Hello wonderful mathematics people, this is Anna Cox from Kellogg community college L'Hopital's rule, suppose that f(a) equal g(a) which equals zero that f prime of a and g prime of a exist and that g prime of a does not equal zero then the limit as x approaches a of the function of f(x) divided by g(x) is really just equal to f prime of a divided by g prime of a looking at the proof, we see that f prime of a divided by g prime a our official definition of a derivative is just the limit as x goes to a of f(x) minus f(a) over x minus a and that's all divided by the official definition of a derivative for the g prime of a which is, limit as x goes to a of g(x) minus g(a) over x minus a remember that's really just the slope as x and a are getting closer and closer and closer to each other so if we simplify that up, we get the limit as x goes to a of f(x) minus f(a) over g(x) minus g(a) when limit as x goes to a, in the original we were told that f(a) and g(a) are gonna be zero so we get f(x) minus zero divided g(x) minus zero or the limit as x goes to a of f(x) divided by g(x) Cauchy's mean value theorem, suppose functions f and g are continuous on a closed and bounded interval [a,b] and differentiable throughout the open interval (a,b) and suppose g prime of x doesn't equal zero throughout (a,b) Then there exists a number c in the interval at which f prime c divided by g prime c is really just the slope of the minus f(b) minus f(a) over g(b) minus g(a) So we're gonna use the mean value theorem. First we're gonna show that g(a) doesn't equal g(b). If they did equal each other, then the mean value theorem would give the that g prime c would equal zero for some c between a and b, which can't happen because the original condition said that g prime of x could never equal zero within the interval next we would use the mean value theorem on f(x). We know that f(x) equals f(x) minus f(a) minus f(b) minus f(a) g(b) minus g(a), g(x) minus g(a). This is really really just the equation of a line. If you thought about two points where your points were g(x) g(x) comma f(x) and g(a) f(a) Now this might seem a little difficult because g and f are both functions now, but they're still really points. We have our x value somewhere on the g. We have our y-value somewhere on the f we have g(a) f(a). So the point slope form, the difference of the y's equals the slope, which is the difference of the y's over the difference of the x's times the difference of the x's now we're gonna say what happens when the slope the slope when x is the value b. So if we're looking for the slope when x is b. We're gonna just replace b right here for the slope portion so f(b) minus f(a) divided by g(b) minus g(a) now what we're gonna look at as we're gonna look at this new function, this capital f(x). We know it's continuous and differentiable where f and g are continuous and differentiable. So we're gonna look at specific points now the point b. If we substitute x for the point b f(b) minus f(a) minus, this was our given slope at the point b remember g(b) minus g(a) the g(b) minus g(a)'s are gonna cancel and we'd end up with f(b) minus f(b) if we distribute that negative and negative f(a) minus a negative f(a), which is gonna also cancel so we just showed that f(b) really does equal zero next if we look at the other endpoint a we'd end up with f(a) minus f(a) because remember I am putting it in for the x. f(a) minus f(a) which we know is zero. Minus this is the given slope at when x is b. Here is g(a) minus g(a) cause we're sticking that a in for this x. That's zero anything times zero is zero so f(a) is also zero. There's a number c then between a and b for which the first derivative of this new function equals zero so if we have our a and our b point, we connect them we can see that there's got to be some point somewhere along the way where that would equal zero the equation then becomes f prime c equals f prime of c minus f(b) minus f(a) divided by g(b) minus g(a) times g prime of c equaling zero or a different way to think of that is the f prime c divided by g prime g c as f(b) minus f(a) over g(b) minus g(a). Just taking this equation here and truly solving getting the primes on one side by themselves. So we're going to add this over and then we're gonna divide so f(x) equals f(x) minus f(a) minus f(b) minus f(a) over g(b) minus g(a), g(x) minus g(a) or f prime x is really just f prime x minus the slope times the g prime x Let's look at an example. If we have limit as x goes to zero of sine 5x over x if we sticks zero in there, we know sine of five times zero is zero, and we know that x is zero. So this is gonna be an indeterminate form which is zero over zero. So we're gonna take the derivative of the top over the derivative the bottom derivative of the top is gonna be five cosine 5x. Derivative of the bottom is one Now when we stick zero in for x, we know that that the cosine of five times zero is really just one So this, according to L'Hopital's rule is just gonna be equivalent to five indeterminate forms can come in different looks it could be zero over zero, infinity over infinity, infinity times zero, infinity minus infinity, just to name just a few so if we look at this next example, we have limit as x goes to pi halves. One minus sine x divided by one plus cosine 2x is zero over zero when I stick in the pi halves so, we're gonna take the derivative of the top and then the derivative of the bottom so if we look at the derivative of the top we get negative cosine x, the derivative of the bottom, negative two sine 2x. If I stick in pi halves again, I'd get zero over zero. So, I'm going to do L'Hopital's rule again. Sine of x going to pi halves. Sorry, limit as x goes to pi halves of sine x over negative four cosine 2x now when I stick and pi halves, I don't get zero over zero, we get one fourth. So we know that the limit of that original function is equivalent to one fourth looking at another one. Limit as x goes to one from the right hand side of one divided by x minus one, minus one over ln(x) if we stick in one from the right hand side we can see that we get one divided by something really really small which is really really big and then minus one divided by something really really small. So we basically are getting infinity minus infinity which is an indeterminate form Now we can't use L'Hopital's rule on it looking like that, so we need to get it to be a fraction, something divided by something so we're gonna get a common denominator here ln(x) minus x minus one divided by x minus one times ln(x) those are quantities for the x minus one portion. Now if I put one going from the right hand side, I realize I get zero divided by zero which is still an indeterminate form so now we're gonna take L'Hopital's rule, and we're gonna get one over minus one divided, on the bottom we're gonna use the product rule so we're gonna get ln(x) plus x minus one divided by x. That's gonna equal, if we simplify it up by multiplying by x over x it is sometimes easier to see So if I just multiply the x over x, we'd get one minus x divided by xln(x) plus x minus one now if I stick one in from the right again, I'm gonna get zero divided by zero so we're gonna use L'Hopital's rule another time. Now we get the limit as x approaches one from the right of negative one divided by ln(x) plus x over x plus one, using product rule once again so when we stick in one from the right hand side this time we can see that our solutions gonna be negative one half if the limit as x goes to a of the natural log of some function equals l, then we know that the limit as x goes to a of the actual function is just equivalent to the limit as x goes to a of e to the ln(f(x)) or e to the l an example of this would be limit of x goes to infinity of one plus 2x all to the one over 2ln(x) so when we look at this example we're gonna start by letting our f(x) be that one plus 2x to the one over 2ln(x) thus the natural log of that function will take the natural log on one side and the natural log of the other and then we're gonna use the product rule, so we're gonna bring down that or not the product rule, the exponential rule. We're gonna bring down the exponent one over the power rule one divided by 2ln(x). And I'm gonna write that as a complex fraction. So I'm gonna have limit as x goes to infinity of the natural log of f(x) equaling the limit as x goes to infinity of the natural log of one plus 2x all over 2ln(x) when I put in the natural log of one plus two times a really really big number that's basically going to be a really really big number we're gonna get an indeterminate form of infinity over infinity So we're gonna use L'Hopitals rule. We're gonna find the derivative of the top over the derivative the bottom and simplify which is gonna give us limit as x goes to infinity of x divided by one plus 2x that is now an indeterminate form again of infinity over infinity. So we're gonna do L'Hopital's rule again and we're gonna get one half so the limit of this is really one half. Therefore when we look at the limit as x goes to infinity of the original function, we know that that's really equal to limit as x goes to infinity of e to the ln of the original function while ln of the original function we just proved had a limit of one half. So our final answer is e to the one half thank you and have a wonderful day