clara1223
  • clara1223
given that f(x)=(x^2)-4x, evaluate the limit as x approaches 2 of (f(x)-f(3))/(x-2)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
first you have to input f(x) into the equation, can you do that?
clara1223
  • 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)
clara1223
  • clara1223
f(3)=-3 not -1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
that is correct, now you must factor the top, can you do that?
anonymous
  • anonymous
CORRECTION: the top portion is actually x^2 - 4x + 3 because you're subtracting f(3)
amistre64
  • amistre64
are you sure its f(3) ?
jdoe0001
  • jdoe0001
hmmm
clara1223
  • clara1223
oops, didn't notice that it was minus f(3)
clara1223
  • clara1223
so that leaves me with (x-1)(x-3)/(x-2)
jdoe0001
  • jdoe0001
well... hmm f(3) is -3, -f(3) is +3
clara1223
  • 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?
jdoe0001
  • jdoe0001
but I gather that's what amistre64 meant, using f(3), doesn't give us much relief on cancelling x-2
clara1223
  • clara1223
it's definitely f(3)
anonymous
  • 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
anonymous
  • anonymous
How about the lim = infinity?
clara1223
  • clara1223
That's what I thought, thanks for confirming everyone!
amistre64
  • amistre64
some foreshadowing tho: \[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\] is something that will come up alot later on
anonymous
  • anonymous
there is a vertical asymptote at x=2 and it is non-removable if you need an explanation
anonymous
  • anonymous
@meagannlaurell the limit exists but at infinity, it is different from "does not exist"
clara1223
  • clara1223
In my class we interchange infinity and does not exist
anonymous
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.