Question

Question below

Answer 1

@SolomonZelman @Loser66 @triciaal

Answer 2

Have you looked at a graph of f(x)?
Have you plugged in x=0? or x=-4?
Have you looked at limits as we approached these numbers?

Answer 3

we're not allowed to use graphing devices.

the answer is, 0 is the V.A. but -4 is not a V.A.

but one of the zeros in the denominator is -4, so it should be a V.A.
this is the part i'm not sure

Answer 4

at x=-4 you should get 0/0 when you plug in the -4
which means it could be a hole at x=-4

at x=0 you should get -2/0 which tells us we have a va at x=0


if you ever get something/0 where the top something isn't 0 then yep it is a va
but if you get 0/0 you have more work

Answer 5

how to prove it using limits?

Answer 6

cot(-4) clearly exists
but the problem is the other function evaluated at x=-4
\[\lim_{x \rightarrow -4} \frac{\sqrt{x^2+9}-5}{x^2+5x+4} \cdot \frac{\sqrt{x^2+9}+5}{\sqrt{x^2+9}+5}\]
try evaluating this limit
notice I multiplied that one function by a fancy one

Answer 7

okay. I get that part now. thanks..

how about the 0?

Answer 8

you should see this limit exists
which tells us we don't have a va

Answer 9

at x=-4

Answer 10

anyways if you plug in x=0 you get something/0
where the top something isn't 0

Answer 11

this means you have a va at x=0

Answer 12

consider evaluating this limit if you are not sure
\[\lim_{x \rightarrow 0}\cot(x)\]

Answer 13

I think if you just multiple the term as what freckles did, and then simplify, x-4 no more in denominator. Hence, no need to take the lim

Answer 14

And that is enough to conclude that x =4 is not a V.A

Answer 15

(x+4) ?
and x=-4? @Loser66 :p

Answer 16

Yes, hehehe.. sorry.

Answer 17

ok thank you both :)