Limits and End Behavior - Problem 3 2,768 views
Recall that the limit of a sum is a sum of the limits. So, in order to find oblique asymptotes, you can apply the procedure for finding limits of rational functions to the rational part of the function, and then add that to the other part of the function. Then As a result, you will get some polynomial, the line of which will be the oblique asymptote of the function as x approaches infinity.
Let’s take a look at the harder problem. Consider the function r of x equals 3x plus 1 plus the quantity 12 over x plus 2. So this is the kind of a rational function. What happens as x goes to infinity? As x goes to infinity the 12 over x plus 2 part is going where?
The denominator is going to infinity so this thing is going to 0. But what about the 3x plus 1? That’s going to infinity. So it’s pretty ambiguous that, the sum 3x plus 1 plus 12 over x plus 2, this piece is going to 0 and this piece is going to infinity. So this is going to infinity plus 0 you could say.
And that means, the whole function is going to infinity. Let’s see what happens as x approaches negative infinity. I can still say that as x goes to negative infinity, 12 over x plus 2 goes to 0. The denominator is getting large in magnitude but negative.
And so this is going to 0. Now 3x plus 1. that first part of the function. is going to go to negative infinity. Thing about what happens is x goes it looks like a -1000 – 1000000 this is going to get very large in magnitude but negative.
So r of x the sum of these two guys, this piece is going to 0. this piece is going to negative infinity. So it’s going to negative infinity plus 0. And that means the limit is negative infinity. Now let’s pause for a second. You know that when there is limit x approaches plus or minus infinity is some constant value, you get a horizontal asymptote. But we didn’t get that here. Here, the limit as x approaches positive infinity is positive infinity. And the limit as x approaches negative infinity is negative infinity. There’s no horizontal asymptote here. But there is a kind of asymptote.
It’s an oblique asymptote. So let me write down the function again just to remind you r of x equals 3 x plus 1 plus 12 over x plus 2. So remember as x approaches infinity, this piece is going to 0. And what that means is that r of x approaches the value of 3x plus 1 intersects.
That means that y equals 3x plus 1 is the equation of the oblique asymptote. I mean you can basically think of this function. You could think of this function as y equals 3x plus 1 plus a little remainder. And this remainder gets very small as x goes to infinity.
It also gets very small as x goes to negative infinity. So the further we get away from 0, as x goes to be very big or very small this thing dwindles away to 0. And the function behaves more and more like the line y equals 3x plus 1. So that makes 3x plus 1 and oblique asymptote.