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## Calculus 1

### Course: Calculus 1>Unit 1

Lesson 7: Limits using algebraic manipulation

# Limits by factoring

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.E (LO)
,
LIM‑1.E.1 (EK)

In this video, we explore the limit of (x²+x-6)/(x-2) as x approaches 2. By factoring and simplifying the expression, we discover that the function is undefined at x = 2, but its limit from both sides as x approaches 2 is in fact 5.

.
Created by Sal Khan.

## Want to join the conversation?

• Does l'Hopital's rule apply here?
• L'Hopitals rule is applicable here

L= lim x->2 for x^2+x-6/(x-2)
L= lim x->2 for f(x)/g(x) where f(x)=x^2+x-6, g(x)=x-2
since lim x->2 f(x)=0 and lim x->2 g(x)=0
and 0/0 is one of the inderminant forms we can apply L'Hopitals rule
f'(x)=2x+1
g'(x)=1
L= lim x->2 for f'(x)/g'(x)=5/1=5
we obtained the same answer when we used factoring to solve the limit

In my opinion, it is easier to use L'Hopitals here than factoring (many will disagree with me).

However, you typically need to know limits before you learn calculus, and you need to know
calculus before you use L'Hopitals rule, so the 1st time people learn how to solve these
types of limits they will use factoring.
• I'm still confuse, is it possible that calculus and algebra can be the same? Some problems i guess can be done through algebra like simple substitution of x=any number.
• Well, in the end, calculus is ultimately based on algebra. It's simply taking a few very useful algebraic functions, making a couple of modifications (like the addition of limits), and expanding on them until they've effectively become their own field of study, in the same way that algebra is ultimately based on arithmetic, but with the added concept of variables. You'd use arithmetic to solve an algebra problem, and you'd use algebra to solve a calculus problem (although, hopefully, you wouldn't use calculus to solve arithmetic). One of the most beautiful parts of mathematics is that it's all connected like that, with one step leading right to the next. So, while they are technically different areas of mathematics, in essence, yes, they are both a part the same thing. One is just a step above the other.
• How did Sal come up with a slope of 1?
• Consider y=mx+c where m is the slope of the line
Now see that when y=x+3 the value of m is 1 which means that slope is 1.
• One question, if you use algebraic rules on it, like sal did and you cancel the (x-2) out, after that the function f(x) is not the same as before or? Because then you dont have that gap anymore?
• Excellent question! And you're right, it's not exactly the same function, because the domain is different -- it now includes x=2, whereas the original function didn't. We can say that it's an extension of the original function, though, because it behaves exactly the same at all the points which were included in the original function.
• Say I have lim of 5/x-7 as x -> 7 I know it is undefined but say I have lim 5/(x-7)^2 as x -> 7 why is it +infinity rather than undefined?
• When you're stuck like this, you need to see the limit when approaching to 7 for the right side and for the left. At the first example, lim x->7 (5/(x-7)), if you approach by the left you got -infinite, because you will have a number really close to 7, but still minor than it. And you approach by the right, you got + infinite, because you will have a number really close to 7, but a little higher than it. That's why it's undefined, because the two limits are different. If you do the same thing on the other one, you will get +infinite in the both sides, and the limit is defined as +infinite.
• I didn't get it...when we have f(x) = (x^2 + x -6)/x-2..then our limf(x)as x approaches 2 is not defined..but when w simplified the unction to x+3 then we got a different value of limit as x approaches 2 ..can we have two values for same limit..of the same function?
• the limit for the original equation is the same thing as the simplified version. we simply cant directly solve the limit in the original equation that easily so we simplify it first
• if
lim f(x) = lim g(x)
x -> 1 x -> 1

can you say that
lim f(x)/g(x) = 1
when x -> 1
is always true?

or it would be correct ONLY as long as lim g(x) does not equal to 0 when x -> 1?

thanks
• You can't say that. Take for example f(x)=x-1 and g(x)=x^2 -1. Then

lim f(x) = lim g(x) = 0
x -> 1 x -> 1

However, lim f(x)/g(x) = lim (x-1)/(x^2 -1) = lim (x-1)/(x-1)(x+1) = lim 1/(x+1) = 1/2.
x -> 1 x -> 1 x -> 1 x -> 1
• If we factor x^2+x-6, f(x)=x+3. And why f(x) is still undefined at x=2?
• You can't just say that f(x) = x + 3 after canceling out (x - 2). That's still a part of the function, and if you just stay f(x) = x + 3 you lose out on the information that (x - 2) was in the denominator. That's still very much a part of the function, so (x+3)(x-2) / (x-2) is not equal to x + 3. It is equal to x + 3 everywhere but where x in the original function would make it undefined, and we need to show that as well by saying that f(x) = x + 3 everywhere but x = 2, where f(x) is undefined.
• Can someone give me some intuition as to why factoring works? I read somewhere that (x-2) is a "culprit" factor, which we need to factor out or something along those lines? Is this true, and if so, is there more to it than that?
• It is kind of because the function is undefined at x=2. But at a value of x infinitely close to 2(like 2.001 or1.999), we can evaluate the function. Hence we can calculate the limit of that function which is exactly that value which we can get very very close to but never exactly evaluate.
• Ok I looked over comments and there is one thing I do not understand. I get how to solve this for 5. I get how the graph starts with the X+3 and the 2 as the undefined location. I do not understand why he puts the circle at where it is. I mean I understand how it will become 5 because that is the solution to this limit question. i do not understand why he places the circle there, he has not solved the equation yet, how does he know to place the circle there? He doesn't really say anything, just draws it there then says its 5 later on. Is there a trick or rule I am missing?
• The circle shows that the function is not defined when x=2. Sal knows this because the denominator is x-2 and anything divided by zero is undefined.

## Video transcript

Let's say that f of x is equal to x squared plus x minus 6 over x minus 2. And we're curious about what the limit of f of x, as x approaches 2, is equal to. Now the first attempt that you might want to do right when you see something like this, is just see what happens what is f of 2. Now this won't always be the limit, even if it's defined, but it's a good place to start, just to see if it's something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator, we get 2 squared plus 2 minus 6. So it's going to be 4 plus 2, which is 6, minus 6, so you're going to get 0 in the numerator and you're going to get 0 in the denominator. So we don't have, the function is not defined, so not defined at x is equal 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, then the limit would be whatever the function is, but that doesn't necessarily mean the case. But we see very clearly the function is not defined here. So let's see if we can simplify this and also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we can rewrite the top expression. And this just goes back to your algebra one, two numbers whose product is negative 6, whose sum is positive 3, well that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all X's except for x is equal to 2. So that's another way of looking at it. Another way we could rewrite our f of x, we'll do it in blue, just to change the colors, we could rewrite f of x, this is the exact same function, f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is, that is not anywhere near being a straight line, that is much better. So let's call this the y-axis call it y equals f of x. And then let's, over here, let me make a horizontal line, that is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3 and then the slope is 1. And it's defined for all X's except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So when x is equal to 2 it is undefined. So let me make sure I can, so it's undefined right over there. So this is what f of x looks like. Now given this, let's try to answer our question. What is the limit of f of x as x approaches 2. Well, we can look at this graphically. As x approaches 2 from lower values in 2, so this right over here is x is equal to 2, if we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5 our f of x is right over there. If we get even closer to 2, our f of x is right over there. And once again, we look like we are approaching this value. Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x, if we write this line from the negative direction, from values less than 2, we seem to be approaching this value right over here. And this is essentially the value of x plus 3 if we set x is equal to 2. So this is essentially going to be, this value right over here, is equal to 5. If we just look at it visually, if we just graphed a line with slope 1 with the y-intercept of 3, this value right over here is 5. Now we could also try to do this it numerically, so let's try to do that. So if this is our function definition, completely identical to our original definition, then we just try values as x gets closer and closer to 2. So let's try values less than 2. So 1.9999, and this is almost obvious. 1.9999 plus 3, well, that gets you pretty darn close to 5. If I put even more 9s here, get even closer to 2, we'd get even closer to 5 here. If we approach 2 from the positive direction, and then, we once again, we're getting closer and closer to 5 from the positive direction. If we were even closer to 2, we'd be even closer to 5. So whether we look at it numerically, or we look at it graphically, it looks pretty clear that the limit here is going to be equal to five.