### Video Transcript

Find the limit of π₯ squared
plus three all over eight π₯ cubed plus nine π₯ plus one as π₯ approaches
infinity.

Our first thought might be to
use the fact that the limit of a quotient is the quotient of the limits. This gives us the limit of π₯
squared plus three as π₯ approaches infinity over the limit of eight π₯ cubed
plus nine π₯ plus one as π₯ approaches infinity. But we run into problems
because neither limit is defined. In the numerator, as π₯
approaches infinity, π₯ squared plus three doesnβt approach any real value, it
just gets bigger and bigger without bound. And the same thing happens in
the denominator. As π₯ increases without bound,
the cubic eight π₯ cubed plus nine π₯ plus one also increases without bound.

Or maybe you think both limits
should be infinity. And so, the limit on left-hand
side is infinity over infinity. But this, just like a zero over
zero, is an indeterminate form. And it doesnβt tell us the
value of our limit. We need to use a different
approach.

The trick to this question is
to find the highest power of π₯ that appears in the numerator or
denominator. Thatβs π₯ cubed here. And having found this highest
power, we divide both numerator and denominator by it. What do we get? π₯ squared divided by π₯ cubed
is π₯ to the negative one. And three divided by π₯ cubed
is three π₯ to the negative three. And in the denominator eight π₯
cubed divided by π₯ cubed is just eight. Nine π₯ divided by π₯ cubed is
nine π₯ to the negative two. And one divided by π₯ cubed is
π₯ to the negative three.

So, now, we just have negative
powers of π₯ and a constant in the numerator and denominator. And as a result, when we apply
this limit law, we find that the limits in the numerator and denominator do now
exist. Letβs find their values. We can use the fact that the
limit of a sum of functions is the sum of their limits. This allows us to find the
limit of each term separately. We can also take the
coefficient outside the limits.

And now, apart from one limit,
which is the limit of a constant function, and whose value must therefore be
eight, all the other limits have the form the limit of π₯ to the power of
negative π as π₯ approaches infinity. Where π is, of course, greater
than zero. And we know the value of such
limits. The value is always zero.

So, this is zero, and this is
zero, and this is zero, and this is zero. Simplifying then, our answer is
zero over eight, which is, of course, just zero.