 Welcome back to our lecture series Math 1210, Calculus I for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Miseltine. In the last video for lecture 34 in our series, we introduced a very powerful technique for computing limits known as L'Hopital's Rule. We learned that L'Hopital's Rule applies only when we have a limit of the form 0 over 0 or infinity over infinity. If we ever have one of these indeterminate forms, what we can do is we can take the derivative of the numerator and the denominator of the function in the limit, and then reevaluate the limit. If this limit turns out to be a number, that was the original limit. If it doesn't exist, it turns out that was the original limit as well. L'Hopital's Rule is a great technique to use when you get 0 over 0 or infinity over infinity. In this first video for lecture 35, I wanted to do some more exercises practicing L'Hopital's technique here. Imagine we have the limit as x goes to infinity of the natural log of x over the cube root of x. Now, notice that if we just start off by plugging in x as infinity, we're going to get the natural log of infinity, so to speak, over the cube root of infinity. This will just become infinity over infinity. We have this indeterminate form, but it's important to check this because L'Hopital's Rule doesn't apply if you don't have an indeterminate form. It only applies when you have 0 over 0 or infinity over infinity. What we're going to do is, since we have the form infinity over infinity, we can apply L'Hopital's Rule for which we're still going to take the limit as x approaches infinity, but now we're going to take the derivative of the natural log, divide that by the derivative of x to one-third power. Since I am going to take derivatives, I want to switch from the cube root to the one-third power there. In which case, as we've seen many times before, the derivative of the natural log is 1 over x. The derivative of x to the one-third by the usual power rule is going to be one-third x to the negative two-thirds power. We take the limit as x approaches infinity. Because we have x to the negative one-third power, we can actually bring it above. Since we're dividing by 3, we can actually bring it above. Since we're dividing by x, it actually comes down below. This is just a topsy-turvy mess right now. But at the end, it's going to become 3 times x to the two-thirds power over x, if we simplify that thing. Notice we now have powers of x on top and bottom. We have x to the two-thirds on top, x to the one-third on the bottom. If we simplify the exponents, this would then give us the expression that we're taking the limit here of 3 over x to the one-third power. Take the limit as x approaches infinity. Now, in this situation, as we allow x to go towards infinity, we're going to end up with 3 over the cube root of infinity again. But this just becomes 3 over infinity, which this is going to squish down to be zero. So by Lopital's rule, since this limit is zero, that implies the original limit was zero as well. Let's do one more example. Let's take the limit as x approaches zero of tangent of x minus x all over x cubed. Now, notice if we plug in x equals zero, we're going to get tangent of zero minus zero all over zero cubed. Getting a zero in the denominator is concerning because dividing by zero is typically a problem, but you'll also notice that since tangent of zero is itself zero, it's the internment form zero over zero. So I can just hear Princess Leia telling me right now that there's still hope, right? The limit could still be something, right? Division by zero is only a problem if you're not also multiplying by zero. Since we have a zero over zero, we have this clash of titans going on here. Who's more powerful? Multiplication by zero or division by zero. What it means to us is that we can apply Lopital's rule and so we're going to take the derivative of the top and the bottom. So we take the limit as x approaches zero. We take the derivative of tangent of x minus x and then we'll take the derivative of the denominator separately. We're not going to use the quotient rule. We're taking the derivative of the top separately from the derivative of the bottom. So the derivative of tangent is secant squared. The derivative of x is going to be one here. It's negative one. By the power rule in the denominator, we're going to get the derivative is going to be three x squared. If we plug in x equals zero at this moment, we see we're going to get secant squared of zero minus one over three times zero squared. The denominator again is going to be zero. But what happens in the numerator? Well, we have to compute secant of zero in this situation, which recall of course that secant of, secant zero is just one over cosine of zero squared, which cosine is going to be one. One over one is still one you square that. You're going to get this one over one in the numerator. And so this still looks like zero over zero. Okay, what does that mean for us? It just means we're going to do L'Hopital's rule one more time. We keep on doing it as long as the function has this indeterminate form zero over zero or infinity over infinity. So we're going to take the limit still as x approaches zero here. We now have to take the derivative of secant squared x minus one. And then the denominator, we're going to take the derivative of three x squared. In the numerator, when we take the derivative, well, the derivative of the negative one is going to be, since it's constant, it's going to be zero. So using the chain rule for secant, we do the outer derivative first. We're going to get a two secant theta. That's our outer derivative. Then we have to times that by the inner derivative, which is going to be a secant tangent. And that's going to be our numerator. The denominator, we're going to get a six x. And so let's see what happens going forward here. Well, I noticed there's a factor of two common to top and bottom. So there's a factor of two. The two goes into six, three times. Then we're going to plug in x equals zero. For which case we're going to get secant of zero. There's actually a secant squared, right? So secant is zero squared. Then you're going to get tangent of zero all over three times zero. What do we have here? We have, holy cow, zero over zero again, right? This is going to look like a one squared times zero over three times zero. This is just zero over zero. Well, guess what? We can do it again. Do it as many times as necessary to get this thing done. So we do L'Hopital's rule one more time. Take the limit as x approaches zero. We're taking this time the derivative of secant squared x times tangent of x. This it's above the three x, which we're going to take the derivative of here. For which we can see that what's going on here is that every time we take the derivative of the denominator, we lose a power of x. So in this situation, the derivative of three x is going to be a three. So in terms of the denominator, there's no longer going to be a zero in the denominator here, okay? For which case then, if we take the derivative of the top by the product rule, we're going to take the derivative of secant squared, which as we saw before is a two secant squared tangent. But then as there's still a tangent sticking around, we're going to get a tangent squared. But then we have to take, for the product when we take the second part, we take the derivative of tangent, which the derivative of tangent is a secant squared. So we're actually going to get a secant to the fourth power there. And so then as we plug in x equals zero, we end up with, we have our resolution finally, we're going to get, for the first part, if you plug in zero into a tangent, you're going to get a zero. So that disappears. We're going to end up with a secant to the fourth zero over three. So we're going to get one third is the limit. So holy cow, we did it. So in the end, we eventually found the limit to be one third. We did use L'Hopital's rule like three times. Now it turns out L'Hopital's rule is not the only technique that one could use. There are situations that alternative strategies that could be used to avoid L'Hopital's rule. For example, right, let's take, let's kind of go back in this situation right here. What if we, if we wanted to, we could have actually avoided L'Hopital's rule. We had this limit as x approaches zero of secant squared x tangent x over three x. With a little bit of manipulation, we could actually factor this thing, taking the limit as x approaches zero of, we're going to get a secant squared x over three times cosine of x. Where did the cosine come from? Well, we also have a sine of x over x. Basically what we're doing here is we're recognizing that tangent is sine over cosine. We're going to separate the sine from the cosine because when x goes to zero, secant and cosine both go to one. So no big deal there. What about sine x over x? Hey, that's also equal to one. We've learned that previously. The limit as x approaches zero of sine of x over x is just one. And so we see this turns out to be a one times one over three times one, which is one third as well. So we could use other techniques combined with L'Hopital's rule. And in fact, if you are curious about that, take the limit as x approaches zero of sine of x over x. Even if you ever forgot that this is equal to one, you could actually use L'Hopital's rule because this thing has the indeterminate form zero over zero. When you take the derivative, you end up with the limit of cosine of x over one, which then is going to give you one over one, which is equal to one. Now, we originally approved this statement using the squeeze theorem honestly. And you can use L'Hopital's rule, but by remembering some of these limits we've already computed like sine x over x going to one, you can actually avoid some of the subsequent calculations of derivatives. So I don't want you to throw everything out that we've learned previously how to compute limits. No, we should just think of L'Hopital's rule, not as like one, you know, the magic tool, the silver bullet that will solve all problems. No, this is just one more tool in our box. We can use the other tools we've already developed to compute limits of these indeterminate forms.