Question

Evaluate the limit, it it exists.
lim_x->2 (x - sqrt(3x-2)) / (x^2-4)

Answer 1

Neep help with algebra in canceling part of the denominator

Answer 2

\[\lim_{x \rightarrow 2} \frac{ x-\sqrt{3x-2} }{ x^2 -4}\]
multiply numerator and denominator by `x+ sqrt{3x-2}

Answer 3

by the conjugate

Answer 4

I get limit_x->2 \[(x^2-3x+2)/(x^2-4)\]

Answer 5

And also denominator I forgot times conjugate

Answer 6

limx>-2 \[\frac{ x^2-3x+2 }{ (x+2)(x-2)(x-\sqrt(3x-2) }\]

Answer 7

The numerator I don't think is factorable

Answer 8

hmm yea denominator looks mssy

Answer 9

it's factorable. :=))

Answer 10

Oops nvm it is

Answer 11

I get the limit as 1/16. Thanks

Answer 12

hmm

Answer 13

Nnesha without any steps than substitute the 2 in place of x will get like rezult 0/0

Answer 14

yes i know the fraction is undefined with denominator zero

Answer 15

do you know how to apply l'hopital's rule ?

Answer 16

but with your first step will get squarroot in denominator what i think not is right

Answer 17

sorry but i ve learned this rule 30 years ago

Answer 18

it's okay to get the square root at the denominator but after solving it i noticed that its still 0

Answer 19

oh no that question was for `anthonyym`

Answer 20

yes i know

and from this

x2−3x+2
--------------------- will result for x->0 so will result 2/0 yes ?
(x+2)(x−2)(x−(√3x−2)

Answer 21

??x->2

Answer 22

ohhh yes

Answer 23

sorry

Answer 24

so than will get 0/0

Answer 25

That's correct it's 1/16 good job

Answer 26

I wrote negative but its x+sqrt (3x-2)