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Limit of a Sequence: Definition and Examples

Lata Agarwal Mathpedia, Sequences

Limit of a Sequence:

If a sequence { ${s_n}$ } tends to a limit $l$ , then we say that $l$ is the limit of a sequence { ${s_n}$ }, and write

$\mathop {\lim }\limits_{n \to \infty } {s_n} = l$ .

In other words,

“A real number $l$ is the limit of a sequence { ${s_n}$ }, if for every $\varepsilon > 0$ , there exists a positive integer $M$ (or $M \in$ N) such that,

$n \ge M$ implies that $\left| {{s_n} - l} \right| < \varepsilon$ .”

The number $l$ is called the limit of the sequence { ${s_n}$ }, and we write, ${s_n} \to l$ as $n \to \infty$ , or

$\mathop {\lim }\limits_{n \to \infty } {s_n} = l$ , or simply, $\lim {s_n} = l$ .

Precise Definition of a Limit of a Sequence at Infinity –

“If for every $\varepsilon > 0$ , there exists a positive integer $M$ (or $M \in$ N) such that

$n \ge M$ implies that $\left| {{s_n} - l} \right| < \varepsilon$ ,

then the number $l$ is called the limit of the sequence $\left\langle {{s_n}} \right\rangle$ , and we write,

$\mathop {\lim }\limits_{n \to \infty } {s_n} = l$ , or simply, $\lim {s_n} = l$ .”

(Note: When it is said that this sequence has a limit, it means that the limit is unique, that is, a finite and definite real number.)

Limit of a Sequence Examples:

Example 1:

Show that the sequence { $1/n$ } has the limit 0.

Proof:

Here, ${s_n} = \frac{1}{n}$ , and $l = 0$ .

In this example, we have to show that the sequence { $1/n$ } has the limit 0, that is,

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0$ .

To prove this limit: $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0$ ; we consider $\varepsilon > 0$ , and show that there exists a positive integer $M$ such that,

$n \ge M$ implies that $\left| {{s_n} - l} \right| < \varepsilon$ , that is,

$n \ge M$ implies that $\left| {\frac{1}{n} - 0} \right| < \varepsilon$ .

To show this, we must find a positive integer $M$ such that $n \ge M$ implies that,

$\left| {\frac{1}{n} - 0} \right| < \varepsilon$ , or

$\left| {\frac{1}{n}} \right| < \varepsilon$ , or

$\frac{1}{n} < \varepsilon$ , or

$\frac{1}{\varepsilon } < n$ , or

$n > \frac{1}{\varepsilon }$ .

Hence, if we take $M > \frac{1}{\varepsilon }$ , then for every $\varepsilon > 0$ , there is a positive integer $M$ such that $n \ge M$ implies that $\left| {\frac{1}{n} - 0} \right| < \varepsilon$ .

Therefore, $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0$ , or the sequence { $1/n$ } has the limit 0.

Example 2:

If ${s_n} = c$ , where c is a real number, prove that $\lim {s_n} = c$ .

Proof:

Here, ${s_n} = c$ , and $l = c$ .

To prove this limit: $\lim {s_n} = c$ ; we consider $\varepsilon > 0$ , and show that there exists a positive integer $M$ such that,

$n \ge M$ implies that $\left| {{s_n} - c} \right| < \varepsilon$ .

Now, given, $\varepsilon > 0$ , we must find a positive integer $M$ such that $n \ge M$ implies that,

$\left| {{s_n} - c} \right| < \varepsilon$ , or

$\left| {c - c} \right| < \varepsilon$ , or

$\left| 0 \right| < \varepsilon$ , or

$0 < \varepsilon$ .

Hence, if we take $M = 1$ , then for all $n \ge 1$ , we have $\left| {{s_n} - c} \right| < \varepsilon$ .

Therefore, $\lim {s_n} = c$ , or the sequence $\left\langle {{s_n}} \right\rangle$ has the limit $c$ .

Example 3:

Use the definition of the limit of a sequence to show that the sequence $\left\langle {{s_n}} \right\rangle$ where,

${s_n} = \frac{{3n}}{{n + 5{n^{1/2}}}}$ ,

has the limit 3.

Proof:

Here, ${s_n} = \frac{{3n}}{{n + 5{n^{1/2}}}}$ , and $l = 3$ .

In this example, we have to show that the sequence $\left\langle {{s_n}} \right\rangle$ has the limit 3, that is,

$\mathop {\lim }\limits_{n \to \infty } \frac{{3n}}{{n + 5{n^{1/2}}}} = 3$ .

To prove this limit, we consider $\varepsilon > 0$ , and show that there exists a positive integer $M$ such that,

$n \ge M$ implies that $\left| {{s_n} - l} \right| < \varepsilon$ , that is,

$n \ge M$ implies that $\left| {\frac{{3n}}{{n + 5{n^{1/2}}}} - 3} \right| < \varepsilon$ .

To show this, we must find a positive integer $M$ such that,

$\left| {\frac{{3n}}{{n + 5{n^{1/2}}}} - 3} \right| < \varepsilon$ , …… (1)

for all $n \ge M$ .

This inequality (1) is equivalent to

$\left| {\frac{{3n}}{{n + 5{n^{1/2}}}} - 3} \right| < \varepsilon$

$\left| {\frac{{3n - 3\left( {n + 5{n^{1/2}}} \right)}}{{n + 5{n^{1/2}}}}} \right| < \varepsilon$

$\left| {\frac{{3n - 3n - 15{n^{1/2}}}}{{n + 5{n^{1/2}}}}} \right| < \varepsilon$

$\left| {\frac{{ - 15{n^{1/2}}}}{{n + 5{n^{1/2}}}}} \right| < \varepsilon$

$\frac{{15{n^{1/2}}}}{{n + 5{n^{1/2}}}} < \varepsilon$ . …… (2)

Now,

$\frac{{15{n^{1/2}}}}{{n + 5{n^{1/2}}}} < \frac{{15{n^{1/2}}}}{n} < \frac{{15}}{{{n^{1/2}}}}$ ,

then inequality (2) will hold if,

$\frac{{15}}{{{n^{1/2}}}} < \varepsilon$

$\frac{{15}}{\varepsilon } < {n^{1/2}}$

${\left( {\frac{{15}}{\varepsilon }} \right)^2} < n$

$\frac{{{{15}^2}}}{{{\varepsilon ^2}}} < n$

$\frac{{225}}{{{\varepsilon ^2}}} < n$

$n > \frac{{225}}{{{\varepsilon ^2}}}$ .

Hence, if we take $M > \frac{{225}}{{{\varepsilon ^2}}}$ , then for every $\varepsilon > 0$ , there is a positive integer $M$ such that $n \ge M$ implies that $\left| {\frac{{3n}}{{n + 5{n^{1/2}}}} - 3} \right| < \varepsilon$ .

Therefore, $\mathop {\lim }\limits_{n \to \infty } \frac{{3n}}{{n + 5{n^{1/2}}}} = 3$ , or the sequence $\left\langle {{s_n}} \right\rangle$ where ${s_n} = \frac{{3n}}{{n + 5{n^{1/2}}}}$ has the limit 3.

Tags: define the limit of the sequences, how to find limit, finding limits of sequences

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About Lata Agarwal 268 Articles

M.Phil in Mathematics, skilled in MS Office, MathType, Ti-83, Internet, etc., and Teaching with strong education professional. Passionate teacher and loves math. Worked as a Assistant Professor for BBA, BCA, BSC(CS & IT), BE, etc. Also, experienced SME (Mathematics) with a demonstrated history of working in the internet industry. Provide the well explained detailed solutions in step-by-step format for different branches of US mathematics textbooks.

Limit of a Sequence: Definition and Examples

Limit of a Sequence:

Limit of a Sequence Examples:

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Limit of a Sequence:

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