Chat Thread

Cam wrote:

Aguerotelli wrote:

Can you do me a favour and check if Soccer A.M are in the full game? i know they was in the demo but are they in the full game? Max Rushden looked beast up front

They are in the full game

SBSP wrote:

NZG wrote:Who has a very high knowledge of sequences and limits? Can't do a question

Mr Leiva wrote:http;//radiohead,com/msgboard/

Most bizarre forum layout I have seen.

SBSP wrote:St. Louis Cardinals win the World Series.

#idgaf

Cam wrote:

SBSP wrote:

1,3,5,7,9,_

What is next?

SBSP wrote:I copied your sig, Rei.

ahlycotc wrote:

Cam wrote:

1,3,5,7,9,_

What is next?

42

Cam wrote:

SBSP wrote:

1,3,5,7,9,_

What is next?

BladeGunner14 wrote:

Cam wrote:

1,3,5,7,9,_

What is next?

Kevin Bacon?

Cam wrote:

ahlycotc wrote:

42

It's 11.

BladeGunner14 wrote:

Cam wrote:

1,3,5,7,9,_

What is next?

Kevin Bacon?

SBSP wrote:How does Aguerotelli have 74 posts on his first day?

Rei Andros wrote:

NZG wrote:Who has a very high knowledge of sequences and limits? Can't do a question

I drunkenly stumbled back onto the forum. Can't say I'll be able to solve it but I'll give it a crack.

Just saw the HP deleted screen with Lupin and Tonks.

SBSP wrote:How does Aguerotelli have 74 posts on his first day?

Aguerotelli wrote:

BladeGunner14 wrote:

Kevin Bacon?

Hello blade.

NZG wrote:

Rei Andros wrote:

I drunkenly stumbled back onto the forum. Can't say I'll be able to solve it but I'll give it a crack.

Just saw the HP deleted screen with Lupin and Tonks.

Find the limit of the sequence as n goes to infinity

n=(ln(n+1)^2)/(n^2)


There are no typos and best thing is.

No answer for it..

BladeGunner14 wrote:Why is the American math system so focused on limits? I was absolutely shocked when I first saw it. It's completely different to everything I did in the IB in high school.

Zlatan wrote:Aguerotelli, were you anyone else on a different forum?

SBSP wrote:

BladeGunner14 wrote:Why is the American math system so focused on limits? I was absolutely shocked when I first saw it. It's completely different to everything I did in the IB in high school.

I haven't done limits in maths since our first test.

BladeGunner14 wrote:

SBSP wrote:I haven't done limits in maths since our first test.

But it's all calculus stuff, no? What are you doing now?

SBSP wrote:

NZG wrote:

Find the limit of the sequence as n goes to infinity

n=(ln(n+1)^2)/(n^2)


There are no typos and best thing is.

No answer for it..

0. It has to be 0.

SBSP wrote:

BladeGunner14 wrote:

But it's all calculus stuff, no? What are you doing now?

Optimization.

NZG wrote:

Rei Andros wrote:

I drunkenly stumbled back onto the forum. Can't say I'll be able to solve it but I'll give it a crack.

Just saw the HP deleted screen with Lupin and Tonks.

Find the limit of the sequence as n goes to infinity

n=(ln(n+1)^2)/(n^2)


There are no typos and best thing is.

No answer for it..

Rei Andros wrote:

NZG wrote:

Find the limit of the sequence as n goes to infinity

n=(ln(n+1)^2)/(n^2)


There are no typos and best thing is.

No answer for it..

lim n->inf (ln(n+1)^2)/(n^2)

It's been a long time but I think you should apply L'Hospital's Rule. If you plug in the limit you get what's called infinity/infinity form. L'Hospital's Rule allows you to take the derivative of the function in this case.

By taking the derivative you get:

lim n->inf (ln(n+1)^2)/(n^2) = [2(n+1)]/(n+1)^2/2n

Simplify

lim n->inf (ln(n+1)^2)/(n^2) = 1/(n^2+1)

Divide the top and bottom by n^2

lim n->inf (ln(n+1)^2)/(n^2) = 1/n^2/(1+1/n^2)

Now plug in the limit again. Both 1/n^2 terms go to zero so the answer should be zero.

SBSP wrote:

NZG wrote:

Find the limit of the sequence as n goes to infinity

n=(ln(n+1)^2)/(n^2)


There are no typos and best thing is.

No answer for it..

0. It has to be 0.

Rei Andros wrote:

NZG wrote:

Find the limit of the sequence as n goes to infinity

n=(ln(n+1)^2)/(n^2)


There are no typos and best thing is.

No answer for it..

lim n->inf (ln(n+1)^2)/(n^2)

It's been a long time but I think you should apply L'Hospital's Rule. If you plug in the limit you get what's called infinity/infinity form. L'Hospital's Rule allows you to take the derivative of the function in this case.

By taking the derivative you get:

lim n->inf (ln(n+1)^2)/(n^2) = [2(n+1)]/(n+1)^2/2n

Simplify

lim n->inf (ln(n+1)^2)/(n^2) = 1/(n^2+1)

Divide the top and bottom by n^2

lim n->inf (ln(n+1)^2)/(n^2) = 1/n^2/(1+1/n^2)

Now plug in the limit again. Both 1/n^2 terms go to zero so the answer should be zero.