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Essential Calculus: Early Transcendentals
Found in: Page 435
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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

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,

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


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

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