Question

solve this limit:

Answer 1

\[\lim_{x \rightarrow -1}\frac{ 2x^2+5x+3 }{ x-1 }\]

Answer 2

can u factor
2x^2+5x+3 ?

Answer 3

Just plug in -1 silly burh! :)

Answer 4

2x^2+2x+3x+3

Answer 5

just do a direct substitution.....

Answer 6

i get 0/0 if i do direct subs.

Answer 7

Oh maybe you wrote something down wrong then, cause it looks like you get -2 in the bottom

Answer 8

the denominator does not end up with 0....

Answer 9

now take 2x common from 1st 2 terms of 2x^2+2x+3x+3

Answer 10

does x->1 ??

Answer 11

\[\lim_{x \rightarrow -1} \frac{ 2x^2+5x+3 }{x+1 }\]
sorry for the typo :$

Answer 12

ok, factor the numerator

Answer 13

2x^2+2x+3x+3

Answer 14

then?

Answer 15

i dont know what's next. Substitution ? :S

Answer 16

take 2x common from 2x^2+2x
what u get ?

Answer 17

2(x^2+x)

Answer 18

*** 2x(x+1)

Answer 19

take 3 common from 3x+3

Answer 20

+3(x+1)

Answer 21

so
2x(x+1) + 3(x+1) = ??

Answer 22

(2x+3)(x+1)

Answer 23

\(\huge \checkmark\)
now put that in numerator

Answer 24

can u see anything cancelling ?

Answer 25

yes (2x+3) would remain after cancelling, you would substitute next to get 1 as your answer

Answer 26

thanks :D

Answer 27

welcome ^_^