= 1 x2r)lim (x+2)5 = (1+2)5 = (3)5 = 81*3 = 243 x1s)lim (x3+x2+x+1)/(x2-1) = (33+32+3+1)/(32-1) = (27+9+3+1)/(9-1) = (40)/(8) = 5 x3t)lim (x2+x)/(x3+x2+x+1) = (02+0)/(03+02+0+1) = (0)/(1) = 0 x0u)lim 3/(2x+1) = 3/(2(1)+1) = 3/(3) = 1 x1v)lim (3x2+x+1)/2 = (3(-1)2+(-1)+1)/2 = (3(1)-1+1)/2 = 3/2 x-1x)lim (x3+2x2+3x+2)/(2x3+x2+2x+4) = (03+2(0)2+3(0)+2)/(2(0)3+(0)2+2(0)+4) = (2)/(4) = 1/2 x02)Calcule os seguintes limites indeterminados:a)lim (x2-9)/(x-3) = ((x+3)(x-3))/(x+3) = x-3 = 3-3 = 0 x3b)lim (49-x2)/(7+x) = ((7+x)(7-x))/(x+7) = 7-x = 7-(-7) = 7+7 = 14 x-7c)lim (5-x)/(25-x2) = (5-x)/((5+x)(5-x)) = 1/(5+x) = 1/(5+5) = 1/10 x5d)lim (x2+x)/(x2-3x) = (x(x+1))/(x(x-3)) = (x+1)/(1-3) = (0+1)/(0-3) = 1/-3 = -1/3 x0e)lim x3/(2x2-x) = x3/(x(2x-1)) = x2/(2x-1) = 02/(2(0)-1) = 0/-1 = 0 x0f)lim (49+14x+x2)/(7+x) = ((7+x)(7+x))/(7+x) = 7+x = 7+(-7) = 0 x-7g)lim (x2-6x+9)/(x-3) = ((x-3)(x-3))/(x-3) = x-3 = 3-3 = 0 x3h)lim (x2-4x+3)/(x-1) = ((x-1)(x-3))/(x-1) = x-3 = 1-3 = -2 x1i)lim (x2-7x+12)/(x-4) = ((x-3)(x-4))/(x-4) = x-3 = 4-3 = 1 x4j)lim (x-1)/(x2-3x+2) x11, vamos encontrar o produtos notavel de (x2-3x+2) encontrando raiz utilizando baskarax2-3x+2 Delta = b2-4acDelta = 9-8Delta = 1x'=(-b+1)/2ax'=(3+1)/2x'=4/2x'=2x"=(-b-1)/2ax"=(3-1)/2x"=2/2x"=1Com : ax2+bx+c = a(x-x')(x-x") entao, (x2-3x+2) e igual a 1(x-1)(x-2)lim (x-1)/(x2-3x+2) = (x-1)/((x-1)(x-2)) = 1/(x-2) = 1/(1-2) = 1/-1 = -1 x1l)lim (x2-2x+1)/(x-1) =((x-1)(x-1))/(x-1) = x-1 = 1-1 = 0 x1m)lim (x-2)/(x2-4) = (x-2)/((x-2)(x+2)) = 1/(x+2) = 1/(2+2) = 1/4 x23) Calcule os seguintes limites indeterminados:a)lim -x = -(-4) = 4 x-4b)lim x3= (-4)3 = 16*-4 = -64 x-4c)lim -x4= -(3)4= -81 x-3d)lim 3x8 = 3(-1)8 = 3(1) = 3 x-1e)lim (-x2-3x) = -(5)2-3(5) = -25-15 = -40 x5f)lim (3x+2)/x2 = (3(-2)+2)/(-2)2 = (-6+2)/4 = -4/4 = -1 x-2g)lim (8x+x2) = 8(-1)+(-1)2 = -8+1 = -7 x-1h)lim (5x-x2)/(x-3) = (5(2)-(2)2)/((2)-3) = (10-4)/(2-3) = 6/-1 = -6 x2i)lim (x2-2x-15) = ((-3)2-2(-3)-15) = 9+6-15 = 0 x-3j)lim (2x3+3x2-7x+3) = (2(-1)3+3(-1)2-7(-1)+3) = 2(-1)+3(1)+7+3 = -2+3+7+3 = -2+13 = 11 x-1l)lim (4x3-x2-4x2+1) = 4(1)3-(1)2-4(1)2+1 = 4-1-4+1 = 0 x1m)lim (x3-2x2-3x)/(x2-8) = ((3)3-2(3)2-3(3))/((3)2-8) = (27-2(9)-9)/(9-8) = (27-18-9)/1=(27-27) /1= 0/1=0 x3n)lim (4x3+2x2+x+2) = 4(-1)3+2(-1)2+(-1)+2 = 4(-1)+2(1)-1+2 = -4+2-1+2 = -5+4 = -1 x-1o)lim (x4-x3+x2+x+1) = 04-03+02+0+1 = 1 x0p)lim (-2x2-x+2) = -2(3)2-3+2 = -2(9)-1 =-18-1 = -19 x3q)lim (x3-x2-3)7 = ((-1)3-(-1)2-3)7 = (-1-1-3)7 = (-5)7 = -25*25*25*5 = -78125 x-1r)lim (x2+1)4 = (22+1)4 = (4+1)4 = (5)4 = 625 x2s)lim (7x8-x7)/(-2x3+5x2-x+6) = (7(0)8-(0)7)/(-2(0)3+5(0)2-0+6) =0/6 = 0 x04) Calcule os seguintes limites indeterminados:a)lim (x2-16)/(x+4) = ((x-4)(x+4))/(x+4) = x-4 = -4-4 = -8 x-4b)lim (64-x2)/(8-x) = ((8-x)(8+x))/(8-x) = 8+x = 8+8 = 16 x8c)lim (9-x)/(81-x2) = (9-x)/((9-x)(9+x)) = 1/(9+x) =1/(9+9) = 1/18 x9d)lim (3x2+x)/(x2-7x) = (x(3x+1))/(x(x-7)) = (3x+1)/(x-7) = (3(0)+1)/(0-7) = 1/-7 = -1/7 x0e)lim x5/(5x2-3x) = x5/(x(5x-3)) = x4/(5x-3) = 04/(5(0)-3) = 0/-3 = 0 x0f)lim (16+8x+x2)/(4+x) = ((4+x)(4+x))/(4+x) = 4+x = 4+(-4) = 0 x-4g)lim (x2-20x+100)/(x-10) = ((x-10)(x-10))/(x-10) = x-10 = 10-10 = 0 x10h)lim (x2-9x+20)/(x-5) = ((x-4)(x-5))/(x-5) = x-4 = 5-4 = 1 x5i)lim (x2+13x+42)/(x+6) = ((x+6)(x+7))/(x+6) = x+7 = -6+7 = 1 x-6j)lim (x-4)/(x2-7x+12) = (x-4)/((x-4)(x-3)) = 1/(x-3) = 1/(4-3) = 1/1 = 1 x4l)lim (x2-8x+7)/(x-1) = ((x-1)(x-7))/(x-1) = x-7 = 1-7 = -6 x1m)lim (x-11)/(x2-121)=(x-11)/((x-11)(x+11)) = 1/(x+11) = 1/(11+11) = 1/22 x115) Calcule os limites abaixo:a)lim (x2+2x-1) = ((2)2+2(2)-1) =