Finding Limits Algebraically - Online Tutor, Practice Problems & Exam Prep

Finding Limits by Direct Substitution

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. The value of 6, so that's what my limit is. Now also looking at 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 here? Well, the reality is 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 want to find the limit of this function 6x^3 + 3x^2 - x + 5 as x approaches 2. Now I definitely don't have a graph here, and I also don't want to make 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, I get 6 times 2 cubed plus 3 times 2 squared minus 2 plus 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 7 x squared plus 4 x plus 16 as x approaches 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 7 times 0 squared plus 4 times 0 plus 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 function as x approaches 0 is 4. Now let's look at one final example here.

Here we want to find the limit of x^2 + 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 in just plugging this x value in. So evaluating this function here, I get 0 squared plus 3 times 0 plus 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 approaches 0. Now this specifically worked here because our denominator was not equal to 0.

Now I'm gonna show you what to do if your denominator is 0 coming up in our next couple of videos. But now that we know how 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.

Finding Limits by Direct Substitution Example 1

Hey, everyone. We just learned that for many functions, our limit will actually just be the same as our function value. So let's keep this in mind as we work through these examples. Now this first example asks us to find the limit of 7 as x approaches 2. Now seeing this might be a bit confusing because how can I find the limit of 7 as x approaches 2?

There's nowhere to plug that value into this function. But because this is a constant function, I don't have to do any plugging in here because my function value the whole time is 7. As x approaches 2, as x approaches 5, as x approaches a million, it's always going to be 7. So here, the limit of 7 as x approaches 2 is just 7. Now let's look at our next example.

Here, we're asked to find the limit of 2x2+3x as x approaches negative 1. So here we just want to plug negative 1 into our polynomial in order to get this limit. So I'm going to take 2*(-1)2+3*(-1), and this is going to be my answer here. Now actually doing this algebra, 2*(-1)2 is going to give me 2 and then 3*(-1) is negative 3. So here I have 2 minus 3 which will give me negative one for the limit of this function as x approaches negative one.

Now let's look at our final example here and here we have a root, the limit of the square root of 3x2-2 as x approaches 3. So with this basic root we want to go ahead and plug in our x value of 3. So doing that here I get the square root of 3*32-2. Now 3*32-2 will end up giving me 25 so this is just the square root of 25 or fully simplified taking that square root, 5, as the limit of this function as x approaches 3. Feel free to let me know if you have any questions here; I'll see you in the next video.

Find the limit.

$\lim_{x\rarr0}x^3+5x^2-7x+3$

• A. $0$
• B. $2$
• C. $3$
• D. $5$

Find the limit.

$\lim_{x\rarr2}\sqrt{x^2+5}$

• A. $0$
• B. $2$
• C. $3$
• D. $5$

Find the limit.

$\lim_{x\rarr3}\frac{x^2+2x-3}{x-2}$

• A. $0$
• B. $3$
• C. $12$
• D. DNE

Limits of Rational Functions: Denominator = 0

We just saw that for functions like polynomials and basic roots, our limit is always the same as our function value. And we also saw that this was true for rational expressions. The limit is the same as the function value as long as our denominator is not equal to 0. But from working with rational functions in the past, we know that this won't always happen. Sometimes our denominator will end up being 0.

So, how do we find the limit in these cases? Well, we're going to have to do a couple of extra steps, but I'm going to walk you through exactly what this process looks like. So let's go ahead and jump in. Now with our rational functions where our denominator is not equal to 0, like this one here, finding the limit of x2+3x+2x+1 as x approaches 0. Since that 0 does not make our denominator 0, we can just plug it in everywhere and evaluate our function there and end up with our final answer.

But what if we want to take the limit of this same function, x2+3x+2x+1, but this time as x approaches negative 1? Well, the first thing we want to do here is just plug that value into our denominator. So if I plug that negative one into my denominator, I get negative one plus 1. Now negative one plus 1 is 0. And because my denominator is 0 here, I can't just plug this negative one in and evaluate my function there.

I have to do something else. So whenever we're faced with this, whenever our denominator is 0, the first thing that we want to do is factor the top and the bottom of our rational function. So let's go ahead and do that here. Let's go ahead and factor this numerator x2+3x+2. Now remember when factoring, we want to look for 2 numbers that are multiplying to 2 and to 3.

So that ends up factoring to x+1 times x+1, so it remains the same x+1. And I have fully factored this rational function. So from here, what we want to do is cancel out our common factor. We are always going to end up with a common factor and here we have x+1 in the top and x+1 in the bottom so I can go ahead and cancel those out. Now I'm just left to find the limit of x+2.

So here I can proceed in evaluating my function as normal, just plugging this negative one in. So that ends up giving me negative 1 plus 2, giving me my final answer of 1 for the limit of this rational function as x approaches negative one. So when you're faced with finding the limit of a rational function the first thing you want to do is check if that x value is going to make your denominator 0. If it's not going to make your denominator 0 you can just evaluate your function plugging that x value in. But if it does make your denominator 0 you want to start first by factoring the top and the bottom of your rational function.

Then you can go ahead and cancel out the common factor that you end up with. And then finally, you can proceed in evaluating as normal. So with this in mind, let's work through a couple more examples here. Now looking at this first example, we want to find the limit of this function x2+2x-15x-3 as x approaches 3. Now remember the first thing we want to do here is check if this value will make our denominator 0.

If I plug this into my denominator, I get 3 minus 3, which definitely is 0, which means that I need to go ahead and start here by factoring. Now we're still finding the limit as x approaches 3, but having factored this numerator, remember that I'm looking for 2 values that multiply to negative 15 and add to positive 2. Now something that you may notice here is that because we are always going to end up with a common factor, I already know that one of my factors is just going to be this x minus 3. So that can make it sort of easier to factor and be kind of a shortcut for you here. Now I have this x minus 3 factor.

And the other factor here is going to be x+5 for that numerator. Now in my denominator that's staying the same, it's still x minus 3. And my second step here is to cancel that common factor, which of course here is x minus 3. So that will all cancel out and now I'm just left to find the limit as x approaches 3 of x+5 because that's all I'm left with here. Now I can evaluate as normal just plugging in this x value of 3 to give me 3+5 for my final answer of 8 for the limit of this rational function as x approaches 3.

Now let's look at one final example here. Here, we want to find the limit of x+2x2-x-6 as x approaches negative 2. So again, the first thing we want to do here is just plug in that negative 2 to our denominator and check if it's going to make it 0. Now doing that here, I get negative 2 squared minus negative 2 minus 6, which ends up giving me 4 +2 minus 6 which is equal to 0. So from here, we can't just plug that negative 2 in everywhere, we need to go ahead and factor.

So my numerator here actually cannot be factored. It's just going to remain that x+2. So keeping my numerator the same there but factoring that denominator x2 x minus 6, I'm looking for two numbers that multiply to negative 6 and add to negative 1. But remember, I can kind of cheat a little bit here, take a shortcut, because I know that one of my factors is going to be x plus 2 because it will end up canceling out with that x plus 2 in the numerator. Now my other factor here is going to be x minus 3 in order to multiply back to what my original denominator was.

Now from here, I have my common factor, that x+2, in the top and the bottom. That's going to end up canceling out here. And I am left to find the limit still as x approaches negative 2, but all I'm left with here is 1 over x minus 3. Now this negative 2 will not make this denominator 0, so I can go ahead and just plug this in as normal, giving me 1 over negative 2 minus 3, which ends up giving me a final answer of negative one-fifth. So this is the limit of this rational function as x approaches negative 2.

So now that we know how to find the limit of any rational function, whether a denominator becomes 0 or not, let's continue getting some practice. Thanks for watching. Let me know if you have any questions.

Limits of Rational Functions: Denominator = 0 Example 1

Hey, everyone. We just learned how to find the limit of a rational function whether the denominator is 0 or not. So let's go ahead and work through these examples here. Now the first limit that I'm asked to find is the limit of x cubed minus 2x squared plus x over x minus 1 as x approaches 1. Now looking at what x is approaching here, 1, if I plug that into my denominator, I get 1 minus 1, which gives me 0.

Now because my denominator would be made 0 by that x value of 1, that means that I need to go ahead and factor this rational function. So let's go ahead and do that here. Now in order to find this limit, I need to fully factor my function.

Considering the limit as x approaches 1, I am going to factor that numerator. The first thing that I see is that each of these terms has an x in it, so I can go ahead and factor that x out. Now that will give me x times (x squared minus 2x plus 1).

Since my denominator cannot be factored any further, that remains as x minus 1. I can continue to factor this further because this polynomial here ends up being a perfect square. I can further factor that as x and then it becomes (x minus 1)². Now this is all still divided by that denominator, x minus 1, and I am still trying to find the limit as x approaches 1.

Here I see that I can cancel some things out because I have this x minus 1 on the bottom and an x minus 1 in the numerator. I actually have two of them on the top. So this no longer squared in my numerator. What I'm left with here is to find the limit as x approaches 1 of x times (x minus 1), having canceled only one of those x minus 1 terms on the top with my x minus 1 on the bottom. From here, I can evaluate this by plugging in 1. This will give me 1 times (1 minus 1), which is 0.

So this just ends up being 0, the limit of this function as x approaches 1. Now looking at our next example, we want to find the limit of the same function, but this time as x approaches 2. Now if I were to plug 2 into my denominator here, that gives me 2 minus 1, which is just equal to 1, not 0. So I can actually just go ahead and plug 2 into my function here and directly substitute that in, not having to worry about factoring or canceling anything out because my denominator is not 0.

So plugging in 2 here, my limit is going to be 2³ minus 2 times 2², lots of twos, plus 2, over 2 minus 1. Now doing this algebra here, 2³ gives me 8. 2 times 2² also gives me 8, so that's a minus 8 plus 2, and then that's all divided by 2 minus 1, which is just 1. Now 8 and 8 cancel each other out and I'm left with 2 over 1 or just 2 for the limit of this function as x approaches 2.

So we now know how to find the limit of any rational function whether the denominator is made 0 or not. If our denominator is 0, we know that we need to factor and then cancel our common factor out and then plug our value in. If our denominator is not 0, we can just go ahead and plug that value in from the beginning. Let me know if you have any questions. I'll see you in the next one.

Find the limit.

$\lim_{x\rarr0}\frac{3x^2+7x}{x}$

• A. $0$
• B. $7$
• C. $-3$
• D. DNE

Find the limit.

$\lim_{x\rarr2}\frac{x^2-7x+12}{x-3}$

• A. $-1$
• B. $-2$
• C. $0$
• D. DNE

Find the limit.

$\lim_{x\rarr-2}\frac{x^2-5x-14}{x+2}$

• A. $-9$
• B. $-2$
• C. $0$
• D. Does not exist

Find the limit.

$\lim_{x\rarr-\pi}\frac{\sin x}{x}$

• A. $-\frac{1}{\pi}$
• B. $0$
• C. $\pi$
• D. Does not exist

Limits of Rational Functions with Radicals

Recall that when working with rational functions, the limit is going to be the same as the function value as long as your denominator is not equal to 0. But even if your denominator is equal to 0, we still know how to find that limit because we can factor our rational function and then cancel out a common factor, allowing us to then evaluate our function and come to a final answer. But what if we're given a rational function in which our denominator is still equal to 0, but it doesn't look quite so easy to factor because it contains radicals? Well, this is actually a common problem that you'll encounter, a rational function where a denominator is equal to 0, and our function contains a radical. Now we're still going to end up canceling out a common factor and then evaluating as we did before.

But instead of starting out by factoring, we're going to start by multiplying the top and the bottom of our function by the conjugate. Now here I'm going to walk you through exactly what that process looks like. So let's jump right into this first example here. Here we're asked to find the limit of x-2x-2 as x→2. Now if I were to just plug this 2 into my denominator, that would end up giving me a root 2 minus root 2, which we know is equal to 0.

So we know that we can't just evaluate this function for 2. We need to go ahead and do something else. Now if I wanted to try and factor here, that might be your first instinct, but I would look at this and not really be sure where to start because typically we factor polynomials. We don't factor these problems that have radicals in them. So we don't want to start by factoring here.

We want to start by multiplying the top and the bottom by the conjugate. Now here we're going to take the conjugate of whatever term contains our radicals. So here, that's our denominator. Now remember when finding the conjugate, we're just reversing the sign in between our two terms. So here, since I have the square root of x minus the square root of 2, the conjugate of this is going to be the square root of x plus the square root of 2, having just changed that negative sign to a positive.

Now I'm multiplying the top and the bottom, so we're also multiplying that numerator, x-2, by that same conjugate, x+2. Now here we need to go ahead and do some multiplication, but we only want to fully multiply out the conjugates being multiplied by each other, which in this case is in our denominator. So that means that I want to leave my numerator factored as is, x-2 times the square root of x+ 2. Now we're gonna see exactly why we don't wanna multiply that out in just a second. But let's go ahead and multiply out this denominator.

Now if I multiply these two conjugates by each other, the square root of x minus the square root of 2 times the square root of x plus the square root of 2, I'm going to end up with x-2. Now from here, I see in my numerator I have x-2, and I have that same term in my denominator. So that's exactly why we didn't want to fully multiply that numerator out, because now we can go ahead and cancel that common factor the same way we did when we factored. Now all I'm left with here is the square root of x plus the square root of 2.

Limits of Rational Functions with Radicals Example 3

In this problem, we want to find the limit of the function 4-x-2 over x as x approaches 0. Now, whenever we see a rational function that contains a radical, we should automatically be thinking about the conjugate. Now here, if I were just to plug 0 in, that would make my denominator 0. So, I need to go ahead and multiply by the conjugate here and proceed in that way before I plug anything in for x.

So here, I want to look at the conjugate of where my radical is, which happens to be in my numerator. And the conjugate of this, 4-x-2 is going to be 4-x+2 because remember we want to change the sign between our two terms here. Now we're not changing the sign that is under the radical; we're just changing the sign in between our two terms. So that minus between the 4-x and 2 changes to a positive. Now remember, if I'm multiplying the numerator, I need to also be multiplying my denominator by the same thing, so that conjugate 4-x+2.

Now here actually multiplying this numerator fully out still finding the limit as x approaches 0, but having multiplied that out, a bunch of stuff is going to cancel. And I'm going to end up with -x in the numerator. So I can rewrite this as the limit as x approaches 0 of -xx(4-x+2). Now here, I see that that negative x will cancel that x on the bottom. So now my numerator is just negative one. So, fully rewriting this limit before I do anything else.

I have the limit as x approaches 0 of negative one in that numerator. And now my denominator is just that conjugate 4-x+2. Now from here, if I go ahead and plug 0 in, it's not going to make my denominator 0. So I'm going to go ahead and plug that 0 in to evaluate this limit. So that will give me negative one over 4-0+2.

Now the 4-0 is just the square root of 4 which is 2 so this ends up being negative one over 4 for the final answer to this limit problem. Let me know if you have any questions here and let's continue practicing together. Thanks for watching.

Find the limit.

$\lim_{x\rarr3}\frac{\sqrt{x}-\sqrt3}{x-3}$

• A. $2\sqrt3$
• B. $0$
• C. $\frac{\sqrt3}{6}$
• D. Does not exist

Find the limit.

$\lim_{x\rarr0}\frac{\sqrt{x+1}-1}{x}$

• A. $0$
• B. $\frac12$
• C. $1$
• D. Does not exist

Find the limit.

$\lim_{x\rarr5}\frac{x-5}{\sqrt{x}-\sqrt5}$

• A. $\sqrt5$
• B. $2\sqrt5$
• C. $\frac{\sqrt5}{10}$
• D. Does not exist