Results 1 to 2 of 2

Thread: Find d such that f(x) converges

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    151

    Find d such that f(x) converges

    EDIT: I see now to late that the title of the thread says: "Find d..." , but it is supposed to be: "Find c..."

    Let:
    $\displaystyle f(x)=x-(4+x^4+x^6)^c$

    The question is, what is c such that when $\displaystyle x\to\infty$ $\displaystyle f(x)\to k$
    where k is some finite number.

    Now what I did is something like this:
    $\displaystyle \lim\limits_{x\to\infty}x-(4+x^4+x^6)^c=\lim\limits_{x\to\infty}x-\lim\limits_{x\to\infty}(x^6(\frac{4}{x^6}+\frac{1 }{x^2}+1))^c$
    $\displaystyle =\lim\limits_{x\to\infty}x + \lim\limits_{x\to\infty}x^{6c} \cdot \lim\limits_{x\to\infty}(\frac{4}{x^6}+\frac{1}{x^ 2}+1)^c$
    $\displaystyle =\lim\limits_{x\to\infty}x-\lim\limits_{x\to\infty}x^{6c}=\lim\limits_{x\to\i nfty}x-x^{6c}=k$ where $\displaystyle k\in\mathbb{R}$

    Now the only way such a limit is to converge to k is that x have equal powers, that is
    $\displaystyle c=\frac{1}{6}$, then $\displaystyle k=0$.

    Now the main problem is: How do I prove there is no other c that such that f(x) converges when $\displaystyle x\to\infty$ ? I just need you to point me in the right direction.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2009
    Posts
    151
    Nothing ?

    Any way, I think I am supposed to express $\displaystyle (4+x^4+x^6)^c$ as a taylor-series.
    I just started to learn Taylor series today, so I have no idea how I am supposed to do that.
    Can someone please clarify?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 12th 2011, 08:43 AM
  2. Replies: 1
    Last Post: May 18th 2010, 10:31 PM
  3. Replies: 1
    Last Post: Jan 16th 2010, 03:44 PM
  4. Replies: 2
    Last Post: Oct 25th 2009, 04:52 PM
  5. converges
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 4th 2008, 05:55 PM

Search Tags


/mathhelpforum @mathhelpforum