# Thread: Find d such that f(x) converges

1. ## 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.

2. 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.