If $‎\lim\limits_{x\to\infty}‎\frac{f(x)}{g(x)} = 1‎$,‎ then $\lim\limits_{x\to\infty}(f(x) - g(x)) = 0‎$. ‎Suppose ‎$‎f‎$ ‎and ‎‎$‎g‎$‎ ‎‎are real functions such that $‎‎‎\displaystyle{\lim_{x\to\infty}}‎\frac{f(x)}{g(x)} = 1‎$‎‎. ‎My ‎question ‎is‎:
‎‎‎
‎‎What other condition is required ‏‎that $‎‎‎\displaystyle{\lim_{x\to\infty}}(f(x) - g(x)) = 0‎$‎‎‎‎‎‎‎‎‎?‎
‎‎ 
‎Generall‏‎y,‎ $‎‎‎\displaystyle{\lim_{x\to\infty}}‎\frac{f(x)}{g(x)} = 1‎\nRightarrow‎‎‎‎‎\displaystyle{\lim_{x\to\infty}}(f(x) - g(x)) = 0‎$‎. ‏‎For ‎example, ‎‎$‎f(x) = x^2 + x‎$ ‎and ‎‎$‎g(x) =‎ ‎x^2‎$‎.‎
 A: You need $f(x) = g(x) + h(x)$ where both $h(x) = o(1)$ and $h(x) = o[g(x)]$.
... or stated more precisely by Mark Viola, these are necessary and sufficient conditions. 
For example, if  $f(x) = e^{-x}+ 1/x$ and $g(x)= e^{-x}$, then the second condition is violated. In your example, the first condition is violated.
A: The conditions (A) $\lim_{x\to\infty}f(x)/g(x)=1$ and (B) $\lim_{x\to\infty}[f(x)-g(x)]=0$ are pretty much unrelated. For example, $f(x):=2/x$ and $g(x):=1/x$ satisfy B but not A, while your example shows independence in the other direction. Any "additional" condition that implies B will need to be, essentially, a statement of B itself (or something stronger). Condition A won't be making a useful contribution.  
A: Notice that from $\lim\limits_{x\to\infty} \frac{f(x)}{g(x)}=1$ you get
$$\lim\limits_{x\to\infty} \frac{f(x)-g(x)}{g(x)}=0.$$
For example, if you have additional information that the limit $\lim\limits_{x\to\infty} g(x)=L$ exists and is finite, then you get
$$\lim\limits_{x\to\infty} (f(x)-g(x)) = \lim\limits_{x\to\infty} \frac{f(x)-g(x)}{g(x)}\cdot g(x) = 0\cdot L = 0.$$
In fact, even a bit weaker assumption that $g(x)$ is bounded is enough. Take any $\varepsilon>0$. You have $|g(x)|\le M$ for some real number $M$ and each $x$ and
$$\left|\frac{f(x)-g(x)}{g(x)}\right| \le \varepsilon$$ for every large enough $x$. Then you get
$$|f(x)-g(x)| = \left|\frac{f(x)-g(x)}{g(x)}\right| \cdot |g(x)| \le \varepsilon M$$
for $x\ge x_0$ (where $x_0$ depends on $\varepsilon$), i.e.,
$$\limsup_{x\to\infty} |f(x)-g(x)| \le \varepsilon M.$$
Since this is true for every $\varepsilon > 0$ (and $M$ is fixed), you get $\limsup\limits_{x\to\infty} |f(x)-g(x)| = 0$ and
$$\lim\limits_{x\to\infty} (f(x)-g(x)) = 0.$$ 
A: $‎‎‎\displaystyle{\lim_{x\to\infty}}(f(x) - g(x)) = 0‎$ is stronger than $‎‎‎\displaystyle{\lim_{x\to\infty}}‎\frac{f(x)}{g(x)} = 1‎.$  If $g(x) \neq 0$ for $x\geq x_0$ and $g(x) \neq o(1)$ we have
$$\displaystyle{\lim_{x\to\infty}}(f(x) - g(x)) = 0 \implies \displaystyle{\lim_{x\to\infty}}‎\frac{f(x)}{g(x)} = 1.$$
The other implication is not true all the time; see that 
$$\displaystyle{\lim_{x\to\infty}}‎\frac{f(x)}{g(x)} = 1 \iff f(x) = g(x) + o(g(x))$$
and
$$\displaystyle{\lim_{x\to\infty}}(f(x) - g(x)) = 0 \iff f(x) = g(x) + o(1).$$
A: For this to be true we need the $\lim_{x\to \infty}f(x)$ and $\lim_{x\to \infty}g(x)$ to exist and be finite or the functions can be such that $\lim_{x\to \infty}f(x)=\lim_{x\to \infty}g(x)=\pm\infty$ but then we need that $f=g$.
$$\lim_{x\to\infty}(f(x) - g(x)) = 0‎$$
$$\lim_{x\to\infty}(f(x)) - \lim_{x\to\infty}(g(x)) = 0‎$$
$$\lim_{x\to\infty}(f(x)) = \lim_{x\to\infty}(g(x))‎$$
