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Limits by factoring

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.

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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.