 In this example, we're going to evaluate the limit as x approaches zero of the function e to the x minus one minus x over x squared Now this does have an indeterminate form It looks like zero over zero right because as x goes to zero we get a one right here One minus one is zero and then if you put them at zero you get a zero all over zero squared So this thing does have this indeterminate form zero over zero and while Lopital's rule is actually a quite a quite slick way Of computing this one. I want to present an alternative method that will make some sense Why we're doing it in just a moment and it's not because the point of this one's not actually to compute the limit But to kind of see why the limit is what it is right Lopital's rule It's doing to get so good at computing that sometimes think they forget why it actually works So I want to offer another perspective here and we're going to do this using the Maclaurin series for e to the x Which looks like the sum n equals zero to infinity of x to the n over Infactorial and if you write this an expanded form this is going to look like Well, I'll do it down below here. So this limit, right? We're taking the limit as x approach to zero now e to the x is equal to its Maclaurin series for any value of x Particularly when you go close to zero and as such you could replace e to the x with its Maclaurin series So you get one plus x plus x squared over two plus x cubed over six plus x to the fourth over 24 Etc. And then you subtracting from that one subtracting from that x over x squared right and so look at some cancellation there We have a minus one minus one. That's the first term in the the Maclaurin series You have a negative x minus x right here and so when you cancel those things from the numerator you end up with the limit as x approach to zero of x squared over two Plus x cubed over six plus x squared over 24, right? And then we get that should be a four right there. We get x to the fifth over 120 Continue on this all sits above x squared All right, so notice here when you subtract the one and subtract the x Right the thing on top it looks like it's a Maclaurin series that starts with a quadratic term And so when you look at this thing as x approaches zero, right? We've done many limits where x approaches infinity I'm like oh you look at the the biggest term on top and bottom But when x is approaching zero you actually kind of want to look at the smallest term on top and bottom which The dominant term as you approach x equals zero be x squared on top and x squared on bottom So with this perspective using the Maclaurin series We actually see that as x approaches zero this function is a balanced rational function x squared on top and bottom And to help us compute this thing, right? We're gonna factor out an x squared from the top You factor out an x squared that leaves behind one half plus x over six Plus x squared over 24 plus x cubed over 120. That's enough go on We have an x square on the bottom this factor out x squared on top cancels with the one on the bottom And so we just get one half plus x over six plus x square over that now as x goes to zero all of these terms that have Multals of x to it will cancel and we see that this thing is equal to one half And so that gives us the correct calculation the limit here is equal to one half Had we tried to do this using L'Hopital's rule we would have done instead with the top We would have gotten e to the x minus one over 2x as x goes to zero And as you plug in zero you would still get one minus one over zero which is still in determined forms You have to do L'Hopital's rule twice In which case then you get the limit as x approaches zero You're gonna get e to the x over two and now as x approaches zero you're gonna get one half So I definitely say that the L'Hopital approach gives you a much cleaner approach, right? It's a shorter calculation I agree with you on that, but one thing you don't see from L'Hopital's rule is that e to the x Minus one minus x is essentially a quadratic power series That is this is a balanced rational function, which you don't necessarily see when you do L'Hopital's rule And so when you start realizing that function these continuous functions like e to the x are these infinite polynomials You start to realize that in some ways they behave like polynomials And so this actually is a rational function and the way we handle this would have been exactly like we handled any other rational function And so that's a quite interesting Observation right there and we'll see some other things like this in future videos as well