Prove that the greatest integer function defined by $f(x) = \left[ x \right],0 < x < 3$ is not differentiable at x = 1 and x = 2.

Hint: To show that the functions are not differentiable at the given points we need to prove that they not continuous at the given points using the condition $\mathop {\lim }\limits_{x \to {x^ - }} f(x) = \mathop {\lim }\limits_{x \to {x^ + }} f(x)$

Complete step-by-step answer:
We know that a function is differentiable only if it is continuous .
So it is enough if we prove that the function is not continuous at x = 1 and x = 2
For a function to be continuous the left hand derivative must be equal to the right hand derivative
That is , $\mathop {\lim }\limits_{x \to {x^ - }} f(x) = \mathop {\lim }\limits_{x \to {x^ + }} f(x)$
So let's check the continuity at x = 1
$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} \left[ 0 \right] = 0$
$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} \left[ 1 \right] = 1$
Hence we can see that $\mathop {\lim }\limits_{x \to {x^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {x^ + }} f(x)$
Hence it is not continuous at x = 1
Hence it is not differentiable at x = 2
So let's check the continuity at x = 2
$ \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} \left[ 1 \right] = 1$
$ \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} \left[ 2 \right] = 2$
Hence we can see that $\mathop {\lim }\limits_{x \to {x^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {x^ + }} f(x)$
Hence it is not continuous at x = 2
Hence it is not differentiable at x = 2
Therefore it is proved that the function is not differentiable at x = 1 and x = 2.

Note: Continuity of a function is the characteristic of a function by virtue of which, the graphical form of that function is a continuous wave. A differentiable function is a function whose derivative exists at each point in its domain.
Every differentiable function is continuous but that doesn’t mean all the continuous functions are differentiable