 Hi, and welcome to the session. Let us discuss the following question. The question says, evaluate limit x approaches 0 x into 1 minus square root of 1 minus x square upon square root of 1 minus x square into sine inverse x whole cube. Let's now begin with the solution. We have to evaluate limit x approaches 0 x into 1 minus square root of 1 minus x square upon square root of 1 minus x square into sine inverse x whole cube. Put x as sine theta as x approaches 0 implies theta also approaches 0. So by substituting sine theta in place of x and changing the limit, we get this expression as limit theta approaches 0 sine theta into 1 minus square root of 1 minus sine square theta upon square root of 1 minus sine square theta. Now x equals to sine theta implies sine inverse x is equal to theta. So we have theta cube in the denominator. Now this is equal to limit theta approaches 0 sine theta into 1 minus sine square theta is cos square theta. So we have square root of cos square theta by square root of cos square theta into theta cube. Now this is equal to limit theta approaches 0 sine theta into 1 minus cos theta by cos theta into theta cube. We know that 1 minus cos square theta is 2 sine square theta by 2. So now we have limit theta approaches 0 sine theta into 2 sine square theta by 2 divided by cos theta into theta cube. This is equal to limit theta approaches 0 sine theta by theta into limit theta approaches 0 sine theta by 2 into sine theta by 2 by. Now we can write 2 by theta square as theta by 2 into theta by 2 into 2 into limit theta approaches 0 1 by cos theta at limit theta approaches 0 sine theta by theta is equal to 1. So by using this result, we get 1 into and here we are going to put theta by 2 as y so that approaches 0 then y also approaches 0. As we have limit y approaches 0 sine by pi by into limit y approaches 0 sine by pi by which is 0 1 by 2 into limit theta approaches 0 1 by cos theta.