Welcome back, everyone. So in recent videos, we talked about how to find limits at infinity. And recall when solving these types of problems, we were given either a table or a graph, and we needed to figure out how the function behaved as \( x \) got infinitely big or infinitely small. Now in this video, we'll see situations where we don't have a graph or table to help us, but we simply just have these types of rational functions that look a bit complicated. Now if this sounds scary, don't sweat it, because it turns out this process for solving limits purely algebraically with these types of rational functions is actually quite straightforward, and we're even going to learn a shortcut, which allows us to solve these problems super fast. So without further ado, let's jump right into some examples of this.

Now we'll start over here at example a, where we're given this rational function and that we have the limit as \( x \) is approaching infinity. Now whenever you have to solve these types of limits for these rational functions, what you need to do is take each term and divide it by the highest power of \( x \). So looking at this example right here, I can see that the highest power of \( x \) that we have is this \( x^3 \). So what that means is I need to take every term on top and divide it by \( x^3 \), and I need to take every term on bottom and divide it by \( x^3 \). This is the strategy. So, I'll rewrite this limit down here. We have the limit as \( x \) is approaching infinity, and what we're going to have is this \( \frac{1}{x^3} \) distributed into every single term. So, we'll have \( \frac{3x^2}{x^3} \), which turns out to just be \( \frac{3}{x} \). Then we're going to have \( \frac{4x}{x^3} \), which is \( \frac{4}{x^2} \), and we're going to have \( -\frac{1}{x^3} \), and then this whole thing is going to be divided by \( \frac{x^3}{x^3} \), which is 1, plus \( \frac{27}{x^3} \). Now recall that whenever you have \( x \)'s on the bottom of the fraction, and you don't have any \( x \)'s on top, if \( x \) is approaching infinity, then the whole thing is just going to go to 0 because the bottom is infinitely growing where the top is not. So in this situation, what we can see here is that this is all just gonna go to 0. That's gonna go to 0, that's gonna go to 0, and that's gonna go to 0. So all we're going to end up with is 0 on top and just one on the bottom, and \( \frac{0}{1} \) is 0. So that right there is the solution to this limit. So this is how you can solve these types of problems where you have \( x \) approaching infinity and you have these complicated rational functions.

Now you may have noticed that this process was a bit on the te