__A function__

**Theorem :-****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**.

__Let__

**Proof :-****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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