Looking for something else?

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

## More answers

Looking for something else?

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

- clara1223

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

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- clara1223

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

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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

- 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

f(3)=-3 not -1

Looking for something else?

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

- anonymous

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

- anonymous

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

- amistre64

are you sure its f(3) ?

- jdoe0001

hmmm

- clara1223

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

- clara1223

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

- jdoe0001

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

- 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

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

- clara1223

it's definitely f(3)

- 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

How about the lim = infinity?

- clara1223

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

- 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

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

- anonymous

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

- clara1223

In my class we interchange infinity and does not exist

- 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.