📖 generic · CBSE Class 12th English Medium · MATHEMATCS PART-1 · Page 1question

DIFFERENTIABILITY · Part 10

Chapter 5: CONTINUITY AND DIFFERENTIABILITY · MATHEMATCS PART-1

are similar and left as an exercise to the reader. Remarks (i) As a special case of ( ) 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 = – , the continuity of f implies continuity of – f . (ii) As a special case of ( ) above, if f is the constant function f ( x ) = λ , then the function g λ defined by g g x λ λ is also continuous wherever g ( x ) ≠ . In particular, the continuity of g implies continuity of g .

The above theorem can be exploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not. The following examples illustrate this: Example Prove that every rational function is continuous. Solution Recall that every rational function f is given by , p x q x q x ≠ where p and q are polynomial functions.

The domain of f is all real numbers except points at which q is zero. Since polynomial functions are continuous (Example ), f is continuous by ( ) of Theorem . Example Discuss the continuity of sine function. Solution To see this we use the following facts lim sin We have not proved it, but is intuitively clear from the graph of sin x near .

Now, observe that f ( x ) = sin x is defined for every real number. Let c be a real number. Put x = c + h . If x → c we know that h → .

Therefore c f x = lim sin lim sin( lim [sin cos cos sin ] lim [sin cos ] lim [cos sin ] = sin c + = sin c = f ( c ) Thus lim → f ( x ) =

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