 Hello and how are you all today? My name is Priyanka and the question says if fx is equal to mx square plus n when x is less than 0, nx plus m when the value of x is greater than equal to 0 or less than equal to 1 and nx cube plus m when the value of x is greater than 1. For what integers m and n does both limit x approaches to 0, fx and limit x approaches to 1, fx exists. So here in this question we are required to find out the value of integer m and n. So when x is less than 0 then the function which is given to us is mx square plus n and when the value of x is greater than equal to 0 but less than equal to 1 then the value of the function is given to us as nx plus m. So we have limit x approaches to 0 from the left hand side mx square plus n and on using the limits we have answer as equal to n. And here we have limit x approaches to 0 from right hand side nx plus m and on using the limit we have it equal to m. So we can write and therefore we have n equal to m. Right, this is for the first case. Now for the second case we have when the value of x is greater than equal to 0 but less than equal to 1 then the function which is known to us is nx plus m and when x is greater than 1 then the function which is given to us is nx cube plus m. Therefore limit x approaches to 1 from left hand side nx plus m and on using the limit we have the answer as n plus m. And here we have limit x approaches to 1 from the right hand side nx cube plus m which gives us the answer as n plus m. Right, so we can say that therefore limit x approaches 0 fx is equal to m and limit x approaches 1 fx is equal to 2m for all m which belongs to real number. So the answer to this question is that for limit x approaches 0 fx to exist we need m equal to n. And for limit x approaches to 1 fx exist for any integral value of m and n. Right, so this is a required answer. Hope you understood the whole concept well. Have a nice day.