If f(x) = |x - 3|, then f is | vedclass

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(D) $\mathop {\lim }\limits_{x 	o {3^ - }} f(x) = \mathop {\lim }\limits_{h 	o 0} f(3 - h) = \mathop {\lim }\limits_{h 	o 0} |3 - h - 3| = 0$$\math

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Continuous but not differentiable at $x = 3$

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(D) $\mathop {\lim }\limits_{x 	o {3^ - }} f(x) = \mathop {\lim }\limits_{h 	o 0} f(3 - h) = \mathop {\lim }\limits_{h 	o 0} |3 - h - 3| = 0$$\mathop {\lim }\limits_{x 	o {3^ + }} f(x) = \mathop {\lim }\limits_{h 	o 0} f(3 + h) = \mathop {\lim }\limits_{h 	o 0} |3 + h - 3| = 0$$\because \mathop {\lim }\limits_{x 	o {3^ - }} f(x) = \mathop {\lim }\limits_{x 	o {3^ + }} f(x) = f(3)$Hence,$f$ is continuous at $x = 3$.Now,$L f'(3) = \mathop {\lim }\limits_{h 	o 0} \frac{f(3 - h) - f(3)}{-h}$$= \mathop {\lim }\limits_{h 	o 0} \frac{|3 - h - 3| - 0}{-h} = \mathop {\lim }\limits_{h 	o 0} \frac{h}{-h} = -1$$R f'(3) = \mathop {\lim }\limits_{h 	o 0} \frac{f(3 + h) - f(3)}{h}$$= \mathop {\lim }\limits_{h 	o 0} \frac{|3 + h - 3| - 0}{h} = \mathop {\lim }\limits_{h 	o 0} \frac{h}{h} = 1$$\because L f'(3) 
eq R f'(3)$Hence,$f$ is not differentiable at $x = 3$.Trick: By observing the graph,the function is continuous,but the tangent is not defined at $x = 3$ due to the sharp corner.

If $f(x) = |x - 3|,$ then $f$ is

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