Question

Prove that the following limit exists and evaluate:

lim (as n goes to infinity) ((n^2+3n+5)/(n^2+4n+7))^n

Answer 1

compare leading terms

Answer 2

just an idea

Answer 3

n^2
-----
n^(2n)


the bottom seems to be bigger than the top to me, what would this suggest?

Answer 4

hmm, is that all to the ^n or just the bottom?

Answer 5

it's the whole thing

Answer 6

so it would be ((n^2/n^2))^n

Answer 7

the ^n part messes with us, things change at different rates

Answer 8

the wolf gives us 1/e but how to process it

Answer 9

I have no idea.

Answer 10

1
----------
n^2+4n+7 | n^2+3n+5
-(n^2+4n+7)
------------
-2n-2



soo

\[(1-2(\frac{n+1}{n^2+4n+7})^n\]

Answer 11

so, this is prolly going to be related to:

(1+f(n))^n

Answer 12

\[(1+f(n))^n=\binom{n}{0}f^0(n)+\binom{n}{1}f^1(n)+...+\binom{n}{n}f^n(n)\]

Answer 13

have you come across a thrm yet that says: (1+r/n)^(nt) as n to inf is equal to e^{rt} ?

Answer 14

yes I just came across it.

Answer 15

the less compact form of the binomial might be more useful here, just some ideas that are rolling around in my head

Answer 16

i still don't see how that theorem gets 1/e like wolfram alpha does

Answer 17

\[\binom{n}{k}=nCk=\frac{nPk}{k!}=\frac{n(n-1)(n-2)...(n-k+1)}{k(k-1)(k-2)...3.2.1}\]
ugh, i used to know the process involved .... been along time

Answer 18

would it be worth getting wolfram alpha pro so I could see the process?

Answer 19

not to me, but thats your call

Answer 20

can we make a taylor poly for this?

Answer 21

......I don't know what that is

Answer 22

its when we equate a polynomial to a function by taking derivatives to construct it

Answer 23

I think we learn that in our differential equations class and we haven't covered it yet so I doubt that he would give a question where we had to solve it with that.

Answer 24

im sure it deals with a compound interest limiting to a continuous interest setup

Answer 25

I'm so confused

Answer 26

1 + n f^0 + n(n-1)/2! f(n) + n(n-1)(n-2)/3! f^2(n) + ...

\[(1+2\frac{-n-1}{n^2+4n+7})^n\]
\[1+2n\left(\frac{-n-1}{n^2+4n+7}\right)+\frac{1}{2!}2n(n-1)\left(\frac{-n-1}{n^2+4n+7}\right)^2 \\
~~~~~+\frac{1}{3!}2n(n-1)(n-2)\left(\frac{-n-1}{n^2+4n+7}\right)^3+...\]
so im thinking as n to infinity we deal with the leading terms rules

having a hard time with this new keyboard, might be better on paper

Answer 27

\[1+\frac{(-2)^1}{1!}\frac{n(n)^1}{(n^2)^1}+\frac{(-2)^2}{2!}\frac{n^2(n)^2}{(n^2)^2}+\frac{(-2)^2}{3!}\frac{n^3(n)^3}{(n^2)^3}\]
the n terms are all the same degrees so they go to 1 and we are left with
\[\sum_{n=0}^{\infty}\frac{(-2)^n}{n!}\]

Answer 28

almost .... that gets us 1/e^2 not 1/e

Answer 29

\[\lim_{n\to\infty}\left(\dfrac{n^2+3n+5}{n^2+4n+7}\right)^n=\exp\left(\lim_{n\to\infty}n\ln\left(\dfrac{n^2+3n+5}{n^2+4n+7}\right)\right)\]

\[=\exp\left(\lim_{n\to\infty}\dfrac{\ln\left(\dfrac{n^2+3n+5}{n^2+4n+7}\right)}{\dfrac{1}{n}}\right)\]

now use L'Hospitals rule on the inside part

Answer 30

\[\begin{align*}\lim_{n\to\infty}\left(\frac{n^2+3n+5}{n^2+4n+7}\right)^n&=\exp\left\{\ln\left[\lim_{n\to\infty}\left(\frac{n^2+3n+5}{n^2+4n+7}\right)^n\right]\right\}
\end{align*}\]
where (if you're unfamiliar with the notation, \(\exp[f(x)]=e^{f(x)}\)). Due to the continuity of the logarithmic function for \(x>0\), we can use the property,
\[\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)\]
which in this case would be to say
\[\begin{align*}\lim_{n\to\infty}\left(\frac{n^2+3n+5}{n^2+4n+7}\right)^n&=\exp\left\{\lim_{n\to\infty}\left[\ln\left(\frac{n^2+3n+5}{n^2+4n+7}\right)^n\right]\right\}
\end{align*}\]
(From here on I'll abbreviate the left side to \(L\).)

Apply the exponent property for logarithms, i.e. \(\ln a^b=b\ln a\), giving
\[\begin{align*}L&=\exp\left\{\lim_{n\to\infty}\left[n\ln\frac{n^2+3n+5}{n^2+4n+7}\right]\right\}
\end{align*}\]
Evaluating directly yields the indeterminate form \(0\times\infty\), so rearrange to a more useful expression:
\[\begin{align*}L&=\exp\left\{\lim_{n\to\infty}\frac{\ln\dfrac{n^2+3n+5}{n^2+4n+7}}{\dfrac{1}{n}}\right\}
\end{align*}\]
Now evaluating directly yields the preferred \(\dfrac{0}{0}\) indeterminate form. Apply L'Hopital's rule.
\[\begin{align*}L&=\exp\left\{\lim_{n\to\infty}\frac{\dfrac{\dfrac{(n^2+4n+7)(2n+3)-(n^2+3n+5)(2n+4)}{(n^2+4n+7)^2}}{\dfrac{n^2+3n+5}{n^2+4n+7}}}{-\dfrac{1}{n^2}}\right\}\\\\

&=\exp\left\{\lim_{n\to\infty}\dfrac{-n^2(n^2+4n+1)}{(n^2+3n+5)(n^2+4n+7)}\right\}
\end{align*}\]
Now you can consider the ratio of the coefficients of the leading terms.

Answer 31

Note that this doesn't PROVE the limit exists, but it's just the computation of it. For the proof, it might be sufficient to show that the sequence converges (i.e. is bounded and monotonic).

Answer 32

Seeing as we know the limit to be \(\dfrac{1}{e}\), we can make use of the limit definition of \(e^{-1}\) and draw some sort of conclusion from a comparison.
\[e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\]