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 same as the function value at that point. And this is actually going to be true for many different functions, including polynomials and basic root functions.

The limit will always just be the 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. Now, also looking at this graph, I can see that when x equals 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 many 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 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 that for many functions, 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 going to 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 6x3+3x2-x+5 as x approaches 2. I don't have a graph here, and I also don't want to make a table of values for this long cubic function. But since 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.

Now plugging 2 in here, I get 6×23+3×22+-2+5. Now, this 6×23 will give me 48, and this 3×22 will give me 12, and then minus 2 plus 5. 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 7x2+4x+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 7×02+4×0+16. 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 x2+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 02+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 approaches 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 0 in our next couple of videos. But now that we know how to find limits using direct substitution, let's continue to get 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 1,000,000, 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 2xx2+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 gonna take 2 times negative one-squared plus 3 times negative one, and this is going to be my answer here. Now actually doing this algebra, 2 times negative one squared is going to give me 2 and then 3 times negative one 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 3x squared minus 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 times 3 squared minus 2. Now 3 times 3 squared minus 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 functions. 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 a 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 multiply to 2 and add to 3.

So that ends up factoring to x+1×x+2. 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 by 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-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-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-3. And my second step here is to cancel that common factor, which of course here is x-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 4 + 2 - 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-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+2 because it will end up canceling out with that x+2 in the numerator. Now my other factor here is going to be x-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 1x-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 1-5 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. The first limit that I'm asked to find is the limit as x approaches 1 of x3-2x2+xx-1. Now, looking at what x is approaching here, 1, if I plug that into my denominator, I get 1-1, 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. In order to find this limit, I need to fully factor my function as x approaches 1, and I'm 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 factor that x out. That will give me x(x2-2x+1).

Since my denominator cannot be factored any further, that remains as x-1. I can continue to factor further because this polynomial here actually ends up being a perfect square. So I can factor that as x(x-1)². This is still divided by x−1, and I am still trying to find the limit as x approaches 1. I see that I can cancel some stuff out here because I have x-1 on the bottom and I have x-1. I actually have two of them on the top. So that's no longer squared in my numerator. What I'm left with here is to find the limit as x approaches 1 of x(x-1), having canceled only one of those x-1 terms in the top with my x-1 on the bottom. This will end up giving me 1(1-1), which is 0.

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 2−1, which is 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. Now doing this algebra here, 2x3-2⋅x2+22-1, gives me 2 over 1 or just 2 for the limit of this function as x approaches 2.

We now know how to find the limit of any rational function whether the denominator is made 0 or not. Now, 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. 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 approaches 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.

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, 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 minus 2, by that same conjugate, square root of x plus the square root of 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 minus 2 times the square root of x plus the square root of 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 2 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 minus 2. Now from here, I see in my numerator I have x minus 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. And I can easily find the limit here by now just evaluating for that value of 2.

So from here I get the square root of 2 plus the square root of 2, which simplifies to 2 to 2 root 2, and this is my final answer here. 2 root 2 for the limit of this rational function as x approaches 2. So when you're faced with finding the limit of a rational function that contains radicals, you're going to start by multiplying the top and the bottom by the conjugate. Then you're going to proceed the same way we did for our other rational functions that we factored by canceling out our common factor and then evaluating. So let's work through one more example to make sure that we have this down.

Now here we're asked to find the limit of x+9-3x as x approaches 0. So if I were to plug that 0 into my denominator, since my denominator is just x, I know that that would make my denominator 0. So we need to go ahead and multiply the top and the bottom by the conjugate here. Now remember that we want to take the conjugate of wherever our radicals are, and here our radicals are in the numerator so I want to take the conjugate of this term specifically.

Now taking the conjugate of this, remember we're just reversing the sign in between these two terms, not the sign that's under the radical. So here I'm just going to take the conjugate, which is x+9, keeping that sign the same, but changing that sign in between my two terms into a positive. Now remember we're multiplying the top and the bottom, I want to make sure that I'm also multiplying the bottom by the same thing, the square root of x+9+3. Now from here, we want to go ahead and multiply this out. We are still finding the limit as x approaches 0 and multiplying those 2 conjugates out together I end up getting x+9 minus 9.

Now x+9 minus 9, that plus 9 minus 9, end up canceling with each other, which we'll see in a second. And then in my denominator, remember we do not want to fully multiply these out because we're going to see what happens here. I have x, keeping this the same, square root of x+9+3. Now here in my numerator, since those nines cancel each other out, here I'm just left with the limit as x approaches 0 of x, that's all that's left in my numerator, divided by x times the square root of x+9 plus 3. Now since I have that x in the top and the bottom, that's my common factor.

It's going to cancel out. And now my final limit that I ask to find is the limit as x approaches 0 of all that's left here, 1 over the square root of x+9+3. Now this is not going to make my denominator 0. I can just plug that 0 in and evaluate as normal. So plugging that 0 in, I have 1 on the top.

And then on the bottom, I have 0+9 under that square root plus 3. Now 0+9 is just 9. So this is 1 over root 9+3. Root 9 is just 3. So I end up with 1 over 3 plus 3 or, finally, my final answer, 1 over 6, as the limit of this rational function as x approaches 0.

Now that we know how to find the limit of any rational function, whether it contains radicals or doesn't or whether the denominator is equal to 0 or not, let's continue practicing. Thanks for watching, and let me know if you have questions.

Limits of Rational Functions with Radicals Example 3

In this problem, we want to find the limit of the function the square root of 4 minus x minus 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 to just 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, the square root of 4 minus x minus 2 is going to be the square root of 4 minus x plus 2 because remember we want to change the sign between our two terms here. We're not changing the sign that is under the radical. We're just changing the sign between our two terms. So that minus between the square root of 4 minus 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 minus x plus 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 am going to end up with 4 - x and then - 4 . So I have a 4 and a 4 that will end up canceling here . Now before I get to that I am going to leave my denominator as is because we are going to see some stuff cancel out there as well . So I have x * sqrt x + 2 .

So here in my numerator that positive 4 will cancel with that negative 4, just leaving me in that numerator with negative x. So I can rewrite this as the limit as x approaches 0 of negative x over x times the square root of 4 minus x plus 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 minus 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 the square root of 4 minus 0 plus 2.

Now the square root of 4 minus 0 is just the square root of 4 which is 2 so this ends up being negative 1 over 2 + 2 or 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