Question

How to evaluate for:

lim x->2 2x^2-x-6 / 3x^2 -7x+2

Answer 1

plug in 2. if you get a number, done. if you get 0/0 it means you can factor and cancel and try again. if you get a non-zero number over zero there is no limit

Answer 2

in this case when you factor and cancel you get
\[\frac{2x+3}{3x-1}\]

Answer 3

I already tried factoring and there's nothing to cancel. Without factoring it's indiscriminate so I have to do something . I'm just not sure what: either direct sub, factoring, rationalizing, one sided limits or change of variable

Answer 4

now replace x by 2 to get the answer. i get
\[\frac{2\times 2+3}{3\times 2-1}=\frac{7}{5}\]

Answer 5

How did you get that as a factor? I got (x-4)(x+3)/(x-7)(x-1)

Answer 6

hold on.

Answer 7

did you replace x by 2?

Answer 8

hmm you did get the right answer but I'm not sure how you got that as a factor. Yup I did

Answer 9

ok if you replaced x by 2, did you get 0/0?

Answer 10

yep

Answer 11

ok that means by the 'factor theorem" that both numerator and denominator factor as
\[(x-2)\times something\]

Answer 12

if r is a zero of a polynomial P then P(x) = (x-r) Q(x) so you factored incorrectly

Answer 13

ohhhh I see what I did as a mistake

Answer 14

it was simply an error in factoring

Answer 15

btw it is obvious that you factored incorrectly because there is no way that (x-4)(x+3)= 2x^2 + anything

Answer 16

but don't think too hard. you have plugged in 2. you have 0/0 so you know how to factor. both top and bottom must be (x-2) times stuff

Answer 17

so just factor as
\[\frac{(x-2)(2x+3)}{(x-2)(3x+1)}\]

Answer 18

yeah I see where I messed up!

Answer 19

actually it is
\[\frac{2x+3}{3x-1}\]

Answer 20

but the simple point i am trying to make is this: if you know a zero, you know how to factor. i repeat, you know 2 gives you 0 so don't try to factor in any other way than
(x-2)(stuff)

Answer 21

oh okay, got it. Thanks!