real analysis - Proving that: $lim_{xrightarrow 1}frac{f(x)}{g(x)}=mu Leftrightarrow lim_{xrightarrow 1} frac{f_{2}(x)}{g_{2}(x)}=nu$

I know that:
$f=f_{1}+f_{2}:[0,1[\rightarrow \infty$
$g=g_{1}+g_{2}:[0,1[\rightarrow \infty$
Where the limits: $\lim_{x\rightarrow 1} f_{1}(x)$ and $\lim_{x\rightarrow 1} g_{1}(x)$ exist and $\lim_{x\rightarrow 1} g_{2}(x)=\infty$
I now want to prove that: $$\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\mu \text{ exists }\Leftrightarrow \lim_{x\rightarrow 1} \frac{f_{2}(x)}{g_{2}(x)}=\nu \text{ exists}$$ and then $\mu=\nu$.
I'm having trouble proving this I tried it using the definition:
$$\forall \epsilon>0\exists\delta>0 \text{ s.t } |x-1|<\delta\Rightarrow |\frac{f(x)}{g(x)}-\mu|=|\frac{f_{1}(x)+f_{2}(x)}{g_{1}(x)+g_{2}(x)}-\mu|=|\frac{f_{1}(1)+f_{2}(x)}{g_{1}(1)+g_{2}(x)}-\mu|<\epsilon$$ but that didn't really work. Can anyone help me with this?

Answer

Hint: Write ${{f_1(x)+f_2(x)}\over {g_1(x)+g_2(x)}}={{{f_1(x)}\over {g_2(x)}}+{{f_2(x)}\over{g_2}(x)}\over{1+{{g_1(x)}\over {g_2(x)}}}}$.

$lim_{x\rightarrow 1}{{f_1(x)}\over{g_2(x)}}=lim_{x\rightarrow 1}{{g_1(x)}\over{g_2(x)}}=0$.