Prove the limit of the sequence by definition

[phepo]

Hey guys. How can I prove it?

 

[Deleted member 4993]

Hey guys. How can I prove it?
View attachment 14646

Factor out n^2 from denominator and numerator and take the limit.

 

[pka]

Hey guys. How can I prove it?
View attachment 14646

$\displaystyle \frac{{8{n^3} + 16}}{{4{n^2} + 8}} = \frac{{4({n^3} + 4)}}{{4({n^2} + 2)}} \geqslant \frac{{2{n^3}}}{{{n^2} + 2}} = \frac{{2n}}{{1 + \frac{2}{{{n^2}}}}} \to ?$

 

[Steven G]

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 S_n= $\displaystyle \frac{{8{n^3} + 16}}{{4{n^2} + 8}}$

Now given any number k, you must find N such that S_n>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.