Re: Limits and e

On Sep 7, 2005 5:59 PM, Joseph Fagan wrote:

> "Narcoleptic Insomniac"
> <i_have_narcoleptic_insomnia@xxxxxxxxx> wrote in
> message
> news:16774194.1126127519699.JavaMail.jakarta@nitrogen.
> mathforum.org...
> > On Sep 7, 2005 4:00 PM, Joseph Fagan wrote:
> >>
> >> "Jim Spriggs"
> >> <jim.sprigs@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote
> >> in message
> >> news:431F4FBF.F19C6456@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >> ..
> >> > Joseph Fagan wrote:
> >> >>
> >> >> Limit[(1+1/n)^nx] as n->inf = Limit[(1+x/n)^n]
> >> >> as n->inf
> >> >>
> >> >> Can anyone give me a reference.
> >> >> Both limits are e^nx
> >> >
> >> > I think the "n" in "e^nx" is a typo. Since
> >> > "n->inf" is a variable
> >> > binding operator, "n" can't appear free as well.
> >>
> >> I don't understand this. I just want to write
> >> Limit of the sequence {2^x ,(3/2)^2x , (4/3)^3x ,...}
> >> Joe
> >
> > Note that lim[(1+1/n)^nx] = {lim[(1+1/n)^n]}^x as n --> oo.
> > Thus you can
> > just compute lim[(1+1/n)^n] and raise the result to
> > the x'th power. Jim Spriggs was right about the typo
>
> Sorry. You, Jim right. I was looking at the wrong
> "nx" for the typo.
> Joe
>
> >
> > lim[(1+1/n)^nx] =! e^(nx)
> >
> > ..since lim[(1+1/n)^n] = e as n --> oo, implying that
> >
> > lim[(1+1/n)^nx] = lim[(1+1/n)^n]^x = e^x. Usually
> > you will arrive at e^x via
> >
> > lim[(1+x/n)^n] = e^x as n --> oo.

Just for kicks here's how you do the limit above:

Set y = (1 + x/n)^n so that ln(y) = n * ln(1 + x/n), then

lim[ln(y)] = lim[n * ln(1 + x/n)] as n --> oo.

Notice that as n --> oo, n * ln(1 + x/n) --> oo * 0. This implies that we should rewrite n as...

n = 1/(1/n) = 1/(n^(-1))

...so that we are left with the undeterminant form oo / oo, in which case we can apply l'Hospital's rule to get

lim[ln(y)] = lim[1/(n^(-1)) * ln(1 + x/n)]
= lim[-1/(n^(-2)) * (-x/(n^2)) / (1 + x/n)]
= lim[(-n^2) * (-x/(n^2)) / (1 + x/n)]
= x * lim[1 / (1 + x/n)] = x.

Since lim[ln(y)] = x and y = (1 + x/n)^n we have

e^{lim[ln(y)]} = lim[e^ln(y)] = lim[y] = lim[ln(1 + x/n)^n] = x.

Regards,
Kyle
.

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