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Finding limits by factoring (cubic)

Video transcript
Let's try to find the limit as x approaches 1 of x to the third minus 1 over x squared minus 1. And at first when you just try to substitute x equals 1, you get 0/0 1 minus 1 over 1 minus 1. So that doesn't help us. So let's see if we can try to simplify this in some way. So you might immediately recognize-- so let's rewrite this expression right over here so it's x to the third minus 1 over x squared minus 1. This on the bottom immediately jumps out as a difference of squares. So we know on the bottom that this could be factored as x minus 1 times x plus 1. And so if somehow this thing on the top also has an x minus 1 as a factor, then that x minus 1 will cancel with this, and then we're not going to have an issue of dividing by 0. The reason why I care about the x minus 1 term is that this is what's making our denominator equal 0. When you say x equals 1, you have 1 minus 1 times 1 plus 1. So 0 times 2, it's this 0 that's making our denominator 0. So if we can have an x minus 1 up here, then we can cancel these out for any x not equal to 1. And then we might have a much simpler thing to find the limit of. So let's think about whether x to the third minus 1 is the product of x minus 1 and something else. And to do that we can do a little bit of algebraic long division. Some of you guys might already recognize a pattern here, but we'll try to do-- well, let's divide x minus 1 into it to see whether it divides evenly into x to the third minus 1. So x minus 1-- we just look at the highest degree term-- x goes into x to the third x squared times. Goes x squared times. Actually, let me do it this way so that way we can keep track of the place. So this would be x-- this would be the second degree place, first degree place, and this would be the constant. So x to the third minus 1. x goes into x to the third x squared times. x squared times x is x to the third. x squared times negative 1 is minus x squared. And now we're going to want to subtract this. So we are then left with x squared. x goes into x squared x times plus x. x times x is x squared. x times minus 1 is minus x. And once again we're going to subtract this. We'll swap the signs, negative and positive. And so these cancel out, and we're left with x. And then we bring down a minus 1. x minus 1 goes into x minus 1 exactly one time. 1 times x minus 1 is x minus 1. And then you subtract, and then you have no remainder. So this numerator right over here can be factored as x minus 1 times x squared plus x plus 1. And so we can say that this is the same exact thing. We can have these cancel out if we assume x does not equal 1. So that is equal to x squared plus x plus 1 over x plus 1, for x does not equal 1. And that's completely fine, because we're not evaluating x equals 1. We're evaluating as x approaches 1. So this is going to be the same thing as the limit as x approaches 1 of x squared plus x plus 1 over x plus 1. And now this is much easier to find. You could literally just say, well, what happens as we get right to x equals 1? Then you have 1 squared, which is 1 plus 1 plus 1, which is 3, over 1 plus 1, which is 2. So we get that equaling 3/2.