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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