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
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.
3 answers
is the correct answerr
is this the right answer?
1 answer