Prove the limit of the sequence by definition
- Thread starter phepo
- Start date
D
Deleted member 4993
Guest
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
\(\displaystyle \frac{{8{n^3} + 16}}{{4{n^2} + 8}} = \frac{{4({n^3} + 4)}}{{4({n^2} + 2)}} \geqslant \frac{{2{n^3}}}{{{n^2} + 2}} = \frac{{2n}}{{1 + \frac{2}{{{n^2}}}}} \to ?\)
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,561
What have you tried? What is the procedure to prove a limit?
First you need to know the limit is \(\displaystyle \infty\). Did you get that far?
Define Sn= \(\displaystyle \frac{{8{n^3} + 16}}{{4{n^2} + 8}}\)
Now given any number k, you must find N such that Sn>k \(\displaystyle \forall\) n>N
That is you must find N such that \(\displaystyle \frac{{8{n^3} + 16}}{{4{n^2} + 8}}>k \forall\) n>N
One way to find N is to find N as a function of k, ie find N(k)
Show us your work and we can see if you have it right or need some help.
First you need to know the limit is \(\displaystyle \infty\). Did you get that far?
Define Sn= \(\displaystyle \frac{{8{n^3} + 16}}{{4{n^2} + 8}}\)
Now given any number k, you must find N such that Sn>k \(\displaystyle \forall\) n>N
That is you must find N such that \(\displaystyle \frac{{8{n^3} + 16}}{{4{n^2} + 8}}>k \forall\) n>N
One way to find N is to find N as a function of k, ie find N(k)
Show us your work and we can see if you have it right or need some help.