1. ## asymptotic sequences

which of the following sequences is asymptotic to the sequence

[(e^-n)+sinn-n^(1/3)]/[2cosn-n^(5/6)]?

a) [-e^(-n)]/n^(5/6)]

b) e^-n

c) sinn/2cosn

d) 1/n^(1/2)

I was thinking that the part of the given sequence that "has more weight" is e^-n so..well,...no..uhmm...WHY????

2. Originally Posted by Aglaia
which of the following sequences is asymptotic to the sequence [(e^-n)+sinn-n^(1/3)]/[2cosn-n^(5/6)]? a) [-e^(-n)]/n^(5/6)] ; b) e^-n; c) sinn/2cosn; d) 1/n^(1/2) i was thinking that the part of the given sequence that "has more weight" is e^-n so..well,...no..uhmm...WHY????
For large n:
[e^(-n) + sin(n) - n^(1/3)]/[2cos(n) - n^(5/6)] -> 1/n^(1/2).

Why?
e^(-n) -> 0 and sin(n) and cos(n) become indeterminate, but vary between -1 and 1. However both n^(1/3) and n^(5/6) get very large for large n, so these will be the dominant terms. Thus

[e^(-n) + sin(n) - n^(1/3)]/[2cos(n) - n^(5/6)] -> -n^(1/3) / -n^(5/6) = 1/n^(1/2)

-Dan

3. Originally Posted by Aglaia
which of the following sequences is asymptotic to the sequence

[(e^-n)+sinn-n^(1/3)]/[2cosn-n^(5/6)]?
multiply top and bottom by n^{-1/3} to get:

[n^{-1/3} e^{-n}+n^{-1/3}sin(n)-1]/[2 n^{-1/3} cos(n) - n^{1/2}]

which behaves like 1/[n^{1/2}] for large n (because all the other terms at the top
and on the bottom go to 0 as n becomes large), which is answer d).

a) [-e^(-n)]/n^(5/6)]

b) e^-n

c) sinn/2cosn

d) 1/n^(1/2)

I was thinking that the part of the given sequence that "has more weight" is e^-n so..well,...no..uhmm...WHY????
RonL