Matemática, perguntado por thamirismr99, 1 ano atrás

ME AJUDEM, É MUITO URGENTE!!!

Calcule o limite:

10 ao 16.

Anexos:

Soluções para a tarefa

Respondido por Usuário anônimo
1

10)


Lim (6t²+5t)/(1-t)(2t-3)

t-->-∞


Lim (6t²+5t)/(5t-3-2t²)

t-->-∞


Lim t²(6+5/t)/t²(5/t-3/t²-2)

t-->-∞


Lim (6+5/t)/(5/t-3/t²-2) =(6+5/(-∞))/(5/(-∞)-3/(-∞)²-2) =(6+0)/(0+0-2)

t-->-∞


=6/(-2)=-3


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11)


Lim √(1+4x²) / (4+x)

x-->∞


Lim x√(1/x+4) / x(4/x+1)

x-->∞


Lim √(1/x+4) / (4/x+1) = √(1/∞+4) / (4/∞+1) =√4/1 =2

x-->∞

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12)


Lim √(x²+4x) / (4x+1)

x-->-∞


Lim √[(x)²(1+4/x) ]/ x(4+1/x)

x-->-∞

Observe ==> √(x)² ==>para  x--> -∞    ==> √(x)² =√(-x)² =-x


Lim √(-x)² *√(1+4/x)/ x(4+1/x)

x-->-∞


Lim -x*√(1+4/x)/ x(4+1/x)

x-->-∞


Lim -√(1+4/x)/ (4+1/x)  =-1/4

x-->-∞


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13) 

Lim (1-√x)/(1+√x)

x-->∞


Lim √x(1/√x-1)  /  √x(1/√x+1)

x-->∞


Lim (1/√x-1)  /  (1/√x+1)  =(1/∞  -1)/(1/∞ +1)=(0-1)/(0+1) = -1

x-->∞

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14)

lim √( x²+3x+1)  -x
x-->∞


lim  [√( x²+3x+1)  -x] *  [√( x²+3x+1)  +x]/ [√( x²+3x+1)  +x]
x-->∞


lim  [ x²+3x+1  -x²] / [√( x²+3x+1)  +x]
x-->∞


lim  [ 3x+1 ] / [√( x²+3x+1)  +x]
x-->∞



lim  x[ 3+1/x ] / [x√( 1+3/x+1/x²)  +x]
x-->∞


lim  [ 3+1/x ] / [√( 1+3/x+1/x²)  +1]
x-->∞

= [ 3+1/∞ ] / [√( 1+3/∞+1/∞²)  +1]  =3/2

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15)

Lim √(x²+1) - √(x²-1)
x-->∞

Lim [√(x²+1) - √(x²-1)][√(x²+1) + √(x²-1)]/[√(x²+1) + √(x²-1)]
x-->∞

Lim [x²+1 - x²+1]/[√(x²+1) + √(x²-1)]
x-->∞

Lim [2]/[√(x²+1) + √(x²-1)]
x-->∞

=2 /[∞+∞] =2/∞ =0



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16)

Lim √(1+x) - √(x)
x-->∞

Lim [√(1+x) - √(x)][√(1+x) + √(x)]/[√(1+x) + √(x)]
x-->∞

Lim [1+x - x]/[√(1+x) + √(x)]
x-->∞

Lim [1]/[√(1+x) + √(x)]
x-->∞

=1/∞ = 0

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