derivative

[hotice]

how to slove this problem

ftp://172.25.53.21/image/002.gif

thank you very much

[Dave L. Renfro]

hotice wrote:

After 25-30 seconds of waiting for the page to load,
I got this message:

"The page cannot be displayed"

You should probably just state the problem in a post,
something I would imagine takes much less time to do
than going to all the trouble of making a .gif image
and then making a web page for it.

Dave L. Renfro

[hotice]

> After 25-30 seconds of waiting for the page to load,
> I got this message:
>
> "The page cannot be displayed"
>
> You should probably just state the problem in a post,
> something I would imagine takes much less time to do
> than going to all the trouble of making a .gif image
> and then making a web page for it.
>
> Dave L. Renfro

i'm so sorry , i pasted a intrenet ip address
i don't know how to express it clearly
you can try this
ftp://www.syrtu.com/image/002.gif

thanks a lot

[Dave L. Renfro]

hotice wrote:

> i'm so sorry , i pasted a intrenet ip address
> i don't know how to express it clearly
> you can try this
> ftp://www.syrtu.com/image/002.gif

How about just saying the following?

The limit as x --> 0 of (1/x) times the integral from
t = 0 to t = (sin x) of the function sin(t^2).

This took me about as long to write as it took for me to
get your new web page to appear on my screen after I clicked
on the address (about 15-20 seconds).

It looks like a straightforward L'Hopital's rule problem
to me. Use the Fundamental Theorem of Calculus, in conjunction
with the chain rule, to differentiate the integral with
respect to x. Note that the integral can be written as
F(u) = integral from t = 0 to t = u of the function sin(t^2),
where u = sin(x). Hence, (d/dx)F(u) is equal to the
u-derivative of F, which is sin(u^2), times the x-derivative
of u. Thus, (d/dx)F(u) = sin[(sin x)^2] * cos(x), so the
limit you want seems to be zero. You might want to test
this out numerically by evaluating "integral/x" for
various small values of x, something even a TI-82
calculator can be used for.

Dave L. Renfro

[[Mr.] Lynn Kurtz]

On 22 Mar 2007 00:23:50 -0700, "hotice" <xuxi...@gmail.com> wrote:

>i'm so sorry , i pasted a intrenet ip address
>i don't know how to express it clearly
>you can try this
>ftp://www.syrtu.com/image/002.gif
>
>thanks a lot

d/dx ( lim[x-->0]{ int_0^x sin(t^2) dt / x} )

Dave gave one way to see it; here's another. First, note that the
quantity in () is independent of x so the answer is zero. Secondly, if
we call f(x) = int_0^x sin(t^2) dt then the quantity in braces is:

{ ( f(x) - f(0) ) / (x - 0) } and its limit is f'(0), which is
constant [zero].

Which all makes me wonder if the problem was posted correctly.

--Lynn

[The World Wide Wade]

In article <1174575999.7...@o5g2000hsb.googlegroups.com>,

"Dave L. Renfro" <renf...@cmich.edu> wrote:

> The limit as x --> 0 of (1/x) times the integral from
> t = 0 to t = (sin x) of the function sin(t^2).

For small x > 0, int_[0,sin(x)] sin(t^2) dt < int_[0,x] t^2 dt = x^3/3,
so the above limit is 0 from the right; the argument from the left is
similar.

[Dave L. Renfro]

Dave L. Renfro wrote (in part):

>> The limit as x --> 0 of (1/x) times the integral from
>> t = 0 to t = (sin x) of the function sin(t^2).

Lynn Kurtz wrote:

> d/dx ( lim[x-->0]{ int_0^x sin(t^2) dt / x} )
>
> Dave gave one way to see it; here's another. First, note
> that the quantity in () is independent of x so the answer
> is zero. Secondly, if we call f(x) = int_0^x sin(t^2) dt
> then the quantity in braces is:
>
> { ( f(x) - f(0) ) / (x - 0) } and its limit is f'(0),
> which is constant [zero].
>
> Which all makes me wonder if the problem was posted correctly.

Hummm...I didn't even notice the (very small) prime mark
in the upper right corner of the right-hand parenthesis.
I guess that means d/dx, which is a weird thing to be
asking in this case (assuming the limit exists, and I
suppose this would need to be shown, and then, as you
implied, we have a constant and thus the derivative is zero),
although this might be intentional to see if the students
are paying attention. On the other hand, it might be the
result of someone incorporating a hint of "take the derivative"
into the statement of the problem, and thus accidentally
(although I think "accidentally" understates what the true
situation is likely to have been if this were the case)
changing the statement of the problem.

Dave L. Renfro

[hotice]

i see
thank you

[hotice]

On 3月23日, 上午2时43分, "Dave L. Renfro" <renfr...@cmich.edu> wrote:

the divisor shoud be x^3

[The World Wide Wade]

In article <1174620539....@y80g2000hsf.googlegroups.com>,
"hotice" <xuxi...@gmail.com> wrote:

So can you state the goddamn problem?

[Dave L. Renfro]

hotice wrote:

> the divisor shoud be x^3

Maybe it should be, but I looked at the .gif again
just now and the denominator is definitely x.

Nevertheless, there's no problem with x^3. Just
do what I said and then follow up with two more
applications of L'Hopital's rule before plugging
in x = 0.

We have:

The limit as x --> 0 of (1/x)^3 times the integral

from t = 0 to t = (sin x) of the function sin(t^2)

One application of L'Hopital's rule gives the limit
x --> 0 of { cos(x) * sin[ (sin x)^2) ] } / 3x^2.
It suffices, by the limit-of-a-product theorem,
to only consider the limit x --> 0 of
sin[ (sin x)^2) ] / 3x^2. Now it's just
straightforward differentiation. However, I can
see that you're going to get 1/3, since the dominate
term in the expansion of sin[ (sin x)^2) ] is (sin x)^2
[put u = (sin x)^2 in the expansion sin(u) = u - (1/3)u^3 + ...],
and (sin x)^2 / 3x^2 = (1/3)*[(sin x)/x]^2 approaches
1/3 as x --> 0.

Dave L. Renfro

[The World Wide Wade]

In article <1174660855.8...@p15g2000hsd.googlegroups.com>,

"Dave L. Renfro" <renf...@cmich.edu> wrote:

Avoiding the awful LHR, note sin(u) = u + O(u^3) as u -> 0 => sin(t^2) =
t^2 + O(t^6). So int_[0,sin(x)] sin(t^2) dt = int_[0,sin(x)] [t^2 +
O(t^6)] dt = sin(x)^3/3 + O(sinx)^7). Divide by x^3 and use sin(x)/x ->
1 to see the limit is 1/3.

[hotice]

i see!
thank you for explaining it so clearly!