Tell what theorems are available for calculating limits of sequences?

Question

Brenton Pennington · Accepted Answer

There are some theorems for calculating limits of sequence.    Let an and bn be sequence of real numbers, and let A and B be real numbers. These rules hold if limn→∞an=A and limn→∞bn=B     Sum rule: limn→∞(an+bn)=A+B    Difference rule: limn→∞(an−bn)=A−B    Constant multiple rule: limn→∞(k×bn)=k×B (any number k)    Product rule: limn→∞(an×bn)=A×B    Quotient rule: limn→∞anbn=AB if B≠0    I can give you some examples:    Constant Rule    limn→∞(−1n)=−1×limn→∞(1n)    −1×0    =0    Difference rule:    limn→∞(n−1n)=limn→∞(1−1n)    limn→∞1−limn→∞1n    =1−0    =0    Product Rule:    limn→∞(5n2)=5×limn→∞(1n)×limn→∞(1n)    =5×0×0    =0    Sum and Quotient Rule:    limn→∞(4−7n6n6+3)    =limn→∞(4n6−71+3n6)    =0−71+0    −7

Addisyn Thompson · Accepted Answer

Theorem 2 is the sandwich Theorem for sequence
Let

a_{n}

,

b_{n}

and

c_{n}

be sequences of real numbers. If

a_{n} \leq b_{n} \leq c_{n}

holds for all n beyond some index N, and if

lim_{n \to \infty} a_{n} = lim_{n \to \infty} c_{n} = L

then

lim_{n \to \infty} b_{n} = L

also.
Consider the n-th formula for the sequence function is

a_{n} = \frac{\sin n}{n}

The sine function lies between

- 1

and

1

- 1 \leq \sin n \leq 1

Divided by n

\frac{- 1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}

If

n \to \infty

then

lim_{n \to \infty} \frac{1}{n} \to 0

and

lim_{n \to \infty} \frac{- 1}{n} \to 0

Apply the sandwich theorem:
If

lim_{n \to \infty} \frac{1}{n} = lim_{n \to \infty} \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 (a_{n}) \to f (L)

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 (x) = \sqrt{x}

and

L = 1

\sqrt{\frac{n + 1}{n}} = \sqrt{1} = 1

Tell what theorems are available for calculating limits of sequences?

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