University-数学讨论区-Calculus, Real Analysis

hihi23

另外，在做这一题时，我发现了一样很奇怪的东西...
lim(x-->infinity) x/ 2^ (ln x） = lim(x-->infinity) 1/ [(ln2) (2^ (ln x）) (1/x)] (L'hospital's rule)
                                           =  lim(x-->infinity) x/ [(ln2) (2^ (ln x）]

lim(x-->infinity) x/ 2^ (ln x）=1/ln 2  *  lim(x-->infinity) x/ 2^ (ln x）
so,  lim(x-->infinity) x/ 2^ (ln x） =0
可是，如果按计算机的话  lim(x-->infinity) x/ 2^ (ln x）会是趋近无限， 不会等于零。
是不是我 apply L'Hospital's rule 时错了？ 还是，我按计算机时错了？

[ 本帖最后由 hihi23 于 11-4-2009 09:54 AM 编辑 ]

DADDY_MUMMY

原帖由 hihi23 于 11-4-2009 09:53 AM 发表
另外，在做这一题时，我发现了一样很奇怪的东西...
lim(x-->infinity) x/ 2^ (ln x） = lim(x-->infinity) 1/ [(ln2) (2^ (ln x）) (1/x)] (L'hospital's rule)
                                           =  li ...

计算机有得按infinity??

hihi23

只要把式子中的  x  换成 9999999 就可以了。

distantstar

Oh yeah, using the comparison test in that case will be much easier.

lim(x-->infinity) x/ 2^ (ln x）=1/ln 2  *  lim(x-->infinity) x/ 2^ (ln x）
so,  lim(x-->infinity) x/ 2^ (ln x） =0.

You can proceed to this step if the limit of x/2^(ln x) as x goes to infinity is a finite number. In short, you have made a very weird mathematical expression just exactly the following:
+(infinity) = 1/ln 2 (=infinity)-----> (+infinity) = 0.

And another problem is as you pulled out the factor 1/ ln 2 in the right hand side, you have preassumed that the limit of x/2^(lnx) exists as x goes to infinity is a finite real number. This is wrong according to the theory of limit, where we can only do so when the limit x/2^(lnx)is finite, though in L'Hospital rule you have otherwise.

About the problem of integration by part, you can see the following:
inte ( exp (-ln2lnx )
= xexp(-ln2lnx) - inte (x(-ln2/x) exp(-ln2lnx)
=xexp(-ln2lnx) + ln2 inte( exp(-ln2lnx)
Now, you can apply the trick that you used just now, and this time we can use this trick because we know for sure the existence of the integration of a continuous function on the interval [1+n+1].

Oh yeah, how to see the limit of x/2^(ln x)
Make good use the following trick, it helps a lot.
x/2^ln(x) = exp( lnx- ln2lnx)
             = exp [lnx (1-ln2) ]
since 1-ln2>0, hence as x goes to infinity, we can see easily from the expression above is going to infinity then.
  
Any mistakes in my explanation pls point them out. Cheers!

[ 本帖最后由 distantstar 于 11-4-2009 04:12 PM 编辑 ]

hihi23

Thanks a lot.  解释到很清楚。

distantstar

小弟需要高手帮忙解答下面的问题。。。。。。

~HeBe~_@

原帖由 distantstar 于 19-7-2009 02:20 AM 发表
小弟需要高手帮忙解答下面的问题。。。。。。

Integrate from 0 to pi     ln (x + cos t) w.r.t.  t
= [t . ln (x + cos t)] (from 0 to pi)  - Integrate from 0 to pi    t.(- sint)/(x + cos t)   w.r.t  t  [Integration by part]
=  [t . ln (x + cos t)] (from 0 to pi) + Integrate from 0 to pi    t.(sint)/(x + cos t)  w.r.t.  t
=    pi   ln ( x + cos pi ) - 0            + Integrate from 0 to pi    t.(sint)/(x + cos t)  w.r.t.  t
=    pi   ln ( x - 1 )                            + Integrate from 0 to pi    t.(sint)/(x + cos t)  w.r.t.  t

我只会做到这里。

接下来，我不知道我的方法是否对。
Since x belongs to ]1,+ infinity [  which is  x belongs to (2,+ infinity),
我正想， 我们是否要用improper integral 来解？但是该如何开始？
你心中有什么方法？


When x=2,
      pi   ln ( 2 - 1 )                            + Integrate from 0 to pi    t.(sint)/(2 + cos t)  w.r.t.  t
=    pi   ln ( 2 - 1 )                            + Integrate from 0 to pi    t.(sint)/[1 + (1 + cos t)]  w.r.t.  t
=    pi   ln 1                                      + Integrate from 0 to pi    t.(sint)/[1 + (1 + cos t)]  w.r.t.  t
=                                                          Integrate from 0 to pi    t.(sint)/[1 + (1 + cos t)]  w.r.t.  t

这只是when x=2.
到这里我就没办法了。。。
各位还有什么看法？