Properties of Limits

Common algebraic rules also apply to limits, namely the following.
If an→aa_n\rightarrow aan​→a
 and bn→bb_n\rightarrow bbn​→b
, then lim⁡n→∞an+bn=a+b.\displaystyle\lim_{n\rightarrow\infty} a_n + b_n=a+b.n→∞lim​an​+bn​=a+b.
​
If an→aa_n \rightarrow aan​→a
​ and bn→b≠0b_n\rightarrow b\neq 0bn​→b=0
​ and bn≠0b_n\neq 0bn​=0
​ for all nnn
​, then lim⁡n→∞anbn=ab.\displaystyle\lim_{n\rightarrow\infty} \dfrac{a_n}{b_n}=\dfrac{a}{b}.n→∞lim​bn​an​​=ba​.
​
Squeeze Lemma
If an≤bn≤cna_n\leq b_n\leq c_nan​≤bn​≤cn​
​, for all n∈Nn\in\mathbb{N}n∈N
, and both the sequences {an}\{a_n\}{an​}
 and {cn}\{c_n\}{cn​}
​ converge to a limit LLL
, then the sequence {bn}\{b_n\}{bn​}
 converges to LLL
​ too.
Squeeze Lemma Proof
Let ε>0\varepsilon > 0ε>0
​ be given. It suffices to show there is N∈NN\in\mathbb{N}N∈N
​ such that
​∣bn−L∣≤ε,for all n≥N.|b_n - L| \leq \varepsilon,\quad \text{for all } n\geq N.∣bn​−L∣≤ε,for all n≥N.
​
Since {an}\{a_n\}{an​}
​ converges to LLL
​, there is N1∈NN_1\in\mathbb{N}N1​∈N
 such that
​∣an−L∣≤ε,for all n≥N1.​|a_n - L|\leq \varepsilon,\quad\text{for all } n\geq N_1.​∣an​−L∣≤ε,for all n≥N1​.​
​
Similarly, there is N2∈NN_2\in\mathbb{N}N2​∈N
 such that
​∣cn−L∣≤ε,for all n≥N2.​|c_n - L|\leq \varepsilon,\quad\text{for all } n\geq N_2.​∣cn​−L∣≤ε,for all n≥N2​.​
​
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).−ε≤an​−L≤bn​−L​≤cn​−L≤ε,for all n≥max(N1​,N2​).
​
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).∣bn​−L∣≤ε,for all n≥max(N1​,N2​).
​
and the desired relation holds, taking N=max⁡(N1,N2).N=\max(N_1,N_2).N=max(N1​,N2​).
 ■\blacksquare■
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Basic Examples
Divergent Sequences
Last modified 3mo ago