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Expert-verified Found in: Page 435 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # 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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## 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$$\mathop {\lim }\limits_{n \to \infty } f({a_n}) = f(L)$$ ### Want to see more solutions like these? 