$f(n)/g(n) \rightarrow 1$ implies $h(f(n))/h(g(n)) \rightarrow 1$? Let $f: \mathbb{N} \rightarrow \mathbb{R}_{>0}$ and $g: \mathbb{N} \rightarrow \mathbb{R}_{>0}$ where $\lim_{n \rightarrow \infty} f(n) = \infty$ and $\lim_{n \rightarrow \infty} g(n) \rightarrow \infty$ and $$\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} = 1.$$ 
Further let $h: \mathbb{R} \rightarrow \mathbb{R}$ be an increasing function. What further conditions do we need on $h$ in order to conclude
$$
\lim_{n \rightarrow \infty}\frac{h(f(n))}{h(g(n))} = 1 \text{ ?}
$$
 A: Apart from the interesting cases mentioned in the comments section (not being exponential or $\lim\limits_{x\to+\infty}h(x)\in\mathbb{R}$ or beng a polynomial), a more generic case would be $h$ being an isometry and moreover, that $h(x)\geq x$ eventually, or, at least $h(g(n))\geq g(n)$, eventually. 
Indeed, let us at first examine what our hypothesis can tell us about our problem. $\lim\limits_{n\to+\infty}\frac{f(n)}{g(n)}=1$ means that for every $\epsilon>0$ there exists a $n_0\in\mathbb{N}$ such that for every $n\geq n_0$ we have that:
$$\left|\frac{f(n)}{g(n)}-1\right|<\epsilon\Leftrightarrow|f(n)-g(n)|<\epsilon|g(n)|=\epsilon\cdot g(n)$$
In the same way $\lim\limits_{n\to+\infty}\frac{h(f(n))}{h(g(n))}=1$ means that for every $\epsilon>0$ there exists a $N\in\mathbb{N}$ such that for every $n\geq N$ we have that:
$$\left|\frac{h(f(n))}{h(g(n))}-1\right|<\epsilon\Leftrightarrow|h(f(n))-
h(g(n))|<\epsilon|h(g(n))|$$
Now, let $n_1\in\mathbb{N}$ such that for every $n\geq n_1$ we have that:
$$h(g(n))\geq g(n)$$
Then, let $N=\max\{n_0,n_1\}$ and note that:
$$|h(f(n))-h(g(n))|=|f(n)-g(n)|<\epsilon\cdot g(n)\leq\epsilon\cdot h(g(n))\leq\epsilon\cdot|h(g(n))|$$
So, the requested has been proved.
Now, let us consider how could we tighten these restrictions. At first, note that $h$ being an isometry means that, for every $x,y\in\mathbb{R}$:
$$|h(x)-h(y)|=|x-y|$$
It is evident that the condition: 
$$|h(x)-h(y)|\leq|x-y|\text{ for every }x,y\in\mathbb{R}$$
is still satisfying, with almost no change in the previous proof. In the same way we could choose $h$ being Lipschitz continuous with any Lipschitz constant $M>0$ (just choose $n_0$ with $\frac{\epsilon}{M}$ instead of $\epsilon$). 
In the same way, we could demand $h(g(n))\geq A\cdot g(n)$ eventually, for some constant $A>0$, or, more generally, for some consant $A>0$:
$$|h(g(n))|\geq A\cdot g(n)$$.
TL;DR:

A set of restrictions for $h$ is:
  
  
*
  
*$h$ is Lipschitz continuous
  
*$h(g(n))\geq A\cdot g(n)$, $A>0$, eventually (or, more generally, $h(g(x))\geq A\cdot g(x)$, eventually).

