# Properties of Limits

Common algebraic rules also apply to limits, namely the following.
• If
$a_n\rightarrow a$
and
$b_n\rightarrow b$
, then
$\displaystyle\lim_{n\rightarrow\infty} a_n + b_n=a+b.$
• If
$a_n \rightarrow a$
​ and
$b_n\rightarrow b\neq 0$
​ and
$b_n\neq 0$
​ for all
$n$
​, then
$\displaystyle\lim_{n\rightarrow\infty} \dfrac{a_n}{b_n}=\dfrac{a}{b}.$
Squeeze Lemma If
$a_n\leq b_n\leq c_n$
​, for all
$n\in\mathbb{N}$
, and both the sequences
$\{a_n\}$
and
$\{c_n\}$
​ converge to a limit
$L$
, then the sequence
$\{b_n\}$
converges to
$L$
​ too.
Let
$\varepsilon > 0$
​ be given. It suffices to show there is
$N\in\mathbb{N}$
​ such that
$|b_n - L| \leq \varepsilon,\quad \text{for all } n\geq N.$
Since
$\{a_n\}$
​ converges to
$L$
​, there is
$N_1\in\mathbb{N}$
such that
$|a_n - L|\leq \varepsilon,\quad\text{for all } n\geq N_1.​$
Similarly, there is
$N_2\in\mathbb{N}$
such that
$|c_n - L|\leq \varepsilon,\quad\text{for all } n\geq N_2.​$
These inequalities imply
$-\varepsilon \leq a_n - L \leq b_n- L​ \leq c_n - L \leq \varepsilon, \quad \text{for all } n\geq \max(N_1,N_2).$
Thus, using the definition of absolute value, the above relation yields
$|b_n - L| \leq \varepsilon,\quad \text{for all } n\geq \max(N_1,N_2).$
and the desired relation holds, taking
$N=\max(N_1,N_2).$
$\blacksquare$