Question

given that f(x)=(x^2)-4x, evaluate the limit as x approaches 2 of (f(x)-f(3))/(x-2)

Answer 1

first you have to input f(x) into the equation, can you do that?

Answer 2

yes, I got f(3)=-1, and i input f(x) and f(3) into the equation and got ((x^2)-4x-3)/(x-2)

Answer 3

f(3)=-3 not -1

Answer 4

that is correct, now you must factor the top, can you do that?

Answer 5

CORRECTION: the top portion is actually x^2 - 4x + 3 because you're subtracting f(3)

Answer 6

are you sure its f(3) ?

Answer 7

hmmm

Answer 8

oops, didn't notice that it was minus f(3)

Answer 9

so that leaves me with (x-1)(x-3)/(x-2)

Answer 10

well... hmm f(3) is -3, -f(3) is +3

Answer 11

how do I get x-2 out of the denominator? or do I leave it there and the answer is that the limit doesn't exist?

Answer 12

but I gather that's what amistre64 meant, using f(3), doesn't give us much relief on cancelling x-2

Answer 13

it's definitely f(3)

Answer 14

It does not exist, if you check the graph you can see that approaching 2, one side goes towards -infinity, and one goes to +infinity

Answer 15

How about the lim = infinity?

Answer 16

That's what I thought, thanks for confirming everyone!

Answer 17

some foreshadowing tho:

\[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]
is something that will come up alot later on

Answer 18

there is a vertical asymptote at x=2 and it is non-removable if you need an explanation

Answer 19

@meagannlaurell the limit exists but at infinity, it is different from "does not exist"

Answer 20

In my class we interchange infinity and does not exist

Answer 21

Well looking at the graph, the two sides approach different infinities, one negative and one positive. So it would still be DNE, and yes, it is debatable as to whether DNE should equal -/+infinity, but many classes do view them as interchangeable