Question

HOW WOULD I DO THE FOLLOWING QUESTION:
find all the values of c for which the following exists...

Answer 1

\[\lim_{x \rightarrow c} \frac{ x^3+3x^2-4x-12 }{x-c}\]

Answer 2

the limit will only exist if you can cancel the \(x-c\) top and bottom, otherwise you will have a zero in the denominator, and not a zero in the numerator
this is another way of saying that \(c\) must be a zero of
\[x^3+3x^2-4x-12\] because if so, then by the factor theorem you can factor
\[x^3+3x^2-1x-12\] as \((x-c)(something)\)

Answer 3

in other words, you solve this question by finding the zeros of the numerator
those are your possible values for \(c\)

Answer 4

am i correct when i say that c cannot equal 2 or -2 or -3 as the limit does not exist

Answer 5

you are asked for values of c that work, not values of c that do not work

Answer 6

if those are the zeros of the numerator, then those are the only values of c that DO work
i.e. the only values of c for which the limit exists
otherwise it will not exist

Answer 7

but none of those values (they are in the numerator) cancel with the denominator.. doesn't the denominator need to cancel in order for the limit to exist?

Answer 8

yes, the question asks which ones give a limit, not which ones do not

Answer 9

set the numerator equal to zero and solve
those are the numbers that will give a limit
any other numbers will not

Answer 10

so all i have to write (this is an assignment)would be..

\[\lim_{x \rightarrow c}\frac{ x^3+3x^2-4x-12 }{x-c}\]

\[x^3+3x^2-4x-12 = 0\]
\[x^2(x+3)-4(x+3)\]
\[(x^2-4)(x+3)\]
\[(x+2)(x-2)(x+3)\]
x=-2, x=2, or x=3

Answer 11

ok so those are the numbers for which you WOULD get a limit

Answer 12

how though? if i plug it in i get 0/0

Answer 13

ok lets go slow
also note that if \(x+3\) is a factor, then the zero is \(-3\)

Answer 14

\[\frac{0}{0}\] is your friend
it means you could get a limit
\[\frac{5}{0}\] means no way you can get a limit

Answer 15

nevermind i got real numbers

Answer 16

so if the numerators is not zero, but the denominator is, then no limit possible

Answer 17

thank you but the way i had it done out here, will that be ok to write out on my assignment or should it written out differently?

Answer 18

you should say that those number WILL give you a limit, and only those numbers
lets compute one
\[\lim_{x\to 2}\frac{(x+2)(x-2)(x+3)}{x-2}=\lim_{x\to 2}(x+2)(x+3)=4\times 5=20\]

Answer 19

Thanks again! don't know what i would've done without your help tonight!!!

Answer 20

yw, hope this one is more clear now