 So when applying L'Hopital's rule, it's important to remember L'Hopital's rule only replaces one limit with another limit. This means that all limit rules previously known remain in place. So I suppose f is continuous with continuous derivative at x equals 3 and our limit is x approaches 3 of f of x is 12 and our limit is x approaches 3 of f prime of x is 8. Let's find the limit of this composition. So we should always check those admission forms. As x goes to 3, our numerator goes to and our denominator goes to and since numerator and denominator both go to 0 as x goes to 3, we can enter the L'Hopital. So we'll differentiate numerator and differentiate denominator and let's take a look. As x goes to 3, our numerator goes to and our denominator goes to so we could apply L'Hopital's rule again but this would require finding the second derivative of f which we don't have. But we can always do some algebra and here the thing to notice is that the numerator has a factor of f prime of x and so does the denominator and we can simplify and numerator and denominator still go to 0 so we can apply L'Hopital's rule again and this time our denominator is not going to be 0 so we might be able to find the limit through direct computation or let's take a look at another limit. So again always check the admissions form and we can apply L'Hopital's rule. However we need an extended L'Hopital stay because our numerator goes to 0 and our denominator also goes to 0 and again differentiating again requires finding the second derivative which we don't have but remember all of our limit rules still apply and the useful one here is that provided the limit exists the limit of a product is the product of the limits and so we note that this thing that we're trying to find the limit of can be written as a product now since L'Hopital's works on indeterminance of the form 0 divided by 0 we might take a look at our numerator and split it into a part that's going to 0 and the leftovers and likewise our denominator will be a part that's going to 0 and the leftovers so we can rewrite this quotient and limits all around since the limit as x goes to 3 of g' of x isn't 0 we can find the second limit directly the other limit requires a return to the L'Hopital and again numerator and denominator are both going to 0 but we can split numerator and denominator into factors that go to 0 and everything else and again the second limit exists and we can find it directly and we can find the first limit by another L'Hopital visit and since g' of x isn't 0 we can evaluate this limit directly our numerator is going to be 0 and everything else is going to be some real number and so our limit is going to be the product of 0 times well it doesn't really matter what