# Thread: find limit by rearrangement or L'Hopital's?

1. ## find limit by rearrangement or L'Hopital's?

Hi, 1st post.
According to my text book the following limit converges to $\displaystyle e^2$:
$\displaystyle \lim_{n\to\infty}(\frac {n + 3}{n + 1})^n$

through rearrangement i can achieve this:

$\displaystyle (\frac {n * (1 + 3/n)} {n * (1 + 1/n)})^n$

giving...

$\displaystyle \frac {(1 + 3/n)^n} {(1 + 1/n)^n}$

The denominator evaluates to $\displaystyle e$ but i can't see how to make the numerator equal to $\displaystyle e^3$ in order to achieve
agreement with the textbook answer. I had a look at using L'Hopital's but
the resulting expression after differentiation looked no simpler.

thanks

2. Originally Posted by framer
Hi, 1st post.
According to my text book the following limit converges to $\displaystyle e^2$:
$\displaystyle \lim_{n\to\infty}(\frac {n + 3}{n + 1})^n$

through rearrangement i can achieve this:

$\displaystyle (\frac {n * (1 + 3/n)} {n * (1 + 1/n)})^n$

giving...

$\displaystyle \frac {(1 + 3/n)^n} {(1 + 1/n)^n}$

The denominator evaluates to $\displaystyle e$ but i can't see how to make the numerator equal to $\displaystyle e^3$ in order to achieve
agreement with the textbook answer. I had a look at using L'Hopital's but
the resulting expression after differentiation looked no simpler.

thanks
Suppose the limit converges to L then

$\displaystyle L=\lim_{n\to\infty}(\frac {n + 3}{n + 1})^n$

taking the natural log of both sides we get

$\displaystyle \ln(L)=\lim_{n\to\infty} \ln [(\frac {n + 3}{n + 1})^n]$

$\displaystyle \ln(L)=\lim_{n\to\infty} n\ln [(\frac {n + 3}{n + 1})]$

$\displaystyle \ln(L)=\lim_{n\to\infty} \frac{\ln [(\frac {1 + 3/n}{1 + 1/n})]}{\frac{1}{n}}$

now finally using L'H rule we get after a ALOT of algebra

$\displaystyle \ln(L)=\lim_{n\to\infty} \frac{2n^2}{(n+3)(n+1)}$

$\displaystyle \ln(L)=2 \iff L = e^2$

I hope this helps.

3. Originally Posted by framer
...
giving...

$\displaystyle \frac {(1 + 3/n)^n} {(1 + 1/n)^n}$

The denominator evaluates to $\displaystyle e$ but i can't see how to make the numerator equal to $\displaystyle e^3$ in order to achieve
agreement with the textbook answer. ...
You already know that

$\displaystyle \lim_{t \mapsto \infty}\left(1+\frac1t\right)^t = e$

Now set $\displaystyle \frac1t = \frac3n~\implies~n = 3t$

Then

$\displaystyle \lim_{n \mapsto \infty}\left(1+\frac3n\right)^n~\buildrel {n=3t} \over \longrightarrow~\lim_{3t \mapsto \infty}\left(1+\frac1t\right)^{3t} = \lim_{3t \mapsto \infty}\left(\left(1+\frac1t\right)^t\right)^3 = e^3$

4. Thanks to both of you for such prompt replies.
Wondering if you might point out where my line of reasoning went wrong.
I thought that if numerator has to evaluate to $\displaystyle e^3$
then I would be satisfied if I could show that

$\displaystyle ((1+1/n)^n)^3 = (1+3/n)^n$

so I used the exponent identity: $\displaystyle a^n . b^n = (a . b)^n$

to get

$\displaystyle ((1+1/n) . (1+1/n) . (1+1/n))^n$

$\displaystyle = ((1 +1/n) . (1 + 2/n + 1/n^2))^n$

$\displaystyle = (1 + 3/n + 2/n^2 + 1/n^3)^n$

$\displaystyle != (1+ 3/n)^n$

Why does this approach not work? Have I made an algebraic error?

5. Originally Posted by framer
Thanks to both of you for such prompt replies.
Wondering if you might point out where my line of reasoning went wrong.
I thought that if numerator has to evaluate to $\displaystyle e^3$
then I would be satisfied if I could show that

$\displaystyle ((1+1/n)^n)^3 = (1+3/n)^n$

so I used the exponent identity: $\displaystyle a^n . b^n = (a . b)^n$

to get

$\displaystyle ((1+1/n) . (1+1/n) . (1+1/n))^n$

...
You have (correctly!) noted that $\displaystyle \frac3n = \frac1n+\frac1n+\frac1n$

That means:

$\displaystyle \left(1+\frac3n\right)=\left(1+\frac1n+\frac1n+\fr ac1n\right)$

But now you perform an algebraic salto mortale :

$\displaystyle \left(1+\frac1n+\frac1n+\frac1n\right) \neq \left(1+\frac1n\right) \cdot \left(1+\frac1n\right) \cdot \left(1+\frac1n\right)$

6. Originally Posted by framer
Thanks to both of you for such prompt replies.
Wondering if you might point out where my line of reasoning went wrong.
I thought that if numerator has to evaluate to $\displaystyle e^3$
then I would be satisfied if I could show that

$\displaystyle ((1+1/n)^n)^3 = (1+3/n)^n$

so I used the exponent identity: $\displaystyle a^n . b^n = (a . b)^n$

to get

$\displaystyle ((1+1/n) . (1+1/n) . (1+1/n))^n$

$\displaystyle = ((1 +1/n) . (1 + 2/n + 1/n^2))^n$

$\displaystyle = (1 + 3/n + 2/n^2 + 1/n^3)^n$

$\displaystyle != (1+ 3/n)^n$

Why does this approach not work? Have I made an algebraic error?
You're obviously correct in saying that (1 + 3/n + 2/n^2 + 1/n^3)^n != (1+ 3/n)^n.

However, $\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{2}{n^2} + \frac{1}{n^3}\right)^n$ IS equal to $\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n}\right)^n$ since the 3/n term dominates over the 2/n^2 and 1/n^3 terms in the limit ......

7. thanks for the reply, most appreciated but I have a nagging feeling of dissatisfaction with the explanation because ....

'if'...

$\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n}\right)^n = e^3$

wouldn't...

$\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{2}{n^2} + \frac{1}{n^3}\right)^n =$

a number just slightly larger than $\displaystyle e^3$?

Doesn't this break the definition of a limit of a sequence?

8. Originally Posted by framer
thanks for the reply, most appreciated but I have a nagging feeling of dissatisfaction with the explanation because ....
$\displaystyle 'if'...$\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n}\right)^n = e^3

wouldn't...

$\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{2}{n^2} + \frac{1}{n^3}\right)^n =$

a number just slightly larger than $\displaystyle e^3$? Mr F says: In the limit the difference goes to zero.

Doesn't this break the definition of a limit of a sequence?

By the way, it doesn't actually make a difference in the limit, but it should be $\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{{\color{red}3}}{n^2} + \frac{1}{n^3}\right)^n$

9. Sorry I somehow didn't see your post until now and now I have some questions

Originally Posted by earboth
You have (correctly!) noted that $\displaystyle \frac3n = \frac1n+\frac1n+\frac1n$

I did? Where?

That means:

$\displaystyle \left(1+\frac3n\right)=\left(1+\frac1n+\frac1n+\fr ac1n\right)$

But now you perform an algebraic salto mortale :

I did!? Ok I googled Salto Mortale and it said it was Italian for deadly jump, which sounds bad!

$\displaystyle \left(1+\frac1n+\frac1n+\frac1n\right) \neq \left(1+\frac1n\right) \cdot \left(1+\frac1n\right) \cdot \left(1+\frac1n\right)$
My algebra is clearly crummy but could you please explain where exactly I went awry in an even more obvious way (if that is possible?)

Is it this part of my reasoning that is wrong??:

$\displaystyle \left( \left( 1 + \frac{1} {n}\right) ^n \right) ^3 = \left(\left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n} \right)\right)^n ?$

I'm confused

10. ## Well

Originally Posted by framer
Sorry I somehow didn't see your post until now and now I have some questions

My algebra is clearly crummy but could you please explain where exactly I went awry in an even more obvious way (if that is possible?)

Is it this part of my reasoning that is wrong??:

$\displaystyle \left( \left( 1 + \frac{1} {n}\right) ^n \right) ^3 = \left(\left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n}\right) . \left( 1 + \frac{1}{n} \right)\right)^n ?$

I'm confused
IF you expand your adaptation you will see you do not get that

11. Originally Posted by Mathstud28
IF you expand your adaptation you will see you do not get that
Sure you do.

$\displaystyle \left[ \left(1 + \frac{1}{n}\right)^{n}\right]^{3} = \left[ \left(1 + \frac{1}{n}\right)^{3}\right]^{n} = \left[ \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right)\right]^{n}$

12. Originally Posted by o_O
Sure you do.

$\displaystyle \left[ \left(1 + \frac{1}{n}\right)^{n}\right]^{3} = \left[ \left(1 + \frac{1}{n}\right)^{3}\right]^{n} = \left[ \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right) \cdot \left(1 + \frac{1}{n}\right)\right]^{n}$

Yeah, that's the way I thought it expanded but an earlier poster
(earboth) said that I was making a fatal error somewhere but I am still no closer to understanding where

13. Originally Posted by earboth
You have (correctly!) noted that $\displaystyle \frac3n = \frac1n+\frac1n+\frac1n$

That means:

$\displaystyle \left(1+\frac3n\right)=\left(1+\frac1n+\frac1n+\fr ac1n\right)$

But now you perform an algebraic salto mortale :

$\displaystyle \left(1+\frac1n+\frac1n+\frac1n\right) \neq \left(1+\frac1n\right) \cdot \left(1+\frac1n\right) \cdot \left(1+\frac1n\right)$
What did earboth mean by this?

14. Originally Posted by framer
Yeah, that's the way I thought it expanded but an earlier poster
(earboth) said that I was making a fatal error somewhere but I am still no closer to understanding where
As you should see by now, there's a small salto but no mortale ..... earboth probably misread your working (it happens, especially late at night - I've done it myself many times)

15. Ah ok, thanks Mr F. I'm not confident with my maths so I tend to distrust my own
working which leads to confusion and negative thinking towards maths in general . I'm still having trouble understanding your argument about the last two terms are able to be ignored in the limit

$\displaystyle \lim_{n \rightarrow \infty} \left( 1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3} \right)^n$

I'm thinking I need to review a textbook on this; do you know of any online
content that talks about this sort of thing in depth?

thanks

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