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```        could someone plz clear the confusion i have in left hand /right hand limits n left hand right hand derivatives??
heres what i know:
when a function is defined differently under 3 conditions ( eg, f(x) = 1+x , x>0 ; 1 , x=0 and 1-x ,x>0
then @ 0, lhl is got by 1+0=1 ; l is given to be =1 ; rhl is got by 1-0=0
then we check for differentiabilty at 0 by left hand derivative = right hand hand derivative n check for differentiability..but how exactly do v evaluate lhd n rhd in the above example?
```
8 years ago

```							Dear student,
We will learn about differentiability , and the various factors  affecting differentiability. We have already defined the derivative of a  function f(x) at a particular point 'a' and derivative of f(x) in  general for the variable x as f (a) and f (x) respectively. The restriction in both the cases is that 'the limit must exist'.
If does not exist, then we say that the function is not differentiable.
If the above limit exists, we say the function f(x) is differentiable.
In order to test the differentiability of a function at a point, the  right hand derivative and left hand derivatives are introduced as  follows:
Right Hand Derivative
Let f be a function of x (y=f(x)). Let a be a point in the domain of f. The RHD of f at a is defined as where h>0, provided the limit exists.
Left Hand Derivative
The LHD of f at a is defined as where h>0, provided the limit exists.
In the above definition showiing differentiabilty, substitute a + h = x, then h = x - a as Rf '(a) can be rewritten as Similarly, substitute a - h = x. Lf '(a) can be written as Differentiability at a Point
Diffferentiability at a point 'a' for a function f(x) if
(i) both Rf '(a) and Lf '(a) exists and finite.
(ii) Rf '(a) = Lf '(a)
Consider the function y=|x|. This function is differentiable on ( ,0) and (0, ), but not differentiable at x = 0.  For x>0, we have  Since the limit exists, f(x) is differentiable at x>0. Similarly, we can show that f(x) is differentiable at x<0.
We shall find RHD and LHD of f(x) at x = 0.    = -1 Therefore y = |x| is not differentiable at x = 0.
```
8 years ago
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