Question

Alguém por favor me ajuda ??????

Answer 1

Resposta:

1)

lim  5  =5

x-->-1

2- lim x+2 =1+2=3

x-->1

3-

lim   x²-3x+1 =0-0 +1=1

x-->0

4-

lim   (x²-4)/(x-2)

x-->2

### x²-4 =(x-2)*(x+2)      ==>a²-b²=(a-b)*(a+b)

lim   (x+2)*(x-2)/(x-2)

x-->2

lim   (x+2)= 2+2=4

x-->2

5-

lim (x²-4x+3)/(x-3)

x-->3

P(x)=ax²+bx+c=a*(x-x')*(x-x'')    x' e x'' são as raízes

x²-4x+3=0   ...a=1

x'=[4+√(16-12)]/2=(4+2)/2=3

x''=[4-√(16-12)]/2=(4-2)/2=1

x'=3 , x''=1  e a=1

x²-4x+3=1*(x-3)*(x-1)

x²-4x+3=(x-3)*(x-1)

lim (x-3)*(x-1)/(x-3)

x-->3

lim x-1 =3-1  =  2

x-->3