Video Transcript
Choose the function whose graph has a horizontal asymptote given by the line 𝑦 equals negative two. Is it a) 𝑦 equals negative two cos of 𝑥 over 𝑥? b) 𝑦 equals negative two sin of 𝑥 over 𝑥. c) 𝑦 is two minus one over 𝑥 plus two. Or d) 𝑦 equals negative two plus 𝑥 over 𝑥 squared minus one.
Remember, an asymptote is a line that the curve tends to but never quite reaches. We can therefore find any horizontal asymptote by evaluating 𝑦 as 𝑥 gets either very, very large or very, very small. Specifically, we’re looking for the function that tends to negative two as 𝑥 tends to either positive or negative infinity. Let’s begin with a. Let’s evaluate the limit as 𝑥 tends to positive infinity first of negative two cos 𝑥 over 𝑥.
This is a tricky one. The graph of 𝑦 equals cos of 𝑥 is oscillating and periodic. It has relative maxima and minima at one and negative one. And it has a period of two 𝜋 radians. This means the graph of negative two cos of 𝑥 is also oscillating. It has relative maxima and minima at two and negative two. But what this also tells us is that the limit as 𝑥 tends to positive infinity of negative two cos of 𝑥 does not exist. The same can also be said for the limit as 𝑥 tends to negative infinity.
However, this isn’t too much of a problem. Negative two cos of 𝑥 will take values between negative two and two. But as 𝑥 gets very, very large, we’re going to be dividing this relatively small number by a very large number. And this means the limit as 𝑥 tends to positive infinity of negative two cos 𝑥 over 𝑥 is zero. And in fact, the same can be said for the limit as 𝑥 tends to negative infinity. We’re dividing a number between negative two and two by a number whose magnitude is extremely large. And it should be quite clear that our function therefore tends to zero.
In fact, if we sketch the graph 𝑦 equals negative two cos of 𝑥 over 𝑥 out with our graphical calculator, we can see that as 𝑥 tends to positive and negative infinity, the curve tends to the line 𝑦 equals zero. So the answer is not a. We can perform a similar process to evaluate the limit as 𝑥 tends to both positive and negative infinity of negative two sin of 𝑥 over 𝑥.
The graph of negative two sin of 𝑥 contains many similar features to the graph of negative two cos of 𝑥. We know it has relative maxima and minima at two and negative two. So we’re going to be dividing a value between two and negative two by a number with an extremely large magnitude. It’s tending to remember positive and negative infinity. And that tells us negative two sin 𝑥 over 𝑥 tends to zero. Once again, our curve is going to tend towards the line 𝑦 equals zero, not the line 𝑦 equals two. So it’s not b.
So what about the third function? We want to find the limit as 𝑥 tends to positive infinity to start of two minus one over 𝑥 plus two. Two is independent of 𝑥. But as 𝑥 tends to positive infinity, one over 𝑥 plus two tends to zero. So the limit as 𝑥 tends to positive infinity of two minus one over 𝑥 plus two is two. And in fact, the same can be said for our limit as 𝑥 tends to negative infinity. One over 𝑥 plus two becomes one divided by a number with a very large magnitude, which tends to zero.
And this time, if we sketch the graph out, we see it does indeed have a horizontal asymptote given by the line 𝑦 equals two, but not 𝑦 equals negative two. And it’s safe to assume that our final function 𝑦 equals negative two plus 𝑥 over 𝑥 squared minus one is going to be the correct answer. But let’s double check.
In this function, we notice that, for the fractional part, the degree in the denominator is larger than the degree in the numerator. And this means that as 𝑥 tends to both positive and negative infinity, 𝑥 over 𝑥 squared minus one tends to zero. And this in turn means that the limit of our function as 𝑥 tends to positive and negative infinity is negative two. It has a horizontal asymptote given by the line 𝑦 equals negative two. And we can once again sketch it out. And we do indeed see that it has horizontal asymptotes given by the line 𝑦 equals negative two.
The correct answer is d.
