Proving an Intuitive Result Rigorously I have found that for distinct functions (not constants) $f(x),g(x),h(x)$ that this is true:
$$\lim_{x \to \infty}\frac{f(x)g(x)}{f(x)g(x)+h(x)}=1$$ I want to know: why is that, ie is there a rigorous proof for why that is? Maybe a paper to turn to or is it really simple?
EDIT: I realized I probably should have put my actual problem first. I hope this gives more context. For this limit:
$$\lim_{n\rightarrow\infty}\frac{e^{H(n)}\log H(n)}{e^\gamma n\log(\gamma+\log n)+\frac{n}{\log\log n}}$$
$H(n)$ tends to $e^\gamma n$, which is the same as $\gamma + \log n$. So, you can reword it as this:
 $$\lim_{n\rightarrow\infty}\frac{e^{\gamma + \log n}\log(\gamma + \log n)}{(\gamma + \log n)\log(\gamma+\log n)+\frac{n}{\log\log n}}$$
So how do you continue to prove  that this limit is $1$?
 A: If $f(x)=x$, $g(x)=x^2$, and $h(x)=x^4$, then$$\lim_{x\to\infty}\frac{f(x)g(x)}{f(x)g(x)+h(x)}=0.$$So, the statement is false.
A: This statement is false as pointed out by Jose. 
For example, if I pick $h(x) = f(x)g(x) $, then 
$$  \lim_{x \to \infty} \frac{f(x) g(x)}{ f(x) g(x) + f(x) g(x)} = \frac{1}{2} $$ 
no matter if you have constant or not, it will always be $\frac{1}{2}$ 
A: There are many continuous nonconstant functions whose limit is zero.  For instance, using three of these, let $f(x) = \dfrac{1}{x}$, $g(x) = \mathrm{e}^{-x}$, and $h(x) = 1+ \dfrac{\sin x}{x}$ so that 
$$  \lim_{x \rightarrow \infty} \frac{f(x)g(x)}{f(x)g(x) + h(x)} = \frac{0 \cdot 0}{0 \cdot 0 + 1} = 0  \text{.}  $$

Anticipating where your edit is going.  Use the manipulation
$$  \frac{f}{f+\frac{n}{\ln \ln n}} = \frac{f}{f+\frac{n}{\ln \ln n}} \cdot \frac{\, \frac{1}{f} \,}{\frac{1}{f}} = \frac{1}{1 + \frac{n}{f \ln \ln n}}  \text{.}  $$
Since your $f$ grows slightly faster than linearly in $n$, you get $\dfrac{1}{1+0}$ as your limit.
A: This is not true in general. For example, if $f$, $g$, and $h$ are all the constant function that maps to $1$, the limit is $1/2$. If you want distinct functions, take $f=1, g=2, h=3$ uniformly to see that the limit is $2/5\neq 1$.
On the other hand, if $h$ is $o(fg)$ and $fg$ is nonzero, then $$\lim_{x\to\infty} \frac{f(x) g(x)}{f(x)g(x) + h(x)} = \lim_{x\to\infty} \frac{1}{1 + h(x) / (f(x)g(x))} = 1. $$
In that situation $fg$ is nonzero, the above argument says that the limit is one if and only if $h = o(fg)$.
