Tried to apply the ratio test to determine the convergence interval, but get the limit as a constant. So here is the series: $\Sigma_{n=1}^{\infty} \frac{x^{2n}}{1+x^{4n}}$
$$\left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x(1+x^{4n})}{1+x^{4n+4}} \right| \overset{\text{ n } \rightarrow \infty }{\rightarrow}\left[\frac{\infty}{\infty} \right]= \frac{x+x^{4n+1}}{1+x^{4n+4}} = \lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \\ 0 + \lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \frac{\infty}{\infty} =\left[\text{ apply L'Hospital rule (4n+1)th times } \right] = \frac{1}{(4n+4)(4n+3)(4n+2)x^3} = 0 $$
Where is my mistake? Maybe some other technique would be useful for these series?
$$\lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}}  \\ \text{ if } x = 0 \text{ then limit equals } 0 \\ \text{if x=1 then limit equals to 1 } \\ \text{if } 0<x<1 \text{ then limit is equal to x [ the same goes for } -1<x<0 \\ \text{ if } x> 1 \text{ then limit is equal to} \frac{1}{x^3} \text{ the same goes for x<1} $$
 A: Let's proceed from your work.  Corrections are highlighted in red.  We have
$$h(x) = \left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x^{\color{red}{2}}(1+x^{4n})}{1+x^{4n+4}} \right|.$$  At this point, we need to consider separate cases.  The reason is becuase for $n \ge 1$, the behavior of $|x^n|$ depends on whether $|x| < 1$, $|x| = 1$, or $|x| > 1$.
Case $|x| = 1$:  Then $x^2 = x^4 = 1$ and we have $$h(x) = \frac{1(1+1)}{(1+1)} = 1.$$  Going back to the original sum, $$f_n(x) = \frac{x^{2n}}{1+x^{4n}}$$ implies $$f_n(-1) = f_n(1) = \frac{1}{2} > 0$$ so the sum diverges.
Case $|x| > 1$:  Then $$\lim_{n \to \infty} 1/x^n = 0$$ and we have, after dividing numerator and denominator by $x^{4n+4}$, $$h(x) = \left|\frac{1/x^{4n+2} + 1/x^2}{1/x^{4n+4} + 1}\right|.$$  Hence $$\lim_{n \to \infty} h(x) = \left|\frac{0 + 1/x^2}{0 + 1} \right| = \frac{1}{x^2} < 1$$ and $\sum f_n(x)$ converges.
Case $|x| < 1$:  Then $x^n \to 0$ as $n \to \infty$, hence $$\lim_{n \to \infty} h(x) = \left|\frac{x^2(1 + 0)}{1 + 0}\right| = x^2 < 1$$ and again the sum converges.
Of course, this is not the simplest nor most elegant solution, but it is conforming to how one would apply the ratio test to determine the convergence of the sum.
A: Without ratio test...
For $\vert x \vert \lt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt \vert x \vert^{2n}$$ hence the series converges as $\sum x^{2n}$ converges.
For $\vert x \vert \gt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt 1/\vert x \vert^{2n}$$ hence the series converges for the same reason.
If $x=1,-1$ the series diverges as its term is constant and equal to $1/2$.
A: At some point in this big jumble of equations, you started taking $\lim_{x\to\infty}$ instead of $\lim_{n\to\infty}$. If you had stuck to $\lim_{n\to\infty}$
then you might see that
$$ \lim_{n \to \infty} \frac{x}{1+x^{4n+4}} = x. $$
The limit is not $0$ as you wrote (except of course when $x = 0$).
Also it's completely absurd to apply L'Hópital's rule as you did when taking a limit for $n\to\infty.$ The rule only seems applicable because (as I already noted) you're taking the limit over the wrong variable.
As an alternative, try comparing $\dfrac{x^{2n}}{1+x^{4n}}$
with $\dfrac{x^{2n}}{x^{4n}}.$ Then see where this leads.
Of course this comparison is only good for convergence, but for divergence it is sufficient show that the absolute value of the individual term goes to $\infty$ as $n$ goes to $\infty.$
If you can prove convergence when $|x|<r$ and divergence when $|x|>r,$ for the same $r$ in both cases, you have found the radius of convergence.
Another alternative is you could try the root test.
Or try the ratio test again but this time evaluate the limits correctly for $x <1$ and for $x > 1.$

Speaking of a big jumble of equations, no wonder you get confused: it's hard to say what even "=" means in some places in that mess. 
Don't write like that.
Try writing one equation at a time (or at most a few things that actually are numbers and are equal in one chain of equations) with words in between to explain how and/or why you go from one equation to the next.
You might think it saves a little time and effort to arrange things more compactly, but you lose all that benefit when you actually have to read what you've written.
