Sequence

1.1 Properties of Limits

Properties of Limits

Properties of Limits: 1

Let limn→∞an=Land limn→∞bn=M. Then•lim n→∞can=cL•lim n→∞(an+bn) =L+M•lim n→∞anbn=LM•lim n→∞a nb n=LM forM?= 0.•lim n→∞f(an) =f(L) forfbeing continuous atL.Example1.If lim n→∞an=L, then limn→∞bn=eLwherebn=ean.

Limit of Sequence Defined by Rational Function

Properties of Limits

Letan=αpnp+αp-1np-1+···+α0withαp?= 0, andbn=βqnq+βq-1nq-1+ ···+β0withβq?= 0. Then•Ifp < q, then limn→∞a nb n= 0.•Ifp=q, then limn→∞a nb n=αpβ q.•Ifp > q, then limn→∞a nb ndoes not exist, and the sequenceanb ndiverges.1

Example2.lim

n→∞3n4-2n2+ 1n

5-3n3= 0.limn→∞1-4n7n

7+ 12n=-4.

lim n→∞n

4-3n2+n+ 2n

3+ 7ndoes not exist.

Pinching Theorem

Pinching Theorem

Suppose that for allngreater than some integerN,

a

If lim

n→∞an= limn→∞cn=L, then limn→∞bn=L. n →0,since????cosnn and1n →0.

2 Some Important Limits

2.1 Some Limits

Some Important Limits: 1

lim n→∞1n

α= 0, α >0.

Proof.

?α >0,?p?Ns.t. 1/p < α. Then 0<1n

α=?1n

By the pinching theorem, lim n→∞1n

α= 0, α >0.

Some Important Limits: 2

lim n→∞x1n = 1, x >0.

Proof.

Note that?x,ln?

x1n =1n lnx→0,asn→ ∞.

Sincef(u) =euis continuous at 0, we have

limn→∞x1n = limn→∞e1n lnx= limu→0eu=e0= 1.2

Some Important Limits: 3

lim n→∞xn= 0 if|x|<1. (The limit does not exist if|x|>1 orx=-1.)

Proof.

Note that for anyxs.t. 0<|x|<1, we have ln|x|<0 and ln(|x|n) =nln|x| → -∞,asn→ ∞.

Since lim

u→-∞eu= 0, we have lim n→∞|xn|= limn→∞|x|n= limn→∞enln|x|= limu→-∞eu= 0.

Therefore,

limn→∞xn= 0 if|x|<1. (The sequencexn,|x|>1, is unbounded, thus divergent.)

Some Important Limits: 3

lim n→∞xn= 0 if|x|<1. (The limit does not exist if|x|>1 orx=-1.)

Proof.

Note that for anyxs.t. 0<|x|<1, we have ln|x|<0 and ln(|x|n) =nln|x| → -∞,asn→ ∞.

Since lim

u→-∞eu= 0, we have lim n→∞|xn|= limn→∞|x|n= limn→∞enln|x|= limu→-∞eu= 0. lim n→∞xn= 0 if|x|<1. (The sequencexn,|x|>1, is unbounded, thus divergent.)

Alembert"s Rule

Alembert"s Rule

Let (an)∞n=1,an?= 0?nsuch that limn→∞? ???a n+1a n? ???=ρ. Ifρ <1, then the sequence (an)∞n=1converges to 0, and, ifρ >1, it diverges.

Proof. (ρ <1)

Since 0<1 +ρ2

<1, limn→∞?

1 +ρ2

n = 0. Since limn→∞? ???a n+1a n? ???=ρ,?N1s.t. ?n > N1,????a n+1a n? |an|. Then, forn > N1,3 2 n-N1 |aN1| and therefore lim theorem, we have lim n→∞an= 0.

Proof. (ρ >1)

?δ >0 s.t.ρ >1+δ. Since limn→∞? ???a n+1a n? ???=ρ,?N2s.t.?n > N2,????a n+1a n? ???≥ρ-δ2

Asρ-δ2

>1 +δ2 ,|an+1| ≥?

1 +δ2

|an|. Then, forn > N1, |an| ≥?

1 +δ2

|an-1| ≥?

1 +δ2

2 |an-2| ≥ ··· ≥?

1 +δ2

n-N2 |aN2| and therefore (an) is unbounded, thus divergent.

Some Important Limits: 4

lim n→∞x nn!= 0.

Proof.

Note that????a

n+1a n? ???=|x|n+1(n+ 1)!n!|x|n=|x|n+ 1→0 asn→ ∞

By Alembert"s Rule,limn→∞x

nn!= 0.

Some Important Limits: 5

lim n→∞n px n= 0,|x|>1.

Proof.

Note that????a

n+1a n? ???=(n+ 1)p|x|n+1|x|nn p=1|x|? 1 +1n →1|x|asn→ ∞ Since

1|x|<1, by Alembert"s Rule,

lim n→∞n px n= 0,|x|>1.4

Some Important Limits: 6

lim n→∞lnnn = 0.

Proof.

=1n n 1dtt n

1dt⎷t

=2⎷n (⎷n-1)→0 asn→0.

By the pinching theorem, lim

n→∞lnnn = 0. Alternative proof is done by L"Hˆopital"s

Rule (10.5),

limn→∞lnnn = limu→∞lnuu = limu→∞1u 1 = 0. lim n→∞n1n = 1.

Proof.

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