TITLE: If $‎\lim\limits_{x\to\infty}‎\frac{f(x)}{g(x)} = 1‎$,‎ then $\lim\limits_{x\to\infty}(f(x) - g(x)) = 0‎$.
QUESTION [12 upvotes]: ‎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 ‎‎$‎g(x) =‎ ‎x^2‎$‎.‎

REPLY [3 votes]: 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
$$\lim\limits_{x\to\infty} (f(x)-g(x)) = 0.$$