Asymptotic Expansion $a \sim b$, $f \sim g \implies f + a \sim g + b$ for positive functions This is probably a really dumb question but for positive functions $f,g,a,b$ why does
$$f(n) \sim g(n),\;a(n) \sim b(n) \implies (a + f) \sim (b + g)$$
Taking $$\lim_{n \to \infty} \frac{f}{g} = 1 = \frac{\lim\limits_{n \to \infty}f}{\lim\limits_{n \to \infty}g}$$
the answer becomes trivial, but without relying on that I'm having difficulty.
 A: The statement is false, consider for example $f(x)=x+\cos(x)$, $g(x)=x+\sin(x)$, $a(x)=b(x)=-x$.
Edit: Since $f\sim g$ and $a\sim b$ as $n\to \infty$, there exist $r, s\to 0$ as $n\to\infty$ such that $f = g\cdot (1+r)$, $a= b\cdot (1+s)$.  Some simplifcation yieds:
$$ \frac{f+a}{g+b} = \frac{g\cdot (1+r) + b\cdot (1+s)}{g+b} = 1 + \frac{gr+bs}{g+b} = 1 + \frac{g}{g+b}r + \frac{b}{g+b}s$$
Now, assuming $g, b$ are positive functions, then $\frac{g}{g+b} \le 1$ and $\frac{b}{g+b} \le 1$, and thus the last 2 terms vanish.
A: Assume $f,g,a,b$ are positive functions such that $f(n) \sim g(n)$ and $a(n) \sim b(n)$.

Fix $\epsilon\in (0,1)$.

Let $n_0$ be such that for all $n \ge n_0$ we have
$$
\left\lbrace
\begin{align*}
&1-\epsilon < \frac{f(n)}{g(n)} < 1+\epsilon\\[4pt]
&1-\epsilon < \frac{a(n)}{b(n)} < 1+\epsilon\\[4pt]
\end{align*}
\right.
$$
Then for all $n \ge n_0$ we have
\begin{align*}
\frac{f(n)+a(n)}{g(n)+b(n)}
&=
\frac{f(n)}{g(n)+b(n)}+\frac{a(n)}{g(n)+b(n)}
\\[4pt]
&=\,
\left(\frac{f(n)}{g(n)}\right)\!\left(\frac{g(n)}{g(n)+b(n)}\right)
\;+\;
\left(\frac{a(n)}{b(n)}\right)\!\left(\frac{b(n)}{g(n)+b(n)}\right)
\\[4pt]
&>\,
\bigl(1-\epsilon\bigr)\!\left(\frac{g(n)}{g(n)+b(n)}\right)
\;+\;
\bigl(1-\epsilon\bigr)\!\left(\frac{b(n)}{g(n)+b(n)}\right)
\\[4pt]
&=\,
\bigl(1-\epsilon\bigr)\!\left(\frac{g(n)+b(n)}{g(n)+b(n)}\right)
\\[4pt]
&=\,
1-\epsilon
\\[4pt]
\end{align*}
and similarly, for all $n \ge n_0$ we have
\begin{align*}
\frac{f(n)+a(n)}{g(n)+b(n)}
&=
\frac{f(n)}{g(n)+b(n)}+\frac{a(n)}{g(n)+b(n)}
\\[4pt]
&=\,
\left(\frac{f(n)}{g(n)}\right)\!\left(\frac{g(n)}{g(n)+b(n)}\right)
\;+\;
\left(\frac{a(n)}{b(n)}\right)\!\left(\frac{b(n)}{g(n)+b(n)}\right)
\\[4pt]
&<\,
\bigl(1+\epsilon\bigr)\!\left(\frac{g(n)}{g(n)+b(n)}\right)
\;+\;
\bigl(1+\epsilon\bigr)\!\left(\frac{b(n)}{g(n)+b(n)}\right)
\\[4pt]
&=\,
\bigl(1+\epsilon\bigr)\!\left(\frac{g(n)+b(n)}{g(n)+b(n)}\right)
\\[4pt]
&=\,
1+\epsilon
\\[4pt]
\end{align*}
Thus for all $\epsilon\in (0,1)$ there exists $n_0$ such that for all $n \ge n_0$ we have
$$
1-\epsilon
 <
\frac{f(n)+a(n)}{g(n)+b(n)}
 <
1+\epsilon
$$
hence $(f+a) \sim (g+b)$.
