Hey! more limits help plz:) lim as x->infiniti of (2x^4-x^2-8x)

f(x)->inf as x->infinity polynomials don't have horiztonal asymptotes

\[\lim_{x \rightarrow \infty} (2x ^{4}-x ^{2}-8x)\]

you know when we take x->inf, we are looking for horizontal asymptotes right?

didnt kno :(..but now i do! horizontal asymptotes are what make equation=0?

no you know what f(x)=1/x looks like?

so if we changed the eq up a bit n it was \[\lim_{x \rightarrow \infty} (2x ^{4}-x ^{2}+8x ) \div (-5x ^{4}+7)\]

yea

since the powers r the same..the horiz asymptotes r -2/5...is tht the limit?

omg very good

you catch on quick

lol...thnks:)) i try;)

f(x)=1/x has horiztonal asymptote y=0 because as x gets large f(x) is getting closer to zero another way to say this is \[\lim_{x \rightarrow \infty} \frac{1}{x}=0\]

what if it was x/1?

x is a polynomial it doesn't have a horiztonal asymptote you can say it does not exist or you can say infinity y=x looks like / so as x gets large f(x) gets large as x gets negative large f(x) gets negative large

oh so basically..wen the powers r equal..u jus divide..if the numerator power is larger thn no asym..if denom power is larger..it =0?

yes i think you mean when the degree of the top=degree of the bottom just take the coeifficient of term with highest exponent on top and put it over the coeifficient of term with highest exponent in bottom. and yes to the otehr part if degree higher on top, then no horizontal asymptote if degree higher on bottom, then y=0 is the horiztonal asymptote

ohhok..yeah thanks a lottt..u wer so wonderfully helpful! lol r u indian?

no im american

lol..jus wonderin cuz..most of the ppl here wer

i seen one indian but i don't really know how many there are i don't ask lol

haha..gud policy..i asked lik u n amistre....so r u good at integrals?

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