>> So, it's a limit as x approaches zero from the right and this is a one sided limit, which is another new ones here that we're showing with this particular example. But let's see what do we have, let's see the sign of zero would be zero and x is approaching zero, so this zero to the zero power. So, this is the indeterminate form zero to the zero. We can't apply L'Hopital's rule. So, we jump in to this because that exponent now we might be tipped off that we're going to use this logarithm technique. So, we set this expression equal to y and then we apply natural log on both sides. And now this is the limit of the natural log and now the x becomes the coefficient. And now we use that little technique of forming a fraction here by kicking this down to the denominator. You see we think of this as x to the 1 and then in the denominator it's x to the negative 1 or in the denominator it's 1 over x, alright so that's the idea. Now, I guess we could try to evaluate limit at this point, but it turns out to be an indeterminate form and it is a form where we can actually use L'Hopital's rule, so we come back up here. Now, before we go up there just wait just a second. Now, remember this is the natural log, the derivative of the natural log of sine x, the derivative of l, n, u is u prime over u. So, it's the derivative of sine x over sine x, the derivative of sine x is cosine x. So, this numerator becomes cosine x over sine x and in the denominator we use the same techniques that we did on the previous problem to get minus x to the negative 2. Alright, now this is, let's see cosine x over sine x, well that's just cotangent x. And this x to the negative 2 can kick up into the numerator to become minus x squared cotangent x. Wow, we can't use L'Hopital's rule here what can we do? Well, we can change this into a fractional form by thinking of cotangent as one over tangent. So, now we have minus x squared over tangent x and we think about the limit as x approaches zero from the right. Well, as x approaches zero from the right this numerator is approaching zero. And as x approaches zero from the right this is also tangent of x is also approaching zero, so we have zero over zero. Now, we can apply L'Hopital's rule. So, we have this becomes minus 2x for the numerator, the derivative of tangent x secant squared x and now we can evaluate our limit. The limit as x approaches zero from the right for the numerator is zero. The limit as x approaches zero from the right of let's see secant x, remember now, think graphically real quick, it's the reciprocal of cosine and cosine, the cosine of zero is 1 and the reciprocal is 1, so this is secant -- I'm getting jammed up here. Secant of zero is 1 and 1 squared is 1, so we have zero over 1 which is zero. Now, the natural log of y is zero implies that y has to be 1.