Limit of question

[seal308]

Hi,

I need help with the following limit.

Limit as n goes to infinity of (-1)^n (n^3 + 2)/(3n^2 + 5)

So what I did was factor out the highest power of n from the numberator and the denominator and simplified and I was left with:

(-1)^n * n(1/3).
But not sure what to do at this point. -1 to infinity would keep changing wouldn't it?

Thanks

 

[Ishuda]

Hi,

I need help with the following limit.

Limit as n goes to infinity of (-1)^n (n^3 + 2)/(3n^2 + 5)

So what I did was factor out the highest power of n from the numberator and the denominator and simplified and I was left with:

(-1)^n * n(1/3).
But not sure what to do at this point. -1 to infinity would keep changing wouldn't it?

Thanks

Yes it would so there is no limit [the limit doesn't exist]

 

[seal308]

Yes it would so there is no limit [the limit doesn't exist]

ok thx

 

[pka]

Limit as n goes to infinity of (-1)^n (n^3 + 2)/(3n^2 + 5)

So what I did was factor out the highest power of n from the numberator and the denominator and simplified and I was left with: (-1)^n * n(1/3).

These are known as alternating sequences. Therefore, usual methods do not work such as highest power.

Recall that a sequence has a limit if and only if any subsequence has the same limit.

The sequence \(\displaystyle \dfrac{n^3+2}{3n^2+5}\) does not have a limit. So can the alternating form have a limit?

 

[seal308]

These are known as alternating sequences. Therefore, usual methods do not work such as highest power.

Recall that a sequence has a limit if and only if any subsequence has the same limit.

The sequence \(\displaystyle \dfrac{n^3+2}{3n^2+5}\) does not have a limit. So can the alternating form have a limit?

According to that, no.
Never knew that, thanks.