Baixe limites euler e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! Aueulog.doc 1/6 Limites Envolvendo a Seqüência de Euler Leohnard Euler, 1707-1783, ( 27/08/01 11:50:50, arquivo: aueulog.doc Usar o limite fundamental: e x x x =+ →∝ )11(lim ou ( ) et t t =+ → 1 0 1lim Limites da Seqüência de Euler: x x )11( + 1. x x x 321lim       + ∝+→ = ? ‡ x x x 321lim       + ∝+→ = 3 21lim               + ∝+→ x x x = 3 2 11lim                   + ∝+→ x x x = 3211lim               + ∝+→ t t t = 6 11lim               + ∝+→ t t t = 6e Fazendo t=    +∞→ +∞→ t xx ,2 e x=2t 2. ( ) x x x 5 0 31lim − → = ? ‡ ( ) x x x 5 0 31lim − → = ( ) 51 0 31lim     − → x x x = ( ) 53 0 1lim     + − → t t t = ( ) 151 0 1lim − →     + t t t = 15−e = 15 1 e Fazendo    → → −= 0 0 ,3 t x xt e 3 tx −= 3. x x x 21 4 31lim − ∝−→       − =? ‡ x x x 21 4 31lim − ∝−→       − = x x x 21 3 4 11lim − ∝−→             − + = 2 3111lim t t t + ∝+→       + = 111lim       + ∝−→ tx . 2 3 11lim t t t       + ∝+→ = 2 3 11lim t t t       + ∝+→    +∞→ −∞→ −= t xxt , 3 4 com 4 3tx −= 4. x x x x 21 52 5lim − ∝+→       + =? ‡ x x x x 21 52 5lim − ∝+→       + = x x x x x x x 21 5 5 5 2 5 5 lim − ∝+→         + = x x x 21 1 2 5 1 1lim − ∝+→               + = 1 1 2 5 1 1lim               + ∝+→ x x . x x x 2 1 2 5 1 1lim − ∝+→               + = Aueulog.doc 2/6 2 . 5 2 11 1lim − ∝+→                           + t t t = ( )2. 5 2 11 1lim − ∝+→                         + t t t = . 5 4 11 1lim − ∝+→                     + tt t = 5 4 1 − e = 5 4 e Fazendo t=    +∞→ +∞→ t xx , 2 5 e x= 5 2t . 5. x x x x       − + ∝+→ 3 4lim =? ‡ x x x x       − + ∝+→ 3 4lim = x x x x x x             − + ∝+→ 3 4 lim = x x xx x xx x             − + ∝+→ 3 4 lim = x x x x             − + ∝+→ 31 41 lim = x x x x               − + + ∝+→ 3 11 4 11 lim = 73 4 e e e =− 6. x x ax ax       − + ∝+→ lim =? ‡ x x x ax x ax             − + ∝+→ lim = x x x a x x x a x x             − + ∝+→ lim = x x x a x a             − + ∝+→ 1 1 lim = x x a x a x               − + + ∝+→ 11 11 lim = x x x x a x a x         − +         + ∝+→ ∝+→ 11lim 11lim = ( )saa a e e e −− − = =e 2a . 7. x x x+ → 1lim 0 =? ‡ ( ) x x x 1 0 1lim + → = 1e 8. x x x.31lim 0 − → =? ‡ x x x.31lim 0 − → = t t t       −+ +∞→ 31lim = t t t             − + +∞→ 3 11lim = y y y 3 11lim − −∞→      + = 3 11lim − −∞→               + y y y = 3−e ← fazendo    −∞→ +∞→ −= y tty , 3 , com t=-3y 9. ( ) xg x xtg 2cot2 0 1lim + → =? ‡ fazendo xgt 2cot= e substituindo no limite, temos ( ) xg x xtg 2cot2 0 1lim + → = t t t       + +∞→ 11lim = e Aueulog.doc 5/6 20. 1 3lim 2 0 − − → xx e xx =? ‡ 1 3lim 2 0 − − → xx e xx = x e x xx xx 1 3 lim 2 0 − − → = x e x xx 1 3lim 0 − − → = x e x x x x 1lim 3lim 0 0 − − → → = eln 3− = 3− 21. 1 3senlim 0 −→ xx e x =? ‡ 1 3senlim 0 −→ xx e x = x e x x xx 1 3 3sen.3 lim 0 −→ = x e x x x x x 1lim 3 3sen.3lim 0 0 − → → = eln 1.3 = 3 22. x e x x 4sen 1lim 3 0 − → =? ‡ x e x x 4sen 1lim 3 0 − → = x x x e x x 4 4sen.4 3 1.3 lim 3 0 − → = x x x e x x x 4 4sen.4lim 3 1.3lim 0 3 0 → → − = 1.4 ln.3 e = 4 3 23.       − − → x e x x 2 cos 1lim sen 0 π =? ‡       − − → x e x x 2 cos 1lim sen 0 π = x e x x sen 1lim sen 0 − → = eln =1 24. ( )1 42lim 1 1 − −+ → xtg x x =? ‡ ( )1 42lim 1 1 − −+ → xtg x x = ( )1 22lim 21 1 − −+ → xtg x x = ( )( )1 12.2lim 12 1 − −− → xtg x x = ( ) 1 1 1 12.2 lim 1 2 1 − − − −− → x xtg x x x = ( ) 1 1lim 1 12.2lim 1 1 2 1 − − − − → − → x xtg x x x x = ( ) 1 1lim 1 12lim.2lim 1 1 1 2 1 − − − − → − →→ x xtg x x x xx = 1 2ln.22 = 2ln.4 25. xx ee xx x cossen1 lim 1cossen 0 −+ − −− → =? ‡ xx ee xx x cossen1 lim 1cossen 0 −+ − −− → = ( ) 1cossen 1.lim cossen1 0 +− −−− → xx ee xx x = ( )1cossen. 1lim 1cossen 0 +− −+− → xxe e xx x = ( )1cossen 1.1lim 1cossen 0 +− −+− → xx e e xx x = t e e t t 1.1lim 0 − → = et e t t 1.1lim 0 − → = e e ln.1 = 1−e Fazendo    → → −+= 0 0 ,cossen1 t x xxt 26. ( )x xe x x + −+ → 1ln 1senlim 0 =? ‡ ( )x xe x x + −+ → 1ln 1senlim 0 = ( )x xe x x + +− → 1ln sen1lim 0 = ( ) x x x x x e x x + + − → 1ln sen1 lim 0 = ( ) x x x x x e x x x x + + − → →→ 1lnlim senlim1lim 0 00 = ( )x x x e 1 0 1limln 1ln + + → = eln 2 = 2. Aplicando a Regra de L’Hôspital: ( )x xe x x + −+ → 1ln 1senlim 0 = ? ‡ 1ln 10sen0 −+e = 0 0 Aueulog.doc 6/6 Fazendo ( ) ( ) ( )x xe xg xf x + −+ = 1ln 1sen ‡ ( ) ( )0 0 g f = 0 0 . Derivando separadamente o numerador e o denominador, temos: ( ) ( ) x xe xg xf x + + = 1 1 cos ' ' e ( ) ( ) 01 1 0cos 0' 0' 0 + + = e g f = 1 2 . Logo ( )x xe x x + −+ → 1ln 1senlim 0 = 1 2 =2. 27. ( )ax axeax x + −+ → 1ln 1senlim 0 =? ‡ ( )ax axeax x + − → 1ln sen. 1 1lim 0 = ( )ax x ax x e x x + − → 1ln.1 sen.1lim 0 = ( )ax x ax x + → 1ln.1 senlim.1 0 = ( )ax x ax x + → 1ln.1 senlim 0 = ( ) xx ax ax 10 1ln senlim + → = ( ) x x x ax ax 1 0 0 1lnlim senlim + → → = ( ) x x x ax ax 1 0 0 1lnlim senlim + → → = ( ) ( )x x ax a 1 0 1limln 0.sen + → = ae 0 = 0