• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q49E

Expert-verified
Found in: Page 435

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Prove Theorem 6. (Hint: Use either definition 2 or the squeeze Theorem).

Using the 2nd definition of limit of a sequence, it is proved that$$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$$if$$\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0.$$

See the step by step solution

## Writing the theorem 6:-

If $$\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0,$$then $$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0.$$

## Applying the Definition 2.

A sequence {$${a_n}$$} has a limit$$L$$,if for every $$\varepsilon > 0$$, there is an integer$$N$$such that$$\left| {{a_n} - L} \right| < \varepsilon$$, for $$n > N$$.

From the theorem6, we have

$$\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0$$

Applying definition 2 for $$L = 0.$$

$$\left| {{a_n} - L} \right| < \varepsilon$$for $$n > N$$

Here, $${a_n} = \left| {{a_n}} \right|$$and$$L = 0.$$

$$\Rightarrow \left| {\left| {{a_n}} \right| - 0} \right| < \varepsilon ,$$for $$n > N$$

## Using/Applying the property $$\left| {\left| {{a_n}} \right|} \right| = {a_n}$$:-

$$\Rightarrow \left| {\left| {{a_n}} \right| - 0} \right| < \varepsilon {\rm{ }}$$

$$\Rightarrow \left| {{a_n} - 0} \right| < \varepsilon ,\forall n \ge N$$

Thus, from the definition 2,

$$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$$

Hence,

It is proved that,if $$\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0$$, then $$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0.$$.