Video Transcript
Find the limit, as 𝑥 tends to ∞, of three 𝑥 squared plus seven 𝑥 minus 12 over 15 minus two 𝑥 minus two 𝑥 squared.
Now, in order to find infinite limits, there’s a very useful rule we can use. And that is that the limit, as 𝑥 tends to ∞, of one over 𝑥 is equal to zero. Now, we don’t have any one over 𝑥 terms within our limit. However, we can multiply the numerator and denominator of our fraction by one over 𝑥. We can complete this multiplication in the numerator and denominator term by term. Our limit is therefore equivalent to the limit, as 𝑥 tends to ∞, of three 𝑥 plus seven minus 12 over 𝑥 all over 15 over 𝑥 minus two minus two 𝑥.
With the limit in this form, both three 𝑥 in the numerator and two 𝑥 in the denominator will tend to ∞, as 𝑥 tends to ∞. And, therefore, our fraction will become ∞ over ∞. So let’s try and multiply the numerator and denominator by one over 𝑥 again. Our limit is now in this form. However, we do not yet know what to do with this negative 12 over 𝑥 squared and 15 over 𝑥 squared terms.
Now, if we square both sides of our limit rule using the fact that the product of limits of functions is equal to the limit of the product of the functions, which tells us that the limit, as 𝑥 tends to ∞, of one over 𝑥 all squared is equal to the limit, as 𝑥 tends to ∞, of one over 𝑥 squared. And we arrive at a new rule which tells us that the limit, as 𝑥 tends to ∞, of one over 𝑥 squared is equal to zero.
Our next step in finding the value of our limit is to use the fact that the limit of product quotients and sums of functions is equal to the product of quotients and sums of limits of the functions. And this enables us to rewrite our limit as the limit, as 𝑥 approaches ∞, of three plus the limit, as 𝑥 approaches ∞, of seven over 𝑥 plus the limit, as 𝑥 approaches ∞, of negative 12 over 𝑥 squared all over the limit, as 𝑥 approaches ∞, of 15 over 𝑥 squared plus the limit, as 𝑥 approaches ∞, of negative two over 𝑥 plus the limit, as 𝑥 approaches ∞, of negative two.
Now, due to our two rules about infinite limits, we know that the terms with an 𝑥 and an 𝑥 squared in the denominator will go to zero. And then, we are left with the limit, as 𝑥 tends to ∞, of three over the limit, as 𝑥 tends to ∞, of negative two. Since three and negative two are both constants, the fact that we are taking a limit as 𝑥 tends to ∞ will have no effect on them. Therefore, our limit must be equal to negative three over two.
