Question

why isn't the limit as x approaches -3 of (x^2 +6x + 9)/ (x^2 + 2x -3) equal to 1?

Answer 1

1. Let's write it better .
\[\huge \lim_{x\to (-3)}\;\;\; \frac{x^2+6x+9}{x^2+2x-3}\]

2. Let's see ! O_O

Answer 2

factorizing: (x+3)(x+3)/(x+3)(x-1) = (x+3)/(x-1)
taking limit equals 0

Answer 3

As @myko said ... factorizing we get :

\[\LARGE \lim_{x\to (-3)}\;\;\;\frac{(x+3)(x+3)}{(x-1)(x+3)}\]

\[\LARGE \lim_{x\to (-3)}\;\;\;\frac{(x+3)}{(x-1)}=\frac{-3+3}{-3-1}=\frac{0}{-4}=0\]

why do you think it should be equal to 1 ?

Answer 4

see i didnt factor...i subsituted -3 into the numerator and denominator... i got -18/-18

Answer 5

substituting we get:

\[\LARGE \frac{(-3)^2+6(-3)+9}{(-3)^2+2(-3)-3}=\frac{9-18+9}{9-6-3}\]

\[\LARGE \frac{-18}{9-9}=\frac{-18}{0}\]

and we know that dividing by 0 is not allowed ! ... you made a mistake while calculating O_O

Answer 6

thanks for the help

Answer 7

what if we factored 9 as (x- 3) (x-3)

Answer 8

Glad to help... You're welcome ! ;)

Answer 9

would that mattered

Answer 10

how do you mean ? O_O

Answer 11

((x- 3) (x-3))/ ((x-1) (x+3))

Answer 12

i guess since you are trying to cancel out...you would be force to use x+3

Answer 13

\[\LARGE (x-3)(x-3) \neq x^2+6x+9\]






\[\LARGE (x-3)(x-3) =x^2-3x-3x+9\]

\[\LARGE (x-3)(x-3) =x^2-6x+9\]

Answer 14

ohhhh yeah lol thats right

Answer 15

;)