Proving $g(x)$ is not a rational function 
Show that $\sqrt{1-4x}$ is not a rational function (i.e. there do not exist polynomials $f(x)$ and $g(x)$ so that $\sqrt{1-4x} = \dfrac{f(x)}{g(x)}.$

I tried assuming that $f(x)$ and $g(x)$ exist, but I'm not really sure how to derive a good contradiction. If I write $f(x) = \sum_{n\geq 0} f_nx^n$ and $g(x) = \sum_{n\geq 0} g_nx^n,$ then $f(x)^2 = \sum_{n\geq 0} (\sum_{k= 0}^n f_k f_{n-k})x^n$ and $g(x)^2 = \sum_{n\geq 0} (\sum_{k =0}^n g_k g_{n-k})x^n$. Squaring both sides and multiplying by $g(x)^2$ yields $g(x)^2 (1-4x) = f(x)^2.$ Then $\sum_{n\geq 0} (\sum_{k= 0}^n g_k g_{n-k})x^n (1-4x) = \sum_{n\geq 0} (\sum_{k= 0}^n g_k g_{n-k})(x^n-4x^{n+1}) = \sum_{n\geq 0} (\sum_{k= 0}^n f_k f_{n-k})x^n.$ Equating coefficients gives that $g_0^2 = f_0^2$ and for $n\geq 1, \sum_{k=0}^{n}g_kg_{n-k}-4\sum_{k=0}^{n-1}g_kg_{n-1-k} = \sum_{k=0}^n f_kf_{n-k}.$ Since this holds for $n=1,$ we have $g_0g_1 + g_1g_0 - 4g_0^2 = g_0g_1 + g_1g_0-4f_0^2= f_0f_1 + f_1f_0\Rightarrow g_0g_1 + g_1g_0 = f_0f_1 + f_1f_0 + 4f_0^2.$ Similarly, for $n=2,$ we have $g_0g_2 + g_1^2 + g_2g_0 - 4(g_0g_1 + g_1 g_0) = f_0f_2 + f_1^2 + f_2f_0.$ Continuing in this fashion, we see that the coefficient of $x^n$ in $g(x)$, which is $\sum_{k=0}^n g_kg_{n-k},$ is equal to $4^n f_0^2$ plus some other terms. If $f_0 = 0,$ then from above, we can deduce that $g_0^2 = f_0^2 \Rightarrow g_0 = 0 \wedge g_0g_1 + g_1g_0 = 0\wedge g_1^2 = f_1^2.$ Thus we may assume WLOG that $f_0 \neq 0$ by finding the least $k$ so that $f_k\neq 0$ (there exists such a $k$ since $f(x) \neq 0$ in order to satisfy the equality. But then, this means that as $n\to \infty, 4^n f_0^2 \to \infty,$ so the coefficient of $x^n$ of $g(x),$ being the sum of $4^n f_0^2$ and finitely many terms of the form $4^j f_mf_n$ tends to infinity or is undefined (if $f_1$ is the opposite sign of $f_0$, then the term $4^{n-1} f_1 f_0$ tends to $-\infty$), which contradicts the fact that the coefficients of $g(x)^2$ are finite.

Is this wrong? I believe the argument that not all of the coefficients of $g(x)^2$ would be well-defined may be wrong. Would there be an easier way to show this contradiction?

 A: $g(x)^2 (1-4x) = f(x)^2$ tells us that $deg(f^2) = 1+ deg(g^2)$. However, both degrees are even, since they are squares, and we reach a contradiction.
A: If $\sqrt{1-4x}={{f(x)}\over{g(x)}}$, $(1-4x)g(x)^2=f(x)^2$. Let $p$ be the  nonzero monome of $g$ whose coefficient is not zero of minimal degree, $g(x)=a_px^p+$ terms of degree>p, $(1-4x)g(x)^2=a_p^2x^{2p}-4a_px^{2p+1}$+terms of degree >2p+1, it cannot be a square since non trivial monomes in a square have even degree.
A: Suppose that $h(x)=\frac{f(x)}{g(x)}$ has degree $n$ (that is the degree of $f(x)$ minus the degree of $g(x)$). Then $$\lim_{n\to -\infty} \frac{h(x)}{c x^n} = 1$$ for some constant $c\ne 0$.
But $$\lim_{n\to -\infty} \frac{\sqrt{1-4x}}{c x^n} =
\begin{cases} 
0 & n \ge 1 \\
\pm \infty & n < 1
\end{cases}$$
A: Suppose there exist polynomials $f(x)$ and $g(x)$ with no common factors such that
$$\sqrt{1-4x} = \frac{f(x)}{g(x)}$$
Square both sides
$$1-4x=\frac{f^2(x)}{g^2(x)}\to f^2(x)=(1-4x)g^2(x)$$
Factors of $f^2(x)$ must be perfect squares.
If $g(x)$ does not contain the factor $(1-4x)$, we have an absurd because RHS contains a factor raised to the 1st degree which is impossible.
If $g(x)$ contains the factor $(1-4x)$, we have an absurd because RHS contains a factor raised to the 3rd degree which is again impossible.
We have a contradiction, thus $\sqrt{1-4x}$ is not a rational function.
A: You have$$\lim_{x\to\infty}\frac{\sqrt{1+4x}}{x^{1/2}}=2.$$But if $\sqrt{1-4x}=\frac{f(x)}{g(x)}$, you have $\sqrt{1+4x}=\frac{f(-x)}{g(-x)}$ and, if $n=\deg f(x)-\deg g(x)$, then $n$ is the only real number such that$$\lim_{x\to\infty}\frac{f(x)/g(x)}{x^n}\in\Bbb R.$$Since $n\in\Bbb Z$, you can't have $n=\frac12$ and so there is a contradiction here.
