Broj e proba
Descrição do Produto
Áðîj
e e
jå
an = (1 + n1 )n , n ∈ N jå êîíâåðãåíòàí è »åãîâà ãðàíè÷íà âðåäíîñò jå áðîj e. äîêàçàëè äà jå íèç (an )n∈N êîíâåðãåíòàí òðåáà äà ïîêàæåìî äà jå ìîíîòîí
è
Ìå¢ó ðåàëíèì áðîjåâèìà ïîñòîjå áðîjåâè êîjè èìàjó çíà÷àjíó óëîãó. èðàöèîíàëàí áðîj ÷èjà jå ïðèáëèæíà âðåäíîñò
Òåîðåìà 1:
äîêàç
:
Òàêàâ jå áðîj
e.
Áðîj
2, 718.
Íèç
Äà áè
îãðàíè÷åí.
(1 + h)n ≥ 1 + nh, h > −1, h 6= 0, n ∈ N . ≥ 1 + n(− n12 ) = 1 − n1 . Äàêëå,
Ó òó ñâðõó ïîñëóæè£å íàì Áåðíóëèjåâà íåjåäíàêîñò Àêî óçìåìî äà jå
h=
− n12 ,
áè£å
(1 −
1 n 1 n 1 1− ≥1− n n n 1 n n − 1 1−n 1 n n n−1 ⇔ 1+ ≥ ⇔ 1+ ≥ n n n n−1 1 n n − 1 + 1 n−1 1+ ≥ n n−1 n 1 n−1 1 ≥ 1+ 1+ n n−1
òj.
1+
an ≥ an−1 ,
1 n 1 1−n ≥ 1− n n
1 n n2 )
1+
ïà jå íèç ìîíîòîíî ðàñòó£è.
Äà jå íèç îãðàíè÷åí, äîêàçà£åìî íà èíäèðåêòàí íà÷èí. Ôîðìèðà£åìî íèç áðîjåâà ÷èjè ñó ÷ëàíîâè ñâè âå£è îä ÷ëàíîâà íèçà
(an ),
à òàj íèç áðîjåâà áè£å ìîíîòîíî îïàäàjó£è. Ïî¢èìî îä ÷è»åíèöå,
1 1 < n n−1 1 1 1+ 1
n 2 n n n 1 ≥1+ 2 >1+ 2 =1+ , n>1 2 n −1 n −1 n n n n n 1 · ≥1+ , n>1 n−1 n+1 n n − 1 + 1 1 n 1 · n+1 ≥1+ , n>1 n−1 n n n 1 1 1 1+ · ≥1+ , n>1 n − 1 1 + n1 n 1 1 n 1 · 1+ ≥1+ , n>1 n−1 n (1 + n1 )n 1 n 1 n+1 1+ ≥ 1+ , n>1 n−1 n {z } | {z } | bn+1
bn
Èç
bn ≥ bn+1 ,
n>1
(bn )n∈N \{1} ìîíîòîíî îïàäàjó£è. Çà n = 2 èìàìî, b2 = 4 b. Íà òàj íà÷èí ñìî ïîêàçàëè äà jå íèç a îãðàíè÷åí, òj. òîìå, íèç (an ) jå êîíâåðãåíòàí è »åãîâà ãðàíè÷íà âðåäíîñò
ñëåäè äà jå íèç
øòî ïðåäñòàâ§à íàjâå£ó âðåäíîñò íèçà
2 ≤ an < 4 çà ñâå n ∈ N \ {1}. jå e, ó îçíàöè:
Ïðåìà
lim
n→∞
1+
1
1 n =e n
Ïðèáëèæíó âðåäíîñò áðîjà
Ïðèìåð 1:
Çà
n = 100
e
ìîæåìî äà ðà÷óíàìî ïîìî£ó èçðàçà
èìàìî,
Ëîãàðèòìójó£è îâó jåäíàêîñò,
a100 = 1 + log a100
1 100
100
= log 1 +
1+
1 n
n
,
.
1 100
100
⇔ log a100 = 100 · log
101 100
⇔ log a100 = 0,432
100 · (log 101 − log 100) ⇔ log a100 = 100 · (2, 00432 − 2) ⇔ log a100 = 0, 432 ⇔ a100 = 10 2, 70396. Íàðàâíî, øòî jå âå£å n, òî ñìî áëèæè áðîjó e.
(1230)
⇔ a100 =
Êîðèø£å»åì ñâîjñòâà íåïðåêèäíîñòè åêñïîíåíöèjàëíå ôóíêöèjå äîêàçójå ñå äà, àêî jå
lim an = a, a > 0,
n→∞
è
lim bn = b,
n→∞
òàäà jå
lim abnn = ab .
n→∞
Ïðèìåíîì îâå ÷è»åíèöå èçðà÷óíàòè
ñëåäå£å ãðàíè÷íå âðåäíîñòè:
(à) lim
n→∞
ðåøå»å
2 1 n +1 1+ 3 n→∞ n
1 n n
1−
n − 1 n n −n 1 n 1 1 n = n = = lim = lim = lim n→∞ n→∞ n→∞ n n n−1 n lim n−1+1 n−1 n−1
(á) lim
n→∞
1+
1 3n n2 + 1
2
1−
n + 1 n n→∞ 2n − 1
(â) lim
(ã) lim
:
(à) lim
n→∞
n→∞
= lim
n→∞
3n2
1 1+ n−1
1 n−1
1 1+ n−1
=
(á) lim
1+
n→∞
1 1+ n−1
1 n−1
lim
n→∞
1 1+ n−1
=
1 e·1
=
1 e
.
3n2 +3−3
1 1 = lim 1+ 2 n→∞ n→∞ +1 n +1 2 1 n +1 3 1 −3 = lim 1+ 2 lim 1 + 2 n→∞ n→∞ n +1 n +1
lim
n2
2 h 1 −3 i 1 3(n +1) 1+ 2 = = lim 1+ 2 n→∞ n +1 n +1 2 n +1 3 1 = lim 1 + 2 · 1 = e3 n→∞ n +1
n2 +1 2 3 n2 +1 3 1 n · n 3 1 n +1 1 n n3 (â) lim 1 + 3 = lim 1 + 3 = lim 1+ 3 = n→∞ n→∞ n n n n→∞ 2 n +1 3 lim n3 lim n1 + n13 lim n1 + lim 13 1 n n→∞ = lim 1 + 3 = en→∞ = e n→∞ n→∞ n = e(0+0) = 1 . n→∞ n 2n + 2 n n + 1 n 1 2n − 1 + 3 n 1 3 n = lim = lim n = lim n · lim 1+ = (ã) lim n→∞ 2(2n − 1) n→∞ 2 n→∞ 2 n→∞ 2n − 1 n→∞ 2n − 1 2n − 1 2n−1 3n 1 1 n 1 1 3 · 2n−1 = lim n · lim 1 + 2n−1 = = lim n · lim 1 + 2n−1 n→∞ 2 n→∞ n→∞ 2 n→∞ 3 3 3n " # lim 2n−1 2n−1 3 1 1 3 n→∞ = lim n · lim 1 + 2n−1 = 0 · e2 = 0 . n→∞ 2 n→∞ 3
(Çàäàöè èç ó¶áåíèêà ñòð.261-262) 2.
6.
1 −n lim 1 − ; n→∞ n n + 2 n lim ; n→∞ n − 1
ðåøå»å
3.
7.
Èçðà÷óíàòè ñëåäå£å ãðàíè÷íå âðåäíîñòè
1 n lim 1 − 2 ; n→∞ n n2 + 1 n2 lim ; n→∞ n2 − 1
2 n 3 n 1+ ; 5. lim 1− ; n→∞ n→∞ n n 3 1 5n +3 8. lim 1+ 3 . n→∞ n +4
4.
lim
:
2.
lim
n→∞
1−
n − 1 −n n n n − 1 + 1 n 1 −n 1 n = lim = lim = lim = lim 1 + = n→∞ n→∞ n − 1 n→∞ n→∞ n n n−1 n−1 " # 1 n−1+1 1 n−1 1 = lim 1 + = lim 1+ 1+ = n→∞ n→∞ n−1 n−1 n−1 1 n−1 1 = lim 1 + lim 1 + =e·1= e . n→∞ n→∞ n−1 n−1
2
ðåøå»å 3.
4.
5.
:
" # n 2 − 1 n n − 1 n n + 1 n n − 1 n n + 1 n 1 n = lim lim 1 − 2 = lim = lim lim n→∞ n→∞ n→∞ n→∞ n→∞ n n2 n n n n n −n n − 1 + 1 −n 1 n = lim lim 1 + = lim ·e= n→∞ n − 1 n→∞ n→∞ n n−1 −n " # lim n−1 −n 1 (n−1) n−1 1 n−1 n→∞ · e = lim 1 + = lim 1 + · e = e−1 · e = 1 . n→∞ n→∞ n−1 n−1 #2 " 2 n 1 n2 ·2 1 n2 = e2 . = lim 1 + n lim 1 + = lim 1 + n n→∞ n→∞ n→∞ n 2 2 n − 3 n n −n n − 3 + 3 −n 3 n 3 −n lim 1 − = lim = lim = lim = lim 1 + n→∞ n→∞ n→∞ n − 3 n→∞ n→∞ n n n−3 n−3 −3n # lim n−3 " n−3 −3n n−3 n→∞ 1 1 3 · n−3 3 = lim 1 + n−3 = lim 1 + n−3 = e−3 = e13
n→∞
6.
lim
n + 2 n n−1
n→∞
= lim
n−1
n→∞
" = 7.
lim
n 2 + 1 n2
lim
n→∞
=
8.
lim
n→∞
1+
1 n3 + 4
lim
n→∞
5n3 +3
n→∞
n−1 1 3
# lim
n→∞
1+
= lim
n→∞
lim
= en→∞
1+
n→∞
# lim
5n3 +3 n3 +4
=e
2
1+
2n2 2 −1
n→∞ n
1 n3 + 4
=
= e3
n2 −1 2
n−1 3n 3 · n−1
3n n−1
= lim
n2 −1 1 2
3
3 n 1 1+ = lim 1 + n−1 n→∞ n−1 3
n−1 3
n2 − 1
n→∞
"
1+
= lim
n2 − 1 + 2 n2
= lim
n2 − 1
n→∞
n→∞
3
n − 1 + 3 n
2 n = lim 1+ n→∞ n2 − 1
n +4
5+ 33 n 4 1+ n→∞ n3
n2 −1 2
=
2 2 lim 1 1 lim 2n 2 = en→∞ n −1 = e n→∞ 1− n2 = e2
n3 +4· 5n33 +3
lim
2 n2 −1 2 −1 1 2 · n2n
" =
lim
n→∞
1+
1 n3 + 4
# lim 5n33 +3 n +4 n3 +4 n→∞
=
= e5 ïðîô. È.Jîêñèìîâè£
3
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