Ley 1122/2007

Solución:

2. A) lim (X2 + 4X) = lim (X2) + lim (4X) = 22 + 4 (2) = 4 + 8 = 12
X → 2 x → 2 x → 2


b) lim ( X3 + 4X2 - 2 X + 4) = lim (X3) + lim (4X2) - lim (2X) + lim (4)
X →1 x →1 x→ 1 x→1 x →1

= (- 13) + 4 (-1)2 – 2 (-1) + 4 = - 1 + 4 + 2 + 4 = 9

c ) lim ( t2 – 2t + 1) = (3)2 – 2 (3) + 1 = 9 – 6+1 = 4.
T→ 3


d) lim ( 3x3 – 2x + 7) = 3(-2)3 – 2 (-2) + 7 = 3 (-8) + 4 + 7 = -24 +4+7 = -13
X → -2




e) lim ( m2 + 3) (m-4) = lim ( m2 + 3) . lim ( m – 4)
m→√2 m→√2m→√2

=[( √2)2 + 3] . [√2 – 4] = (2 + 3). (√2 – 4) = 2√2 – 8 + 3√2 – 12

= 2√2 + 3√2 -20 = 5√2 – 20

f) lim [( x2 – 9) (x +3)] = [(1/2)2 - 9] [ ½ + 3] = ( ¼ - 9) (½ -9) = ( 1- 36) ( 1-18) = (-35) ( -17) = 595
X→1/2 4 2 4 2 8


g) lim [(2X2 -3X + 2) ( 4 X2 + 2X + 2)] = Lim ( 2x2 – 3x+2) . lim ( 4x2 + 2 +2) = (2(0)2 -3 (0) +2) .(4(0)2 +2 (0) +2 = 2.2 = 4
x→0 x→0



h) lim (x + 1) ( +3) = lim ( x+1). Lim (x+3) = (-2+1) (-2+3) = (-1) (1) = 1/4
X→-2 x -2x-→-2 (x -2) -2-2 -4



i)lim (2x+1)2 (x-1) = lim (2x+1)2 . lim (x-1) = [2(-1/2)+1]2[(-1/2) -1] = (-1+1)2 . (-3/2) = 0 = 0
x→-1/22x-1 x →-1/2 (2x-1) 2(-1/2)-1 -1-1 -2








J) lim/x→2 (5x+1) (x+1) = lim/x→2 (5x+1). Lim (x+1) =2x -3 lim/x→2 (2x-3)

= (5x(2) +1) (2+1) = ( 10+1)(3) = 33 = 33
2(2)-3 4 -...