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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 the Continuity and Convergence theorem.

Using the 2nd definition of limit of a sequence, it is proved that\(\mathop {\lim }\limits_{n \to \infty } f({a_n}) = f(L)\)if\(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\)and the function\(f\)is continuous at \(L.\)

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Step by Step Solution

Continuity and Convergencetheorem:-

If \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\)and the function\(f\)is continuous at \(L\), then\(\mathop {\lim }\limits_{n \to \infty } f({a_n}) = f(L)\)where

\(L\)is the limit of the sequence {\({a_n}\)},\(f\)is the function continuous at \(L.\)

ApplyingDefinition 2, limit of a sequence :-

For every \(\varepsilon > 0\), there exists a positiveinteger\(N\)such thatfor \(n > N\), \(\left| {{a_n} - L} \right| < \varepsilon .\)

As the function\(f\)is continuous at \(L\), for every \(\varepsilon ' > 0,\)there is a corresponding \(\varepsilon > 0.\)

Such that,

If \(\left| {{a_n} - L} \right| < \varepsilon \) then,

\(\left| {f({a_n}) - f(L)} \right| < \varepsilon '\)

Combining both the inequalities and applying limit of a sequence:-

\(\left| {f({a_n}) - f(L)} \right| < \varepsilon '\)

\( \Rightarrow \mathop {\lim }\limits_{n \to \infty } f({a_n}) = f(L).\)

Therefore, if \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\)and the function \(f\)is continuous at \(L\), then\(\mathop {\lim }\limits_{n \to \infty } f({a_n}) = f(L)\)

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