in this video i'm going to give you a quick overview of l'hopital's rule and when and why we might need to use it okay so l'hopital's rule helps us solve limits questions so this is in topic five calculus and it helps us find these limits where the answers might not be as simple as they as they might have been if it was just a simple limit so we might need to use a rule but i'll get to that in a little bit okay i have three examples let's just try and solve them without even knowing what l'hopital's rule is okay well the limit as x approaches 0 of sine of x over x let's just try our luck let's substitute in zero for x and we will get sine of zero over zero and sine of zero is in fact zero so we get zero over zero okay i'll come back to that that's a little bit of a strange answer let's try example two we have the limit as x approaches infinity of the natural log of x over x okay let's try our luck again let's substitute in infinity in for x well the natural log of infinity is just infinity we will have infinity on the bottom infinity over infinity that's another strange answer okay let's try the third one maybe this will be easier the limit as x approaches infinity of x multiplied by e to the negative x well this will give us infinity multiplied by e to the power of negative infinity and e to the power of negative infinity is an incredibly small number it's essentially a zero because of our negative power here so we are going to get infinity times zero okay we didn't have much luck there with our three examples and this is why l'hopital's rule is very helpful and very important okay so the three answers that we got there if we can have a look up here these are the three indeterminate forms that we need to deal with now what that means is that we don't actually have an answer for our limits here so what we can do is use l'hopital's rule to help us solve these limits now basically all l'hopital's rule is is that if we can't find the limit or if the limit is in one of these three indeterminate forms here zero over zero infinity over infinity now this could be plus or minus infinity over plus or minus infinity or infinity times 0 what we can do is take the derivative of the numerator and the denominator and we can do this multiple times if we need to but then we can try and solve these limits so let's go to the first example if we can't solve this limit straight away just by direct substitution let's find the derivative of the numerator and the derivative of the denominator and the derivative with respect to x of sine of x will be equal to the cosine of x and the derivative of x is just equal to 1. now once we have found the derivative of the numerator and the denominator let's try our luck again can we substitute in x to be equal to 0 and get an answer that's not in one of these forms and we do we get the cosine of 0 is 1 and we have a 1 on the bottom and 1 over 1 that is that is a number that we can actually confirm as our limit the answer is equal to 1. so we have shown that the limit as x approaches 0 of sine of x over x is in fact 1 and we used l'hopital's rule to help us do it okay so l'hopital's rule means we can just take the derivative on the of the numerator and the denominator and then try our limit again okay let's try the second one well using l'hopital's rule let's take the derivative of the numerator and the denominator now the derivative of the natural log of x is one on x and the derivative of x is just one so now we have the limit as x approaches infinity of one over x over one let's substitute infinity we'll get one over infinity over one now one over infinity is in fact zero and 0 over 1 that is an answer this is not one of our three indeterminant forms because 0 over 1 is in fact just 0. so this is our answer and once again we found it this is the answer for our limit we found it using l'hopital's rule where we take the derivative of the of both our numerator and our denominator okay let's try our third example let's now take the the derivative once again but before we do that we can hopefully recognize that we can express this as a fraction because that's what we want to do here have a function over a function and we can do that by making this e to the negative x e to the x on the denominator okay using our indices laws indices rules and laws okay so let's take the derivative of the numerator and the denominator this is l'hopital's rule and the derivative of x becomes one the derivative of e to the x with respect to x is just e to the x and now let's try and substitute in infinity and we will get one over e to the power of infinity and this is in fact infinity 1 over infinity can be determined it's not one of our three indeterminant forms one over infinity is just equal to zero okay so they are the three examples that we've gone through here we've actually found the limits of each of them and just to conclude here what l'hopital's rule is actually telling us it's that if we have the limit as x approaches some value here i'm just going to say a of a function over a function if we can't find this straight away just by doing a direct substitution we know that as assuming that both of these functions are differentiatable we can actually find the derivative we can just we can say that the limit here is equal to the limit as x approaches that same value of the derivative over the derivative now in some challenging questions even if you do use l'hopital's rule once and find the first derivative and the first derivative we still might get an indeterminant form so that's where you might need to do l'hopital's rule twice or three times you can do it multiple times here until you can substitute in your value for a that does actually give you an answer that's not one of these three okay so in conclusion this is l'hopital's rule a very very interesting story about uh l'hopital's rule it was actually discovered by a different mathematician uh johann bernoulli but l'hopital was a very very rich man and he bought the rights to this rule and now it's named after him so interesting story there okay good luck