Prove that a horizontal asymptote can never be crossed

Question

This is a lot simpler of a problem than others posted here, but I was bored in class and decided to work out why a horizontal asymptote exists. Bear in mind that I am still fairly low on the “math ladder.”

So to accomplish this I worked off of one example equation that could be made into a general case later on, with the chosen one being $$y = \frac{2x+6}{x+1}$$ Now, proving any vertical asymptote is simple (by definition it creates a “divide by zero” error when plugged into the equation), but a horizontal asymptote “proof” requires some manipulation. So, doing a little shuffling of the equation to isolate the variables slightly... $$y = \frac{x(2+\frac{6}{x})}{x(1+\frac{1}{x})}$$ The $x$ on the top and bottom of the function will cancel out, and we are left with $$y = \frac{2+\frac{6}{x}}{1+\frac{1}{x}}$$ With a little more rearranging by multiplying the function by the denominator, you get $$y\left(1+\frac{1}{x}\right) = 2+\frac{6}{x}$$ You likely see what is wrong with this, but let’s distribute the $y$ to simplify further: $$y+\frac{y}{x} = 2+\frac{6}{x}$$ Now everything is ready to plug in. Say that there is a value that corresponds to $y = 2$ (the horizontal asymptote of this graph, found by dividing $p$ by $q$, or $2/1$, because $\operatorname{deg} p = \operatorname{deg} q$). Then the equation should be able to be used to find this value. However, if $2$ is substituted for $y$... $$2+\frac{2}{x} = 2+\frac{6}{x}$$ $$\frac{2}{x} = \frac{6}{x}$$ $$2 = 6$$ Therefore $y$ cannot be equal to $2$. I believe that is referred to as a “proof by contradiction” but correct me if I’m wrong. Or if the whole proof is wrong, for that matter. Anyways, I’d love to generalize this but I’m hitting a roadblock due to not knowing of a comprehensive standard form for all rational expressions. I’ve tried it on numerous equations but I’m not sure how to explain what I think is going on. Every time you plug the asymptote into the reworked equation you end up with a contradiction, and any other numbers (minus holes) returns a value that can be found on the graph.

Can anyone help with this? I just felt like sharing this little 10 minute project because it intrigued me, but being able to generalize it to see what’s going on between each variable and coefficient (specifically, how/why the leading terms and degrees affect the H.A) would be helpful for overall understanding of the topic. Thanks!

Answer 1

You indeed proved that your function never crosses the asymptote $y=2$. Specifically, you proved following statement

There does not exists $x\in \mathbb{R}$ such that $$ \frac{2x+6}{x+1} = 2. $$

To so this, you indeed used a "proof by contradiction". If there exists a number $x$ such that $$ \frac{2x+6}{x+1} = 2. $$ then $$ 2x+6 = 2(x+1)\implies 6=2 $$ which is absurd.

But is what you proved sufficient to conclude that the function has a horizontal asymptote described by the line $y=2$? Let's take a look at a different function; $$ y = \frac{x^2+1}{x^2}. $$ For this function, there does not exist a number $x$ such that $$ \frac{x^2+1}{x^2} = 0. $$ But this does not mean that the horizontal asymptote is described by $y=0$. Your horizontal asymptote should instead by the line $y=1$.

Ask yourself the following: how would you got about defining a horizontal asymptote? What properties do you think it should have?

Usually, we want an asymptote to somehow capture the behaviour of the function as $x$ gets very large. If you want more information on how to make this statement precise, let me know in the comments and I can add detail.

Answer 2

Let's stick to rational functions, even if it's not explicitly mentioned that this is what you want to use. So we want to construct such function $f(x)$, that has some asymptote at $\pm\infty$ and that you cross this asymptote. How would be approaching this? First, note that we can choose $f(\pm\infty)=0$. If you want a different asymptote, just add a constant.

So we write $$f(x)=\frac{P(x)}{Q(x)}$$ where $P$ and $Q$ are polynomials. If the degree of $Q$ is greater than the degree of $P$, this asymptotically will go to $0$. For simplicity, just choose $Q(x)=x^n$. Now if we choose $n-1$ different real, non-zero numbers $x_1, x_2,...,x_{n-1}$, we can construct $P(x)=(x-x_1)(x-x_2)...(x-x_n-1)$, a polynomial with degree $n-1$, with $n-1$ real, non-zero roots. With these choices, $f(x)$ defined above has a horizontal asymptote, and the function is crossing the horizontal line at $n-1$ different points.