Thursday, 2 January 2014

Differentiability Theorem




Theorem :-  A function   f    is differentiable  at     x=a     if and only if there exists a number   l    .
                   such that ;

                    f(a+h)-f(a)=lh+hn

                   Where     n    denotes a quantity which tends to    0     as   h-->0     .

Proof :-     Let    f     be differentiable at    x=a    . Then there exists a number   l    
                Such that ;

                lim x-->a  [f(x)-f(a)]/[x-a]=l

               putting ;     x=a+h ,

               lim h-->0  [f(a-h)-f(a)]/h=l

           or;    lim h-->0 [{f(a+h)-f(a)/h}-l]=0

              therefore   [{f(a+h)-f(a)}/h]-l     is equal to    n  

            where    n-->0    as     h-->0    

          Therefore f(a+h)-f(a)=lh+hn 

          where    n-->0    as    h-->0     .

          Thus it is the necessary condition .

As the argument is reversible , the condition is also sufficient .


No comments:

Post a Comment

Our Latest Post

How to Evaluate an Integral

In this video you will see how to evaluate an Integral. This video shows an example, by this example you will see about how to evaluate an ...

Popular Post