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Mathematics 15 Online
OpenStudy (baldymcgee6):

Is there a horizontal asymptote in this function? y = (x+2)/((x−1)(x^2+3))

OpenStudy (anonymous):

Best way is to draw your function.

OpenStudy (baldymcgee6):

I have it graphed

OpenStudy (baldymcgee6):

Just because it appears on the graph that there is a horz. asymptote doesn't mean that it exists.

OpenStudy (anonymous):

vertical asymptote where the denominator is zero, namely \(x=1\)

OpenStudy (baldymcgee6):

right, got that part.. but the graph looks like it might have a horz. asymptote at y=0

OpenStudy (anonymous):

horizontal asymptote is \(y=0\) since the degree of the numerator is smaller than the degree of the denominator

OpenStudy (baldymcgee6):

is that a general rule?

OpenStudy (anonymous):

degree of numerator is 1, degree of denominator is 3 yes

OpenStudy (baldymcgee6):

okay, well then. How do i prove it using limits.. if we take the limit as x->1 it approaches infinity from both sides, but what about the horz. asym.?

OpenStudy (anonymous):

I've only ever used limits with vertical asymptotes but I don't really know... I would go with what satellite is saying because i have also heard that general rule.

OpenStudy (baldymcgee6):

@satellite73, is there a way to prove it using limits?

OpenStudy (anonymous):

it is obvious, but yes, you can think of what you would get if you replace \(x\) by \(10^6\)

OpenStudy (anonymous):

you would have \(10^6\) in the numerator, but you would have \(10^{18}\) in the denoninator

OpenStudy (baldymcgee6):

get a number close to zero

OpenStudy (anonymous):

sure do

OpenStudy (baldymcgee6):

i see.. alrighty thanks.

OpenStudy (anonymous):

yw

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