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2015-03-09T09:38:52+05:30

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As you can see from the graph, It is continuous at x=1 and its value is 0.
But it is Not differentiable at x=1 as the graph is not smooth at x=1. It is changing its slope at that point.
For x<1, slope is -1 and for x>1, slope is +1. So at x=1, slope is not defined. Thus it is not differentiable at x=1.
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2015-03-09T15:01:08+05:30

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   y = f(x) = | x - 1 |
           =  x - 1   , if  x > 1
           =  0      , if x = 1
           =  1 - x  , if  x < 1

For continuity of the function at x = 1,  the right limit of the function as x -> 1⁺ from right side (values a little higher than 1) must be equal to the left limit of the function, as x -> 1⁻  (from values a little less than 1).  The right limit is :
 \lim_{x \to 1+} f(x)=[x-1]_{x=1^+} = 1-1=0\\\\equivalently, Let\ x=1+\Delta x\\\lim_{x \to 1+} f(x)=\lim_{\Delta x \to 0^+}\ f(1+\Delta x)=1+\Delta x-1\\\\=\lim_{\Delta x \to 0}\ [\Delta x]=0\\\\

The left limit is :
\lim_{x \to 1^-} f(x)=[1-x]_{x=1^-} = 1-1=0\\\\equivalently, Let\ x=1-\Delta x\\\lim_{x \to 1^-} f(x)=\lim_{\Delta x \to 0^+}\ f(1-\Delta x)=1-(1-\Delta x)\\\\=\lim_{\Delta x \to 0}\ [\Delta x]=0\\
The values of left and right limits are equal and hence, the function is continuous.
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the definition of derivative is
\frac{df(x)}{dx}|_{x=1} =f '(1)=[ \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} ]_{x=1}\\

The function must be continuous and then the left derivative and right derivative must exist and then be equal.  Then the derivative at the function exists.

Right derivative is :
\frac{df(x)}{dx}|_{x=1^+} =f '(1^+)=[ \lim_{\Delta x \to 0^+} \frac{f(x+\Delta x)-f(x)}{\Delta x} ]_{x=1}\\\\=\lim_{\Delta x \to 0^+}\frac{(x+\Delta x-1)-(x-1)}{\Delta x}\\\\=\lim_{\Delta x \to 0^+}\frac{\Delta x}{\Delta x}\\\\=1

Left derivative is :
\frac{df(x)}{dx}|_{x=1^-} =f '(1^-)=[ \lim_{\Delta x \to 0^+} \frac{f(x-\Delta x)-f(x)}{\Delta x} ]_{x=1}\\\\=[\lim_{\Delta x \to 0^+}\frac{(x-\Delta x-1)-(x-1)}{\Delta x}]_{x=1}\\\\=\lim_{\Delta x \to 0^+}\frac{-\Delta x}{\Delta x}\\\\=-1

Since the left and right derivatives are not equal the function given, is not derivable at x = 1.

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