Question

Limit

Answer 1

@ganeshie8

Answer 2

\[\lim_{n \rightarrow \infty} a_n~~~where~~~a_n = \left( \frac{ n+1 }{ n-1 } \right)^n\]

Answer 3

L'hospital rule doesn't apply since it's not a function, right?

Answer 4

I don't think it would. I recognize that question, and I'm looking for it in my notes right now.

Answer 5

No good. Sorry, I can't find it.

Answer 6

is answer e^(2)

Answer 7

I don't care about the answer, I want to know how to do it.

Answer 8

I'm guessing we have to let f(x) = f(a_n) or something...

Answer 9

if I know the answer i can counter check. thats why i am asking

Answer 10

How did you get e^2?

Answer 11

That's from calc 1? Rings a bell, can't remember it entirely though haha.

Answer 12

Techworm I think you're right

Answer 13

remember this definition of e^x ?

\[\large \lim\limits_{t\to \infty} \left(1+\frac{x}{t}\right)^{t} = e^x \]

Answer 14

\[f(x) = \left( \frac{ x+1 }{ x-1 } \right)^x\]

Answer 15

Notice that n+1 = n-1+2

Answer 16

\[\Large \lim\limits_{n\to \infty} \left(\frac{n+1}{n-1}\right)^{n} = \lim\limits_{n\to \infty} \left(1 + \frac{2}{n-1}\right)^{n} \]

Answer 17

is that clear

Answer 18

I'm not sure why they =?

Answer 19

\[\Large \dfrac{n+1}{n-1} = \dfrac{(n-1)+2}{n-1} = \dfrac{n-1}{n-1} + \dfrac{2}{n-1} = ?\]

Answer 20

Oh derp, haha, that makes sense, thanks :P

Answer 21

Good one. @Astrophysics

Answer 22

are you really sure why it equals e^2 ?

Answer 23

the denominator has n-1 and not n,
doesn't that bother you ?

Answer 24

Yeah it does lol, I was trying to get my head around it.

Answer 25

as n goes to infinity,

n-1 is same as n

Answer 26

n-1000000000000 is same as n

Answer 27

finite numbers don't matter for the limit

Answer 28

Nice!

Answer 29

I get it now, thanks

Answer 30

If you don't have faith in that, you can work it out using substitution :

substitute t = n-1
the limit becomes :
\[\lim\limits_{t\to \infty} \left(1+\frac{2}{t}\right)^{t+1}\]

Answer 31

\[ \lim\limits_{t\to \infty} \left(1+\frac{2}{t}\right)^{t+1} = \lim\limits_{t\to \infty} \left(1+\frac{2}{t}\right)^{t} \times \lim\limits_{t\to \infty} \left(1+\frac{2}{t}\right) \]