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 .


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