Re: Compute two limits
- From: "TCL" <tlim1@xxxxxxxxxxx>
- Date: Sat, 19 Aug 2006 19:02:39 GMT
"David C. Ullrich" <ullrich@xxxxxxxxxxxxxxxx> wrote in message
news:jvtde25drppgs61ia3caq38kkld96lpq3t@xxxxxxxxxx
On Sat, 19 Aug 2006 02:56:01 GMT, "TCL" <tlim1@xxxxxxxxxxx> wrote:
Let g(x)= int_0^x sin( 1/t ) dt and
h(x)= int_0 ^x | sin( 1/t ) | dt.
Compute
lim_{x --> 0} g(x)/x and lim_{x --> 0} h(x)/x .
It seems to me these two limits are not easy.
The first one is not so bad. If you make a change
of variables and then integrate by parts you so
f(x) = int_{1/x}^infinity sin(t)/t^2 dt
= x^2 cos(1/x) + c int_{1/x}^infinity cos(t)/t^3 dt,
so |f(x)| <= cx^2, hence f'(0) = 0. (Here "c" varies
from line to line...)
I'm pretty sure the second one is 2/pi.
For a given x, say N = N(x) is the smallest integer
with N pi > 1/x. Making a change of variable you see
g(x) = int_{1/x}^N |sin(t)|/t^2
+ sum_{n=N}^infinity |sin(t)|/t^2.
Now
int_{1/x}^N |sin(t)|/t^2 <= c x^2
so you can ignore it, when you divide by x and let x
tend to zero the limit is 0.
For the other terms you could estimate the difference
between the integral and the integral with
1/(n pi)^2 in place of 1/t^2. If n pi < t < (n+1)pi
then 1/t^2 - 1/(n pi)^2 <= c/n^3, so the sum
of the errors is <= c/N^2 <= c x^2, again something
you can nneglect.
But if you consider the sum with 1/(n pi)^2 in place
of 1/t^2 you get
(sum_{n=N}^infinity pi/(n pi)^2) int_0^pi sin(t) dt
= 2 sum_{n=N}^infinity pi/(n pi)^2
~ (2/pi) 1/N ~ (2/pi) x,
and if you divide that by x and let x -> 0 you get 2/pi.
************************
David C. Ullrich
Following your line of thought, I also get 2/pi. So that must be correct.
I appreciate very much your quick response.
.
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