Question

Clarification

Answer 1

Someone posted this problem before. was wondering how else we could evaluate it.

\[\lim_{x \rightarrow \infty }\frac{ -2x^{3}+x }{ -4x^{5}+2x^{2}+2 }\]

Don't know if I remember, but could we use L hotails rule

\[\frac{ \frac{ d }{ dx }-2x^{3}+x }{ \frac{ d }{ dx }-4x^{5}+2x^{2}+2} = \frac{ -6x^{2}+1 }{ -20x^4+4x }\]

Answer 2

the degree of the numerator is less than the degree of the denominator is all you need

Answer 3

that's it?

Answer 4

yup

Answer 5

remember in some pre calc class you learned how to find a horizontal asymptote? if the degree of the numerator is less than the degree of the denominator it was \(y=0\)
that is the same question

Answer 6

Totally forgot that. that's why I was thinking something foolish like this:

\[\frac{ 1}{ 20x^{2} } = \frac{ 1 }{ 20(\infty)^{2} } = \frac{ 1 }{ \infty } = 0\]

Answer 7

it is not like that method doesn't work, you could use l'hopital 3 times !! but what a waste of gunpowder

Answer 8

yeah, that's what I was thinking :(

so it's just the fact that because the power of x is larger in the denominator the denominator grows much aster than the numerator and then goes to zero..

oh yeah! I see now

Answer 9

yes

Answer 10

thank you

Answer 11

yw