Find the limit of negative two 𝑥 to the negative four plus eight 𝑥 to the negative three minus 𝑥 to the negative two plus nine 𝑥 to the negative one minus four all over two 𝑥 to the negative four minus six 𝑥 to the negative three plus seven 𝑥 to the negative two plus six 𝑥 to the negative one plus three, as 𝑥 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 they’re 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 𝑥 to the power of some negative number as 𝑥 approaches infinity.
What are the values of such limits? Well, we can write 𝑥 to the negative 𝑛 as one over 𝑥 to the 𝑛. And for 𝑛 greater than zero, the value is zero. All these limits then are zero. And we’re 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 𝑥 is as 𝑥 approaches infinity; it’s zero.