Is the horizontal asymptote of: g(x)=x/(x^2+2) zero or non existant?

Question

myininaya · Answer

y=0 since the degree of the numerator less than the degree of the denominator

myininaya · Answer

is less*

anonymous · Answer

Doesnt y=1/2?

amistre64 · Answer

HA = 0

myininaya · Answer

yes there is a horizontal asymptote at y=0

myininaya · Answer

there is no horizontal aymptote when the degree of the numerator is greater than the degree of the denominator

amistre64 · Answer

$$\frac{x}{x^2+2}*\frac{1/x^2}{1/x^2}=\frac{x/x^2}{x^2/x^2 + 2/x^2} $$
$$\frac{1/x}{1 + 2/x^2}$$
everything with an x in it goes to 0; and we are left with 0/1 = 0

anonymous · Answer

Sorry for the confusion

amistre64 · Answer

HA is for x at infinity; not x at 0

amistre64 · Answer

VA cant be crossed; HA can be crossed because its only concerned with the infity parts and doesnt care about the innards

myininaya · Answer

when the degree of the numerator equals the degree of the denominator
you just take the coefficient of the term that has the largest exponent and put it over the coefficient of the term with the largest exponent in the denominator

myininaya · Answer

and i missed the words "in the numerator" before the conjunction "and"

anonymous · Answer

I'm confused. I thought that horizontal (and verticle) asymptotes were attractors, not if x is oo or zero??

myininaya · Answer

i don't understand your question

anonymous · Answer

I know that the HA can be crossed becuase of the hltest

myininaya · Answer

hltest?

anonymous · Answer

nevermind, my computer is really slow so my messages come really late, but Im confused about where the HA is.

amistre64 · Answer

the HA is at y=0

amistre64 · Answer

it is the x axis

myininaya · Answer

it is what f(x) gets closer to as x gets really large (or negative large)
and if f(x) doesn't get closer to anything then we say there is no horizonal asymptote

anonymous · Answer

ok, that makes sense. so the HA is 0, right?

myininaya · Answer

lol yes my answer doesn't change
y=0 is the horizontal asymptote

anonymous · Answer

I checked this answer and it's not correct (online hmwk). I think that there is no HA.

anonymous · Answer

wait, actually that's not true either. their is an HA but I have not idea as to what it is.

myininaya · Answer

no, y=0 is the horizontal asymptote 
maybe you type your question in wrong

anonymous · Answer

It asks y=  , and I typed y=0