If I gave you this graph and asked you to find the limit of this function as x approaches 2, you would take a look at the behavior of this function as x gets really, really close to 2 from either side. But what if I didn't give you this graph and you also didn't want to spend time creating a table of values? How could you find the limit then? Well, when we were finding limits using tables and graphs, we saw that for a lot of our functions, the limit just ended up being the exact same as the function value at that point. And this is actually going to be true of a lot of different functions, including polynomials and basic root functions. The limit will always just be the exact same as the function value, meaning that we can just take the value of x for which we're trying to find the limit and plug it into our function, evaluating it as we've done a million times before. Now here I'm going to show you exactly how to use this method of direct substitution to find limits and we'll work through some examples with a value of 6, so that's what my limit is. Now also looking a...

...t this graph, I can see that when x is equal to 2, my function value is just 6, the same exact thing as my limit. So if I didn't have this graph and I just took my function and evaluated it for x equals 2 doing a bit of algebra, I would still get my same answer of 6. Now in your textbooks, you're going to see a lot of different limit rules all written down in a big table. But all of those rules together just boil down to this. For many functions, including polynomials and basic roots, the limit will always just be the same as the function value, meaning that all we have to do is plug x in. Now, let's take a step back here for a second. Because when we were finding limits using tables and graphs, I told you that you needed to be really careful when finding your limits because you couldn't just assume that your limit would always be the same as your function value. But here I'm telling you that your limit is the same as your function value. So what's going on her...

...e? Well, the reality is that for many functions, including those that we've already seen here, the limit is going to be the same as your function value. So here we're going to focus specifically on those functions but there still are functions for which this won't be true. The limit won't be the same as your function value and I'm gonna show you how to find those limits coming up in the next couple of videos. But for now, let's focus on these functions that we can use direct substitution for. So let's work through these examples together. Now our first example here, we wanna find the limit of this function f(x)=6x3+3x2-x+5 as x approaches 2. Now I definitely don't have a graph here and I also don't wanna ma...

...ke a table of values for this long cubic function. But because this is just a polynomial, I know that I can use direct substitution here and just plug x equals 2 into my function to get my limit. So let's go ahead and do that. Now plugging 2 in here,y=6∙2+3∙2+2-2+...

...5. Now this 6 times 2 cubed will give me 48, and this 3 times 2 squared will give me 12, and then minus 2 plus 5. Now putting all of that together, I get a final answer here of 63 for the limit of this function as x approaches 2, and we're done here. Let's move on to our next example. Here, we're asked to find the limit of the square root of 7x+4x+16 as x appro...

...aches 0. Now again here, I just have a basic root function, so I know I can use direct substitution here and just evaluate my function at x equals 0. Now plugging 0 in here, I get the square root of 70+4∙0+ 16. Now these two terms will just end up being 0, so I am left here with the square root of 16, which we know is just 4. So the limit of this func...

...tion as x approaches 0 is 4. Now let's look at one final example here. Here we want to find the limit of x+3x+2 over x+1 as x approaches 0. Now the first thing that you might notice here is that this is a rational function. So if this is a rational function, can I still just plug x...

...equals 0 in? Well, for rational functions, the limit will still be the same as the function value, just as we've seen for our other functions here, as long as the denominator is not equal to 0. So let's double check here that if we plug 0 into our denominator, it's not going to make it 0. Now plugging my value of x equals 0 into my denominator here, that gives me 0+1, which is just equal to 1, which is definitely ...

...not 0. So we can proceed just plugging this x value in. So evaluating this function here, I get 0+3∙0+2 over that denominator value that we already found of 1. Now since these two terms will be 0, I end up in my numerator with just 2 and then in my denominator with 1, giving me a final answer of 2 for the limit of this rational function as x appro...

...aches 0. Now this specifically worked here because our denominator was not equal to 0. Now I'm going to show you what to do if your denominator is zero coming up in our next couple of videos. But now that we know to find limits using direct substitution, let's continue getting practice with this. Thanks for watching, and I'll see you in the next one.