Maharashtra State BoardHSC Arts 12th Board Exam
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Algebra of Continuous Functions

theorem

Suppose f and g be two real functions continuous at a real number c. Then 
(1) f + g is continuous at x = c. 
(2) f – g is continuous at x = c. 
(3) f . g is continuous at x = c.
(4) `(f/g)` is continuous at x = c ,
(provided g(c) ≠ 0).

Proof:  We are investigating continuity of (f + g) at x = c.
Clearly it is defined at x = c. We have  
`lim_(x->c)`(f + g)(x) = `lim_(x->c)`[f(x) + g(x)]   
(by defination of f + g)
 `lim_(x->c)`f(x) + `lim_(x->c)`g(x) 
(by the theorem on limits)
= f(c) + g(c)  (as f and g are continuous)
=(f+g)(c)       (by defination of f + g)
Hence, f + g is continuous at x = c. Proofs for the remaining parts are similar and left as an exercise to the reader.

Remarks:  
(i) As a special case of (3) above, if f is a constant function, i.e., f(x) = λ for some real number λ, then the function (λ . g) defined by (λ . g) (x) = λ . g(x) is also continuous. In particular if λ = – 1, the continuity of f implies continuity of – f.

(ii) As a special case of (4) above, if f is the constant function f(x) = λ, then the function `lambda/g` defined by `lambda/g(x)` = `lambda/g(x)` is also continuous wherever g(x) ≠ 0.

In particular, the continuity of g implies continuity of  `1/g`
The above theorem can be exploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not. 

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Shaalaa.com | Continuity and Differentiability part 12 (Continuity composite function)

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Continuity and Differentiability part 12 (Continuity composite function) [00:06:40]
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