 Hello and welcome to this session. In this session we discuss the final question which is evaluate the limit. Limit x tends to 0, 6 to the power x minus 3 to the power x minus 2 to the power x plus 1. This will upon x square. Before moving on to the solution, let's discuss one special limit which says that limit x tends to 0, a to the power x minus 1 upon x is equal to log a to the base e. This is the key idea that we would use in this question. Let's proceed with the solution now. We are supposed to find the limit. Limit x times to 0, 6 to the power x minus 3 to the power x minus 2 to the power x plus 1. And this will upon x square. Now we can write this whole to the power x minus 3 to the power x minus 2 to the power x plus 1. And this will upon x. We have limit x tends to 0 into 2 to the power x minus 3 to the power x minus 2 to the power x plus 1. And this will upon x square. Then we have limit x tends to 0. Now we will return these two terms. So we have 3 to the power x into minus 1 the whole. And this will upon x square. Or you can say we have limit x tends to 0 3 to the power x minus 1 the whole into 2 to the power x minus 1 the whole. This will upon x square which could also be written as limit x tends to 0 upon x. In this idea we know that limit x tends to 0 a to the power x minus 1 upon x is equal to non a to the base e. So using this result we can have that this is equal to log 3 to the base e into log 2 to the base is equal to log 3 into log 2. That is the required limit which is limit x tends to 0 3 to the power x minus 2 to the power x plus 1. This will upon x square is equal to log 3 into log 2. This is the solution of this question.