43 Encontre os limites
ll lim x 0 sen x4 x & lim x +infty(1+1x)^2 x lim x 0 tg 2 x3 x & lim x -infty(1+1x)^x / 3 lim x 0 sen 4 xsen 3 x & lim x +infty(1+1x)^x+2 lim x 0 tg 3 xtg 5 x & lim x -infty(1+4x)^x

Passo 1

O enunciado pede oito limites, quatro quando xto0 envolvendo senos e tangentes, e quatro quando xtopminfty envolvendo a forma l(1+tfraccdotxr)^algo. O conteúdo a usar é:
  1. para xto0, as aproximações sin tsim t e tan tsim t;
  2. para xtopminfty, o limite padrão
limxtopminftyl(1+1xr)^x=e,
e suas extensões.

Passo 2

Limites quando xto0: use sin xsim x e tan xsim x.
limxto0sin x4x =14,limxto0sin xx =14cdot1 =14,
limxto0tan2x3x =23,limuto0tan uu(com u=2x) =23cdot1 =23,
limxto0sin4xsin3x =4x3x,sin4x/(4x)sin3x/(3x) =43,11 =43,
limxto0tan3xtan5x =3x5x,tan3x/(3x)tan5x/(5x) =35,11 =35

Passo 3

Limites exponenciais quando xtopminfty: definimos u=1/x, então uto0 e
l(1+tfrac1xr)^x =l(1+ur)^1/uto e
A partir daí:
limxto+inftyl(1+tfrac1xr)^2x =l[l(1+tfrac1xr)^xr]^2 =e^2,
limxto-inftyl(1+tfrac1xr)^x/3 =l[l(1+tfrac1xr)^xr]^1/3 =e^1/3,
limxto+inftyl(1+tfrac1xr)^x+2 =l(1+tfrac1xr)^x,l(1+tfrac1xr)^2 to ecdot1 =e,
limxto-inftyl(1+tfrac4xr)^x =l[l(1+tfrac4xr)^x/4r]^4 to e^4

Resposta

limxto0sin x4x=14
limxto0tan2x3x=23
limxto0sin4xsin3x=43
limxto0tan3xtan5x=35
limxto+inftyl(1+tfrac1xr)^2x=e^2
limxto-inftyl(1+tfrac1xr)^x/3=e^1/3
limxto+inftyl(1+tfrac1xr)^x+2=e
limxto-inftyl(1+tfrac4xr)^x=e^4

Exercícios relacionados