## clara1223 one year ago given that f(x)=(x^2)-4x, evaluate the limit as x approaches 2 of (f(x)-f(3))/(x-2)

1. anonymous

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

2. clara1223

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)

3. clara1223

f(3)=-3 not -1

4. anonymous

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

5. anonymous

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

6. amistre64

are you sure its f(3) ?

7. jdoe0001

hmmm

8. clara1223

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

9. clara1223

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

10. jdoe0001

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

11. clara1223

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?

12. jdoe0001

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

13. clara1223

it's definitely f(3)

14. anonymous

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

15. anonymous

How about the lim = infinity?

16. clara1223

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

17. amistre64

some foreshadowing tho: $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ is something that will come up alot later on

18. anonymous

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

19. anonymous

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

20. clara1223

In my class we interchange infinity and does not exist

21. anonymous

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