límites de factorización, plis.
Respuestas a la pregunta
Respuesta: 1) -4/3
2) 1
3) -2/5
4) 4
5) 1/7
6) -2/3
7) -1/4
8) 0
Explicación:
1) Lim (x→1) [ (x² + 2x - 3)/(x² - 5x + 4)] . Al evaluar, resulta 0/0.
Entonces, se factoriza el numerador y el denominador.
Lim (x→1) [ (x² + 2x - 3)/(x² - 5x + 4)] = Lim (x→1) [ (x-1)(x+3) ]/(x-1)(x-4) ]
= Lim (x→1) [ (x+3)/(x-4) ]
= (1+3)/(1-4) = -4/3
2) Lim (x→1) [ (x^4 - x^5)/(1 - x) ] . Al evaluar , resulta 0/0.
Entonces , Lim (x→1) [ (x^4 - x^5)/(1 - x) ] = Lim(x→1) [ x^4(1-x) /(1-x) ]
= Lim (x→1) [ x^4 ]
= 1^4
= 1
3) Lim(x→-2) [ (x³+3x²+2x)/(x²-x-6) ] . Al evaluar, resulta 0/0.
Entonces, Lim(x→-2) [ (x³+3x²+2x)/(x²-x-6) ]
= Lim(x→-2)[x(x²+3x+2)/(x²-x-6)]
= Lim(x→-2)[ x(x+2)(x+1) /(x-3)(x+2) ]
= Lim(x→-2) [ x(x+1)/(x-3) ]
= (-2)(-1) / (-5)
= -2/5
4)Lim (x→-2) [ (x³ + 2x²) /(x + 2) ] . Al evaluar , resulta 0/0
Entonces, Lim (x→-2) [ x²(x+2) /(x+2) ] = Lim (x→-2) [ x² }
= (-2)² = 4
5) Lim(x→4) [ (x-4)/(x²-x-12) ]. Al evaluar, resulta 0/0.
Entonces, Lim(x→4) [ (x-4)/(x²-x-12) ] = Lim(x→4) [ (x-4)/(x-4)(x+3) ]
= Lim(x→4) [ 1/(x+3) ]
= 1/7
6) Lim (x→5) [ (x²-2x -35)/(x²+3x-10) ]. Al evaluar, resulta -20/30
Entonces, al simplificar, Lim (x→5) [ (x²-2x -35)/(x²+3x-10) ] = -2/3
7) Lim (x→0) { [( 1/(x+2) ) - (1/2)] }/ x. Al evaluar, resulta 0/0
Entonces, Lim (x→0) { [( 1/(x+2) ) - (1/2)] }/ x = Lim(x→0) [ (2-(x+2))/x(2x+4) ]
= Lim(x→0) [ -x / (x .(2x+4) ) ]
= Lim(x→0) [ -1 / (2x+4) ]
= -1 /(0+4)
= -1/4
8)Lim(x→3) [ (x³ - 6x² + 9x)/(x²-9) ] . Al evaluar, resulta 0/0
Entonces:
Lim(x→3) [ (x³ - 6x² + 9x)/(x²-9) ] = Lim(x→3) [ x.(x²-6x+9)/(x²-9) ]
= Lim(x→3) [ x.(x-3)(x-3)] /[(x-3)(x+3)]
= Lim(x→3) [ x. (x-3) / (x+3) }
= 3 . 0 / 3
= 0