### Video Transcript

Find the limit of negative two
π₯ to the negative four plus eight π₯ to the negative three minus π₯ to the
negative two plus nine π₯ to the negative one minus four all over two π₯ to the
negative four minus six π₯ to the negative three plus seven π₯ to the negative
two plus six π₯ to the negative one plus three, as π₯ approaches infinity.

This is the limit of a quotient
of functions. And we know that the limit of a
quotient of functions is the quotient of their limits. So, we can find the limits of
the numerator and denominator separately, should they exist. And as the limit of a sum of
functions is the sum of their limits, we can find the limits term-by-term.

Now, we have lots of limits to
evaluate but theyβre all of very simple terms. And we can make them simpler by
taking the constants outside the limits. As the limit of a constant
times a function is that constant times the limit of the function. And now, the vast majority of
our limits are of the form the limit of π₯ to the power of some negative number
as π₯ approaches infinity.

What are the values of such
limits? Well, we can write π₯ to the
negative π as one over π₯ to the π. And for π greater than zero,
the value is zero. All these limits then are
zero. And weβre only left with the
two limits to worry about, both of which are limits of the constants four and
three. The limit of a constant
function is just that constant. And so, being careful to
include this minus sign, we see the answer is negative four over three.

Solving this problem was
straightforward because we only had constants and negative powers in the
numerator and denominator. And we know what the limit of a
negative power of π₯ is as π₯ approaches infinity; itβs zero.