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Current time:0:00Total duration:3:56

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.