Algebra of Continuous Functions

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) ${\displaystyle \left(\frac{f}{g}\right)}$ 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  
${\displaystyle \underset{x\to c}{lim}}$(f + g)(x) = ${\displaystyle \underset{x\to c}{lim}}$[f(x) + g(x)]   
(by defination of f + g)
 ${\displaystyle \underset{x\to c}{lim}}$f(x) + ${\displaystyle \underset{x\to c}{lim}}$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 ${\displaystyle \frac{\lambda }{g}}$ defined by ${\displaystyle \frac{\lambda }{g\left(x\right)}}$ = ${\displaystyle \frac{\lambda }{g\left(x\right)}}$ is also continuous wherever g(x) ≠ 0.

In particular, the continuity of g implies continuity of  ${\displaystyle \frac{1}{g}}$
The above theorem can be exploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not.