Asked by Megan

The nth term of a sequence is represented by 2n^4+25n^2+32n−15

____________________

6n^4+2n^3−11n^2−2n+17.

What is the limit of the the nth term as x becomes increasingly large?



A)0

B)1/3

C)3

D)The limit does not exist.

Answers

Answered by Reiny
Post it like this ....
(2n^4+25n^2+32n−15)/(6n^4+2n^3−11n^2−2n+17)

You want:
limit (2n^4+25n^2+32n−15)/(6n^4+2n^3−11n^2−2n+17) , as n ---> ∞
divide each term by the highest power of n , that is , by n^4
= lim (2 + 25/n^2 + 32/n^3 − 15/n^4)/(6 + 2/n − 11/n^2 − 2/n^3 + 17/n^4) as n ---> ∞
as n ---> ∞, all terms that still have n's in them will approach zero, so
lim (2 + 25/n^2 + 32/n^3 − 15/n^4)/(6 + 2/n − 11/n^2 − 2/n^3 + 17/n^4) as n ---> ∞
= lim (2/6)
= 1/3
Answered by hazem
is the correct answerr
Answered by hazem
is this the right answer?
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions