https://www.youtube.com/watch?v=Y5zCUpQy6Rw
Hola
Para x -> -3
NO hay indeterminación,
simplemente
L = (2 - 3(-3) - 2(-3)^2)/(16+6(-3)-(-3)^2)
L = (2+9-18)/(16-18-9)
L = -7/-11
L = 7/11
Resulta que para
x -> -2
SI hay indeterminación,
así que voy a suponer eso
Dividendo
(-1) (2 x^2 + 3 x - 2)
(-1) (2 x^2 + 4 x - x - 2)
(-1) ( 2 x (x + 2) - 1 (x + 2)
(-1) (2 x - 1) (x + 2)
Divisor
(-1) (x^2 - 6 x - 16)
(-1) (x^2 + 2 x - 8 x - 16)
(-1) (x (x + 2) - 8 (x + 2))
(-1) (x - 8) (x + 2)
El cociente queda
L = Lim[(-1) (2 x - 1) (x + 2)]/[(-1) (x - 8) (x + 2)]
x-> -2
L = Lim[(2 x - 1) ]/[(x - 8) ]
L =(2(-2)-1) / ((-2) - 8)
L = -5/-10
L = 1/2
**************
(para x-> -2)
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Answers & Comments
https://www.youtube.com/watch?v=Y5zCUpQy6Rw
Hola
Para x -> -3
NO hay indeterminación,
simplemente
L = (2 - 3(-3) - 2(-3)^2)/(16+6(-3)-(-3)^2)
L = (2+9-18)/(16-18-9)
L = -7/-11
L = 7/11
Resulta que para
x -> -2
SI hay indeterminación,
así que voy a suponer eso
Dividendo
(-1) (2 x^2 + 3 x - 2)
(-1) (2 x^2 + 4 x - x - 2)
(-1) ( 2 x (x + 2) - 1 (x + 2)
(-1) (2 x - 1) (x + 2)
Divisor
(-1) (x^2 - 6 x - 16)
(-1) (x^2 + 2 x - 8 x - 16)
(-1) (x (x + 2) - 8 (x + 2))
(-1) (x - 8) (x + 2)
El cociente queda
L = Lim[(-1) (2 x - 1) (x + 2)]/[(-1) (x - 8) (x + 2)]
x-> -2
L = Lim[(2 x - 1) ]/[(x - 8) ]
x-> -2
L =(2(-2)-1) / ((-2) - 8)
L = -5/-10
L = 1/2
**************
(para x-> -2)