Berasiniz1

2022-01-22

Tell what theorems are available for calculating limits of sequences?

Brenton Pennington

There are some theorems for calculating limits of sequence.
Let ${a}_{n}$ and ${b}_{n}$ be sequence of real numbers, and let A and B be real numbers. These rules hold if $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=A$ and $\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=B$
Sum rule: $\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}+{b}_{n}\right)=A+B$
Difference rule: $\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}-{b}_{n}\right)=A-B$
Constant multiple rule: $\underset{n\to \mathrm{\infty }}{lim}\left(k×{b}_{n}\right)=k×B$ (any number k)
Product rule: $\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}×{b}_{n}\right)=A×B$
Quotient rule: $\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=\frac{A}{B}$ if $B\ne 0$
I can give you some examples:
Constant Rule
$\underset{n\to \mathrm{\infty }}{lim}\left(-\frac{1}{n}\right)=-1×\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}\right)$
$-1×0$
$=0$
Difference rule:
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{n-1}{n}\right)=\underset{n\to \mathrm{\infty }}{lim}\left(1-\frac{1}{n}\right)$
$\underset{n\to \mathrm{\infty }}{lim}1-\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}$
$=1-0$
$=0$
Product Rule:
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{5}{{n}^{2}}\right)=5×\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}\right)×\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}\right)$
$=5×0×0$
$=0$
Sum and Quotient Rule:
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{4-7{n}^{6}}{{n}^{6}+3}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\frac{4}{{n}^{6}}-7}{1+\frac{3}{{n}^{6}}}\right)$
$=\frac{0-7}{1+0}$
$-7$

Theorem 2 is the sandwich Theorem for sequence
Let ${a}_{n}$, ${b}_{n}$ and ${c}_{n}$ be sequences of real numbers. If ${a}_{n}\le {b}_{n}\le {c}_{n}$ holds for all n beyond some index N, and if $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=\underset{n\to \mathrm{\infty }}{lim}{c}_{n}=L$ then $\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=L$ also.
Consider the n-th formula for the sequence function is ${a}_{n}=\frac{\mathrm{sin}n}{n}$
The sine function lies between $-1$ and $1$
$-1\le \mathrm{sin}n\le 1$
Divided by n
$\frac{-1}{n}\le \frac{\mathrm{sin}n}{n}\le \frac{1}{n}$
If $n\to \mathrm{\infty }$ then $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\to 0$ and $\underset{n\to \mathrm{\infty }}{lim}\frac{-1}{n}\to 0$
Apply the sandwich theorem:
If $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{-1}{n}=0$ then the function converges
Thus, the sequence is converge
The limit of the sequence is 0.
Theorem 3. The continuous function theorem for sequences.
Let ${a}_{n}$ be a sequence of real numbers. If ${a}_{n}\to L$ and if $f$ is a functon that is continuous at $L$ and defined at all ${a}_{n}$, then $f\left({a}_{n}\right)\to f\left(L\right)$
For example,
Consider the sequence
$\sqrt{\frac{n+1}{n}}$
The value of the sequence is 1
The value of $\frac{n+1}{n}\to 1$
Consider $f\left(x\right)=\sqrt{x}$ and $L=1$
$\sqrt{\frac{n+1}{n}}=\sqrt{1}=1$

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