 Hi and welcome to the session. My name is Shashi and I am going to help you with the following question. Question says, show that the function defined by gx equal to x minus x is discontinuous at all integral points. Here, x denotes the greatest integer less than or equal to x. Let us now start the solution. We are given gx is equal to x minus greatest integer x. Now, let us consider any integer a. Let us now find out left hand side limit at x equal to a. Now, we can write limit of x tending to a minus gx is equal to limit of x tending to a minus x minus greatest integer x. Now, put x equal to a minus h then as x tends to a minus h tends to 0. So, we can write this limit as equal to limit of h tending to 0 a minus h minus greatest integer a minus h. Now, this is equal to limit of h tending to 0 a minus h minus a minus 1. Now, we know a minus h is less than a and greater than equal to a minus 1. So, therefore, we can write a minus h is equal to a minus 1. So, we have written a minus 1 for a minus h here. Now, this is equal to a minus a minus 1 which is further equal to a minus a plus 1. Now, we can write it equal to 1. So, we get limit of x tending to a minus gx equal to 1. Now, let us find out right hand side limit of the function at x equal to a. So, we can write right hand side limit at x equal to a as limit of x tending to a plus gx is equal to limit of x tending to a plus x minus greatest integer x. Now, put x equal to a plus h then as x tends to a plus h tends to 0. Now, we get this limit as limit of h tending to 0 a plus h minus greatest integer a plus h. Now, this is equal to limit of h tending to 0 a plus h minus a. We know a plus h is less than a plus 1 and greater than equal to a. So, we can write greatest integer a plus h equal to a. Now, this is equal to a plus 0 minus a which is further equal to 0. So, we get limit of x tending to a plus gx equal to 0. So, clearly we can see right hand side limit at x equal to a is not equal to left hand side limit at x equal to a. So, we can write therefore limit of x tending to a minus gx is not equal to limit of x tending to a plus gx since 1 is not equal to 0. So, this implies given function g is discontinuous at all integral points. We know function is discontinuous at point when left hand side limit is not equal to right hand side limit of the function at that point. So, clearly we can see left hand side limit and right hand side limit are unequal. So, function g is discontinuous at all integral points. Hence, proved. This completes the session. Hope you understood the session. Goodbye.