 Hello and how are you all doing today? My name is Priyanka and I shall be helping you with the following question. It says, find limit x approaches to 1 fx where fx is equal to x square minus 1 when x value of x is less than equal to 1 and minus x square minus 1 where the value of x is greater than 1. Now proceeding on with the solution we must remember that limit of a function exists if the left hand limit is equal to the right hand limit. Why the knowledge of this concept is the key idea towards this question. Now here in this question we will be finding out the left hand limit. Now here in this question we will be finding out the left hand limit and the right hand limit. Now here we have the left hand limit of the function x is equal to 1. We have as equal to limit x tends to 1 from the left hand side fx that is limit x approaches to 1 from the left hand side. Now here the fx will be taken as x square minus 1 because when x is less than equal to 1 then we need to have the function x square minus 1. Further putting x is equal to 1 minus h we have as h x approaches to 1 therefore h will also approach to 0. So now we have limit h approaches 0 1 minus x square that is x square minus 1 that can be written as limit h approaches to 0. Now 1 a minus b the whole square is a square minus b square minus 2ab and we have minus 1 that is further equal to limit h approaches to 0. h square minus 2h now on simply substituting the value of h as 0 we have 0 the whole square minus 0 into minus 2 into 0 that is 0. Now we will be finding out the right hand limit of the function at x is equal to 1. It will be limit x approaches to 1 from the right hand side fx that is limit x approaches to 1 from right hand side. Now here the function will be minus x square minus 1 putting x equal to 1 plus h and as x approaches 1 then h will approach to 0. So we have limit h approaches 0 minus 1 plus h the whole square that is minus x square minus 1 and opening the brackets we have limit h approaches 0 minus 8 plus b the whole square is a square plus b square plus 2ab minus h which is limit h approaches to 0 minus 1 minus h square minus 2h again minus 1 which can be written as limit h approaches 0 minus 2 minus h square minus 2h. Now on simply putting the value of h as 0 we have minus 2 minus 0 the whole square minus 2 into 0 that is minus 2. Since limit x approaches to 1 from left hand side of the function is not equal to limit x approaches to 1 from the right hand side so we can see that therefore the limit does not exist at x equal to 1. Right so this is the answer to the solution hope you understood the concept well have a nice day.