In arithmetic there is a property that if $\frac{a}{b}=\frac{c}{d}=\alpha$ then $\frac{a-c}{b-d}=\frac{a+c}{b+d}=\alpha$, with the first we assume $b\neq d$.
With the limits, for example, if $\lim_{n\to\infty }\frac{a_{n}}{b_{n}}=\lim_{n\to\infty }\frac{c_{n}}{d_{n}}=\alpha$, assuming $b_{n}> 0$ and $d_{n}> 0$, then it is easy to prove that $\lim_{n\to\infty }\frac{a_{n}+c_{n}}{b_{n}+d_{n}}=\alpha$. I.e.
$\alpha -\varepsilon < \frac{a_{n}}{b_{n}} < \alpha +\varepsilon$ $\alpha -\varepsilon < \frac{c_{n}}{d_{n}} < \alpha +\varepsilon$
From which
$\left ( \alpha -\varepsilon \right )\cdot b_{n} < a_{n} < \left ( \alpha +\varepsilon \right )\cdot b_{n}$ $\left ( \alpha -\varepsilon \right )\cdot d_{n} < c_{n} < \left ( \alpha +\varepsilon \right )\cdot d_{n}$
And as a result $\left ( \alpha -\varepsilon \right )\cdot \left ( b_{n} + d_{n} \right ) < a_{n}+c_{n} < \left ( \alpha +\varepsilon \right )\cdot \left ( b_{n} + d_{n} \right )$
How about $\lim_{n\to\infty }\frac{a_{n}-c_{n}}{b_{n}-d_{n}}$ ?
Let's ignore the case when $\left \{ a_{n} \right \}$, $\left \{ b_{n} \right \}$, $\left \{ c_{n} \right \}$ and $\left \{ d_{n} \right \}$ converge, this is easy to prove. Stick with $\lim_{n\to\infty }\frac{a_{n}}{b_{n}}=\lim_{n\to\infty }\frac{c_{n}}{d_{n}}=\alpha$ though and (e.g.) $b_{n} > d_{n} > 0$.