Question

lim of (x^4 + x)/(x^6 + x^2 + 3) as x approaches infinity

Answer 1

I'm going to try the limit comparison theorem. f(x) resembles 1/x^2

Answer 2

i think its 0 ha

Answer 3

can you show the steps involve pls

Answer 4

\[\frac{ a _{k} }{ b _{k} } = \frac{ x ^{4}+x }{ x ^{6}+x ^{2}+3 }*x ^{2} \rightarrow \lim_{x \rightarrow \infty} \frac{ x ^{6}+x ^{3} }{ x ^{6}+x ^{2}+3 } \rightarrow \frac{ 1+\frac{ 1 }{ x ^{3} } }{ 1+\frac{ 1 }{ x ^{4} } +\frac{ 3 }{ x ^{6} }}\]

Answer 5

by the limit comparison the limit diverges because it approaches 1 which is not equal to 0

Answer 6

I'm not sure if this is right so let me check it in wolfram.

Answer 7

wolfram says zero is the correct answer.

Answer 8

where does the x^2 come from?
as well as the 1/x^3, 1/x^4, 3/x^6?

Answer 9

im playin erbody chill.

Answer 10

\[\frac{ x ^{4} +x}{ x ^{6}+x ^{2}+3 } \rightarrow \frac{ \frac{ 1 }{ x ^{2} }+\frac{ 1 }{ x ^{5} } }{ 1+\frac{ 1 }{ x ^{4} }+\frac{ 3 }{ x ^{6} }} \rightarrow 0\]

Answer 11

just i need fans :)

Answer 12

this is how I think they got zero. They just divided everything by the highest power of x

Answer 13

plug infinity into x and set the 1 over any power of x equal to zero

Answer 14

you will be left with 0/1 which equals 0

Answer 15

thanks

Answer 16

wait im rite?

Answer 17

no problem.

Answer 18

can you help me with an antiderivative problem

Answer 19

please

Answer 20

\[\int\limits_{0}^{1} x^2 (x^2 + 1)^3 dx\]

Answer 21

find the anti derivative

Answer 22

make a new thread for it and I will.

Answer 23

ok thanks