Limit.

[shapper]

Hello,

I came across a limit which I am having trouble to solve.
I applied L'Hopital rule but I think I am missing something because I
never get to the result.
The strange thing is that I have been creating a few documents with
limits an I had no trouble with any.

But this is driving me crazy ... I know I am missing something but now
idea what.

The limit is:

lim when (x-> 0) of (1/x - cotan x)

cotan is tg^-1 x

Could someone show me the steps?

By the way, the result is 0.

Thanks,
Miguel

[Dave L. Renfro]

Miguel wrote (in part):

> But this is driving me crazy ... I know I am missing
> something but now idea what.
>
> The limit is:
>
> lim when (x-> 0) of (1/x - cotan x)

Since successive derivatives of tangent and cotangent
get messy, convert to sines and cosines. And, of course,
rewrite it as a quotient if you intend to use L'Hopital's
rule -->

[(sin x) - x(cos x)] / [x(sin x)]

Two derivatives suffice, or after the first derivative
you can divide numerator and denominator by x and make
use of (sin x)/x --> 1 and x/(sin x) --> 1 as x --> 0.

Dave L. Renfro

[José Carlos Santos]

On 10-01-2008 14:32, shapper wrote:

> I came across a limit which I am having trouble to solve.
> I applied L'Hopital rule but I think I am missing something because I
> never get to the result.
> The strange thing is that I have been creating a few documents with
> limits an I had no trouble with any.
>
> But this is driving me crazy ... I know I am missing something but now
> idea what.
>
> The limit is:
>
> lim when (x-> 0) of (1/x - cotan x)
>
> cotan is tg^-1 x

I have no better suggestion than the one already provided by David L.
Renfro, but I want to call your atention to the fact that, in the
context of functions, f^{-1} means "inverse of f". The number cotan(x)
is 1/tan(x).

Best regards,

Jose Carlos Santos

[G. A. Edgar]

In article <5umqpdF...@mid.individual.net>, José Carlos Santos
<jcsa...@fc.up.pt> wrote:

> On 10-01-2008 14:32, shapper wrote:
>
> > I came across a limit which I am having trouble to solve.
> > I applied L'Hopital rule but I think I am missing something because I
> > never get to the result.
> > The strange thing is that I have been creating a few documents with
> > limits an I had no trouble with any.
> >
> > But this is driving me crazy ... I know I am missing something but now
> > idea what.
> >
> > The limit is:
> >
> > lim when (x-> 0) of (1/x - cotan x)
> >
> > cotan is tg^-1 x
>
> I have no better suggestion than the one already provided by David L.
> Renfro,

Another would be to write the Laurent series for cotan x at x=0
and then the result will be obvious. How to get that Laurent series?
Probably easiest is to divide the Taylor series for cos x and sin x ...
You only need the first few terms.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

[Ray Vickson]

Following the suggestion of Dave Renfro, write your function as f(x) =
1/x - cos(x)/sin(x) = [sin(x) - x*cos(x)]/[x*sin(x)]. The series
expansion of the numerator is sin(x) - x*cos(x) = (1/3)*x^3-
(1/30)*x^5+O(x^6), while the denominator is x*sin(x) = x^2-
(1/6)*x^4+O(x^6). These are obtained right away from the expansions of
sin and cos. Anyway, we have
f(x) = [(1/3)*x^3 +...]/[x^2 + ...] -> 0 as x -> 0. Of course, if you
don't know the series expansions of sin and cos (or are not allowed to
use them), then straight application of l'Hopital is as good a way as
any.

R.G. Vickson

[David W. Cantrell]

[Miguel: My two comments below are not intended for you, but others may
find them interesting.]

1) The limit in question being 0, we may define

g(x) = 0 if x = 0, 1/x - cot(x) otherwise

thereby removing the singularity of the original function. Then

sum_k=1^oo tan(x/2^k)/2^k = g(x), a result due to Euler.

2) Dave mentioned using the well known fact that sin(x)/x -> 1 as x -> 0
after using l'Hopital's rule once. But some nice limits, normally done with
that rule, can be done without it, using just sin(x)/x -> 1 as x -> 0. As
an example, suppose that we want the limit of (1 - cos(x))/x^2 as x -> 0
without resorting to l'Hopital's rule or series or the definition of limit:

(1 - cos(x))/x^2 = (1 - (1 - 2 sin^2(x/2)))/x^2 = 2 sin^2(x/2)/x^2

= 1/2 ( sin(x/2)/(x/2) )^2 -> 1/2 (1)^2 = 1/2 as x -> 0

The above also shows:
The function h(x) obtained by removing the singularity of (1 - cos(x))/x^2
may be written neatly using the sine cardinal function, sinc(x), defined to
be 1 if x = 0, sin(x)/x otherwise. Namely,

h(x) = 1/2 sinc^2(x/2)

Can someone write g(x) similarly, using sinc, or prove that cannot be done?
I have not found a way to write g(x) using sinc, but I have no idea how one
might prove that such is not possible. OTOH, if we were also allowed to use
the derivative of the sine cardinal function, then our task would be easy:

g(x) = - sinc'(x)/sinc(x)

David W. Cantrell

[Dave L. Renfro]

David W. Cantrell wrote (in part):

> 2) Dave mentioned using the well known fact that
> sin(x)/x -> 1 as x -> 0 after using l'Hopital's rule
> once. But some nice limits, normally done with that
> rule, can be done without it, using just sin(x)/x -> 1
> as x -> 0. As an example, suppose that we want the
> limit of (1 - cos(x))/x^2 as x -> 0 without resorting
> to l'Hopital's rule or series or the definition of limit:
>
> (1 - cos(x))/x^2 = (1 - (1 - 2 sin^2(x/2)))/x^2 = 2 sin^2(x/2)/x^2
>
> = 1/2 ( sin(x/2)/(x/2) )^2 -> 1/2 (1)^2 = 1/2 as x -> 0

A bit simpler is to multiply the numerator and denominator
of [1 - (cos x)] / x^2 by 1 + (cos x) to get
[(sin x)/x] ^ 2 divided by 1 + (cos x), and now it's
easy to see the expression approaches 1/2 as x --> 0.

> The above also shows:
> The function h(x) obtained by removing the singularity
> of (1 - cos(x))/x^2 may be written neatly using the
> sine cardinal function, sinc(x), defined to be 1
> if x = 0, sin(x)/x otherwise. Namely,
>
> h(x) = 1/2 sinc^2(x/2)
>
> Can someone write g(x) similarly, using sinc, or prove
> that cannot be done? I have not found a way to write
> g(x) using sinc, but I have no idea how one might
> prove that such is not possible.

Basically, you're asking (I think) if some sort of
algebraic manipulative devices can be used to find
the limit of 1/x - (cot x) as x --> 0. I worked on
this some before I made my post, and some more just
after I made my post, but I didn't get anywhere.

Trig. books in the 1800s often gave non-calculus
derivations to show (sin x) is roughly equal to
x - (1/6)x^3 for values of x close to zero, so I'm
sure a non-calculus argument exists for the limit
we're dealing with, although it's probably going
to involve some nontrivial trig/geometry steps.

For an example of what I'm talking about, see
"IX. Construction of Trigonometric Tables" in
Isaac Todhunter's book (1882 printing) at

http://books.google.com/books?id=Y3dvVnfqE9AC

Specifically, pp. 82-91 (= pp. 93-102 in the
downloaded .pdf version), especially page 85
(= page 96 in .pdf) and page 90 (= page 101
in .pdf).

You can also find this sort of thing in the more
available (off the internet) trig. books by Hobson
and Durrell/Robson.

Dave L. Renfro