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Properties of Limits

Common algebraic rules also apply to limits, namely the following.
  • If
    an→aa_n\rightarrow a
    and
    bn→bb_n\rightarrow b
    , then
    lim⁡n→∞an+bn=a+b.\displaystyle\lim_{n\rightarrow\infty} a_n + b_n=a+b.
    ​
  • If
    an→aa_n \rightarrow a
    ​ and
    bn→b≠0b_n\rightarrow b\neq 0
    ​ and
    bn≠0b_n\neq 0
    ​ for all
    nn
    ​, then
    lim⁡n→∞anbn=ab.\displaystyle\lim_{n\rightarrow\infty} \dfrac{a_n}{b_n}=\dfrac{a}{b}.
    ​
Squeeze Lemma If
an≤bn≤cna_n\leq b_n\leq c_n
​, for all
n∈Nn\in\mathbb{N}
, and both the sequences
{an}\{a_n\}
and
{cn}\{c_n\}
​ converge to a limit
LL
, then the sequence
{bn}\{b_n\}
converges to
LL
​ too.
Let
ε>0\varepsilon > 0
​ be given. It suffices to show there is
N∈NN\in\mathbb{N}
​ such that
​
∣bn−L∣≤ε,for all n≥N.|b_n - L| \leq \varepsilon,\quad \text{for all } n\geq N.
​
Since
{an}\{a_n\}
​ converges to
LL
​, there is
N1∈NN_1\in\mathbb{N}
such that
​
∣an−L∣≤ε,for all n≥N1.​|a_n - L|\leq \varepsilon,\quad\text{for all } n\geq N_1.​
​
Similarly, there is
N2∈NN_2\in\mathbb{N}
such that
​
∣cn−L∣≤ε,for all n≥N2.​|c_n - L|\leq \varepsilon,\quad\text{for all } n\geq N_2.​
​
These inequalities imply
​
−ε≤an−L≤bn−L​≤cn−L≤ε,for all n≥max⁡(N1,N2).-\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
​
∣bn−L∣≤ε,for all n≥max⁡(N1,N2).|b_n - L| \leq \varepsilon,\quad \text{for all } n\geq \max(N_1,N_2).
​
and the desired relation holds, taking
N=max⁡(N1,N2).N=\max(N_1,N_2).
■\blacksquare
​
​
​