If $\lim\limits_{x \to x_0} f(x) = \lim\limits_{x \to x_0} g(x) = +\infty$ prove that $\lim\limits_{x \to x_0} \frac{f(x)+g(x)}{f^2(x)+g^2(x)}=0$ 
Consider the functions $f, g : \mathbb{R} \rightarrow \mathbb{R}$ for which holds that $\lim\limits_{x \to x_0} f(x) = \lim\limits_{x \to x_0}  g(x) = +\infty$ for some $x_0 \in \mathbb{R}$. Prove that $\lim\limits_{x \to x_0} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}=0$

Is the following proof correct? For some $x \to x_0$ holds that $f(x)>0$ and $g(x)>0$
So it is true that $2f(x)g(x)\geq 0 \\ \Rightarrow (f(x)+g(x))^2 \geq f^2(x)+g^2(x) \\ \Rightarrow \dfrac{1}{f^2(x)+g^2(x)} \geq \dfrac{1}{(f(x)+g(x))^2} \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \dfrac{f(x)+g(x)}{(f(x)+g(x))^2}=\dfrac{1}{f(x)+g(x)}$
But $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}>0$ hence $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}=\Bigg|\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}\Bigg|$
Thus $\Bigg|\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}\Bigg| \geq \dfrac{1}{f(x)+g(x)} \\  \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \dfrac{1}{f(x)+g(x)} \text{OR} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq -\dfrac{1}{f(x)+g(x)} \\ \Rightarrow \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \lim_{x \to x_0}\dfrac{1}{f(x)+g(x)} \text{OR} \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq -\lim_{x \to x_0}\dfrac{1}{f(x)+g(x)} \\ \Rightarrow \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq 0 \text{OR} \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq 0 \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} = 0$
 A: I don't see how you conclude $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} = 0$ from $\lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq 0$. 
In any case, the proof can be simplified. Write
$$
\lim_{x \to x_0} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}= \lim_{x \to x_0} \dfrac{f(x)}{f^2(x)+g^2(x)} + \dfrac{g(x)}{f^2(x)+g^2(x)}
$$
Since $f^2(x), g^2(x) \geq 0$, we have
$$
\lim_{x \to x_0} \dfrac{f(x)}{f^2(x)+g^2(x)} +  \dfrac{g(x)}{f^2(x)+g^2(x)} \leq \lim_{x \to x_0} \dfrac{f(x)}{f^2(x)} +  \dfrac{g(x)}{g^2(x)} = \lim_{x \to x_0} \dfrac{1}{f(x)} +  \dfrac{1}{g(x)} =0
$$
A: $\bigg|\dfrac{f(x)+g(x)}{f(x)^2+g(x)^2}\bigg|=\bigg|\dfrac{r(x)\big(\cos \theta(x)+\sin\theta(x)\big)}{r(x)^2}\bigg|\le\dfrac{2}{r(x)}\to 0$
Since $r(x)=\sqrt{f(x)^2+g(x)^2}\to+\infty$
A: For $x>0$ and $y>0$ we have, 
$$\color{blue}{x+y+\frac{y^2}{x} +\frac{x^2}{y}  =(x^2 +y^2)(\frac{1}{x} +\frac{1}{y}) \ge x+y  \implies \frac{x+y}{x^2+y^2} \le \frac{1}{x} +\frac{1}{y}}$$
Hence, since $f(x)>0, g(x)>0$  for $x$ evry closed to $x_0$ we have
$$0 \le \lim_{x \to x_0} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \le \lim_{x \to x_0} \frac{1}{f(x)} +  \frac{1}{g(x)} =0$$
