Limit question

[Rascuache]

How can I see that the limit of

(1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]

goes to x as n goes to infinity?

The book mentions that L'Hopital's Rule but I can't make it work.

If I let f(n) = (1/2)n and
g(n) = ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]

and compute f'(n)/(1/g(n))' it doesn't seem to go to x.

--
To reply by email replace dots with hyphens

[Piet Vanraad]

Rascuache schreef:

> How can I see that the limit of
>
> (1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> goes to x as n goes to infinity?
>
> The book mentions that L'Hopital's Rule but I can't make it work.
>
> If I let f(n) = (1/2)n and
> g(n) = ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> and compute f'(n)/(1/g(n))' it doesn't seem to go to x.
>

taking the limit of g'(n)/(1/f(n))' works.

[Ignacio Larrosa Cañestro]

Rascuache wrote:
> How can I see that the limit of
>
> (1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> goes to x as n goes to infinity?
>
> The book mentions that L'Hopital's Rule but I can't make it work.
>
> If I let f(n) = (1/2)n and
> g(n) = ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> and compute f'(n)/(1/g(n))' it doesn't seem to go to x.

We have that

Lim(Ln(1 + x)/x, x, 0) = 1

You can apply here Taylor´s series or L'Hôpital.

Then,

Lim((1/2)n*Ln(1 + (2x/n) + ((x^2 + y^2) / n^2)), n, inf)

= Lim((1/2)n*((2x/n) + ((x^2 + y^2) / n^2))*

*Ln(1 + (2x/n) + ((x^2 + y^2) / n^2))/((2x/n) + ((x^2 + y^2) / n^2)), n,
inf)

= Lim((1/2)n*((2x/n) + ((x^2 + y^2) / n^2)), n, inf)*1

= Lim(x + (x^2 + y^2)/(2n), n, inf) = x

--
Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUIT...@mundo-r.com

[Dave L. Renfro]

Rascuache <rascua...@at.the.drive.in.net> wrote:

> How can I see that the limit of
>
> (1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> goes to x as n goes to infinity?
>
> The book mentions that L'Hopital's Rule but I can't
> make it work.
>
> If I let f(n) = (1/2)n and
> g(n) = ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> and compute f'(n)/(1/g(n))' it doesn't seem
> to go to x.

The usual calculus method is to put the simpler
expression in the denominator. So you want to
compute the limit as n --> infinity of
g'(n) / (1/f(n))' = 2*g'(n) / ((d/dn)(n^(-1))).
This cleans up nicely for limit purposes if, after
taking the derivatives, you multiply both the numerator
and the denominator by n^2.

Even simpler is to use the expansion ln(1 + u) = u - ...
for u close to zero.

Dave L. Renfro

[W^3]

In article <J5Wdndg6GN2zvAbU...@giganews.com>,
Rascuache <rasc...@at.the.drive.in.net> wrote:

> How can I see that the limit of
>
> (1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> goes to x as n goes to infinity?

By the definition of a derivative, ln(1+h)/h = [ln(1+h)-ln(1)]/h ->
ln'(1) = 1 as h -> 0. Set h_n = (2x/n) + ((x^2 + y^2)/n^2). Your
expression is then (1/2)n*h_n*[ln(1+h_n)/h_n]. The term inside the []
-> 1, so all you have to do is figure out the limit of (1/2)n*h_n,
which is easy.

[William Elliot]

On Tue, 17 Feb 2009, Rascuache wrote:

> How can I see that the limit of
>
> (1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>
> goes to x as n goes to infinity?
>

ln[1 + (2x/n) + ((x^2 + y^2) / n^2)] / 2/n

l'Hopital's rule.

(-2x/n^2 - 2(x^2 + y^2)/n^3) / (1 + 2x/n + (x^2 + y^2) / n^2)
divided by
-2/n^2

(x + (x^2 + y^2)/n) / (1 + 2x/n + (x^2 + y^2)/n^2)

(-2x/n^2 - 2(x^2 + y^2)/n^3) / (-1/2)e^(-n/2)

[W^3]

In article <2009021722...@agora.rdrop.com>,
William Elliot <ma...@rdrop.remove.com> wrote:

> On Tue, 17 Feb 2009, Rascuache wrote:
>
> > How can I see that the limit of
> >
> > (1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
> >
> > goes to x as n goes to infinity?
> >
> ln[1 + (2x/n) + ((x^2 + y^2) / n^2)] / 2/n
>
> l'Hopital's rule.
>
> (-2x/n^2 - 2(x^2 + y^2)/n^3) / (1 + 2x/n + (x^2 + y^2) / n^2)
> divided by
> -2/n^2
>
> (x + (x^2 + y^2)/n) / (1 + 2x/n + (x^2 + y^2)/n^2)
>
>
>
>
> (-2x/n^2 - 2(x^2 + y^2)/n^3) / (-1/2)e^(-n/2)

Yet another example of how LHR can hide what is really happening and
lead to unnecessary work.

[William Elliot]

On Wed, 18 Feb 2009, W^3 wrote:
> William Elliot <ma...@rdrop.remove.com> wrote:
>> On Tue, 17 Feb 2009, Rascuache wrote:
>>
>>> How can I see that the limit of
>>>
>>> (1/2)n * ln[1 + (2x/n) + ((x^2 + y^2) / n^2) ]
>>>
>>> goes to x as n goes to infinity?
>>>
>> ln[1 + (2x/n) + ((x^2 + y^2) / n^2)] / 2/n
>>
>> l'Hopital's rule.
>>
>> (-2x/n^2 - 2(x^2 + y^2)/n^3) / (1 + 2x/n + (x^2 + y^2) / n^2)
>> divided by
>> -2/n^2
>>
>> (x + (x^2 + y^2)/n) / (1 + 2x/n + (x^2 + y^2)/n^2)
>

> Yet another example of how LHR can hide what is really happening and
> lead to unnecessary work.
>

It this case, it was an effective procedure.
Have you a simple intuitive way?

[oppt]

On Wed, 18 Feb 2009 22:38:01 -0800, William Elliot
<ma...@rdrop.remove.com> wrote:

W^3 posted the following earlier in this thread:
<http://groups.google.com/group/sci.math/msg/f6f4aa276f0ddd99>

Those who won't demonstrate elementary reading skills proclaim
ignorance beyond any help.