Question

what is the limit if x -> negative infinity for f(x) = 1/x

Answer 1

what if it was infinity?

Answer 2

what do you mean

Answer 3

x-> infinity 1/x?

Answer 4

so if x is approaching negative infinity and the domain of 1/x is x<0; the limit is approaching infinity?

Answer 5

nono Im asking you what is 1/x as x approaches infinity

Answer 6

that's what i'm trying to find lol. The equation is a piecewise function: f(x) = 1/x x<0 and -1 x(included)>0. It's asking to find the limit of f(x) as it approaches negative infinity

Answer 7

yes NEGATIVE infinity
im asking if you know the answer for POSITIVE infinity

Answer 8

is it zero?

Answer 9

yes

Answer 10

i never asked for positive infinity lol

Answer 11

it can't because infinity does not comply with the domain

Answer 12

I know ... I did.

you did not say it was a peice wise function in the topic. I was trying to guide you to the answer....lim x->infinity(1/x) = lim x -> negative inifinity (1/x) = 0

Answer 13

lol, regardless. i was just looking for confirmation. thank you :)

Answer 14

np

Answer 15

oh if there is a limit of a constant like f(x) = -1 and x is approaching infinity? does that mean the limit of f(x) will always be a constant (in this case, -1?)

Answer 16

yes

Answer 17

even if it's negative infinity, it will still remain the constant?

Answer 18

yeah even if its lim x->32 it will still be -1

Answer 19

think about y = -1 and what happens to y = -1 as x gets bigger or approaches a number. its alwaus y = -1

Answer 20

oh alright! ok, what if x is approaching 0^-1 for f(x)= 1/x? since i can't plug in zero for the denominator, does that mean there's no finite limit?

Answer 21

*0^- (0 to the left, sorry)

Answer 22

x -> 1/0 I think is the same thing as x->infinity so it would be 0

Answer 23

my book says negative infinity

Answer 24

Did someone ask you that, or are you just asking because you thought of it?

Answer 25

hmm

Answer 26

lol!! i do have hw, but the general rules apply lol. i tried to ask from my notes

Answer 27

it says the lim x-> (1/0) of 1/x = -infinity?

Answer 28

yea... i'm trying to figure out why... is it because you can plug in any number to the left of 0 (which is negative) and divide that by the constant... which gives you a negative number approaching zero?

Answer 29

http://www.wolframalpha.com/input/?i=+lim+x-%3E+%281%2F0%29+of+1%2Fx+

hit the show steps button. I have never studied lim x -> 1/0 so im sort of guessing, but wolfram says its 0 also.

Answer 30

lol... well my calc book says infinity... so i'm kinda conflicted haha

Answer 31

wait

your book says lim x-> 1/0 of 1/x = - infinity

or does it say that limx->1/0 is the same thing as lim x-> -infinity

Answer 32

it says the limit of f(x) = 1/x as x approaches 0^- is negative infinity

Answer 33

that makes no sense to me.

wolfram alpha say = 0
my ti-89 says = 0

books are wrong all the time. but wolfram and ti are never both wrong:)

Answer 34

LOL yea i went on wolfram once... it was wrong. idk, right now, i only have trouble graphing the asymptotes. Not very many people on here helps with calculus

Answer 35

I very much doupt it was wrong, it was prob user error. what were you doing?

Answer 36

and my ti-89 has never been wrong.

Answer 37

i didn't use it, somebody on open study tried to solve a problem with wolfram and it wasn't correct. i'm sure it was user error

Answer 38

it makes no sense for it to be a negative number. first we are not approaching a negative number and there are no negatives in 1/x so why would it be negative. Second 1/0 is the same thing as infinity.If we are going to talk about infinity, which of course does not exist, then we can talk about 1/0 which of course does not exist. I dont see any place for there to be a negative.

Answer 39

the same domain applies (x<included 0) which makes sense that it's approaching negative

Answer 40

i guess?

Answer 41

oooooooooooooooooooooooooooooooh

Answer 42

lollllllllll is that right?

Answer 43

I dont know... i dont see how x can approach 1/0 if it is always negative.

Answer 44

well the domain is x< including 0, which is the only probable way to have x-> 0^-... and since 1/x, you can substitute x for any number to the left of 0 and see that when you divide by the constant, the number is getting relatively smaller and in a negative direction. That's what I thought... lol

Answer 45

post the question as a new question, and maybe some people who know more than me will answer.

Answer 46

lol i doubt it.... it's rare for calculus help. i see a lot of geometry, algebra and some precal. but thank you for you help!

Answer 47

There are lots of people here that are way beyond calc. jsut post it I will help you find someone that can help. I twill be a good post to. I have never seen this.