Now that we have seen how to find limits algebraically, we can now look at finding asymptotes. We will be looking at functions at points where asking for a function value won’t help. Instead, the idea of limit will allow us to explain what these functions look like at these points.

**Asymptotes**

We will want to find all vertical and horizontal asymptotes of

\begin{align*}

f(x)=\frac{x^{2}+3x-4}{2x^{2}+x-3}.

\end{align*}

To begin with, we will look at what an asymptote is. We say that a function has a vertical asymptote if it looks like the graph of \(x=c\) for some constant \(c\). In the same way, we will say the a function has a horizontal asymptote if it looks like the graph of \(y=k\) for some \(k\). If we graph both of these options, they will look like

In order to “look like” \(x=c\) or \(y=k\), we would say that the function approaches these graphs. A graphical example of this would be the following.

What we end up seeing is there is a vertical asymptote if \(\lim\limits_{x \to c}f(x)=\pm \infty\), that is, the function grows without bound in either a positive or negative direction at the point \(x=c\). We also get that there is a horizontal asymptote at \(y=k\) if the function values approach \(k\) as the \(x\) values get arbitrarily large in either the positive direction. That is,

\begin{align*}

\lim_{x \to \infty}f(x)=k \text{ or }\\

\lim_{x \to -\infty}f(x)=k.

\end{align*}

Therefore, if we want to find asymptotes, we have to find points where these things occur.

**Vertical Asymptotes**

Note that at a vertical asymptote, we have that \(\lim\limits_{x \to c}f(x)=\pm \infty\). Since \(\infty\) and \(-\infty\) are not numbers, this implies that the limit as \(x\) goes to \(c\) does not exist. Therefore, we would have a discontinuous function at \(x=c\).

The reason why this is helpful, is that it drastically decreases the number of limits we have to find when checking for vertical asymptotes. In this case, we note that the function is only discontinuous if it is undefined. This rational function is also only undefined if the denominator is \(0\). Therefore, the only potential asymptotes are when \(2x^{2}+x-3=0\). If we solve this, we find

\begin{align*}

2x^{2}+x-3&=0 \\

(2x+3)(x-1)&=0 \\

(2x+3)&=0 \text{ or } x-1=0 \\

x&=\frac{-3}{2} \text{ or } 1.

\end{align*}

Therefore, the only places there could be asymptotes are at \(x=\frac{-3}{2}\) or \(x=1\). However, these may not actually be asymptotes. In order to check if they are, we need to determine what the limit is at each of these points. We will then start with \(x=\frac{-3}{2}\).

In order to find the limit, we first plug the point \(\frac{-3}{2}\) into \(f(x)\) in order to see what happens. Here we would get

\begin{align*}

f(\frac{-3}{2})&=\frac{\left(\frac{-3}{2}\right)^{2}+3\left(\frac{-3}{2}\right)-4}{2\left(\frac{-3}{2}\right)^{2}+\left(\frac{-3}{2}\right)-3} \\

&=\frac{\frac{-25}{4}}{0}.

\end{align*}

Because we get a number over \(0\), we would see that as we approach \(x=\frac{-3}{2}\), the denominator would get extremely small. Since the numerator gets close to \(\frac{-25}{4}\), the whole fraction will get extremely large (since something divided by a very small number is very large). Therefore

\begin{align*}

\lim_{x \to \frac{-3}{2}}f(x)= \infty \text{ or } -\infty.

\end{align*}

That is, while we are unsure if it gets large in the positive or negative direction, we know that our function will get arbitrarily large. Therefore, the limit will undefined in such a way that it will look like either positive or negative infinity. Hence, there will be a vertical asymptote at \(x=\frac{-3}{2}\).

We now need to find the limit as \(x\) goes to \(1\). In this case, if we plug \(1\) into \(f(x)\), we find

\begin{align*}

f(1)&=\frac{1^{2}+3(1)-4}{2(1)^{2}+1-3} \\

&=\frac{0}{0}.

\end{align*}

In this case, the function is still undefined, but the limit may be defined since we arrive at an indeterminate form. We will therefore have to do more work. We can now see

\begin{align*}

\lim_{x \to 1}\frac{x^{2}+3x-4}{2x^{2}+x-3} &=\lim_{x \to 1}\frac{(x-1)(x+4)}{(x-1)(2x+3)} \\

&=\lim_{x \to 1}\frac{x+4}{2x+3} \\

&=\frac{1+4}{2+3} \\

&=1.

\end{align*}

Note that, since the limit exists, there will be a hole discontinuity at \(x=1\). Therefore, there is not a vertical asymptote.

We now see that the only vertical asymptote of \(f(x)\) is \(x=\frac{-3}{2}\).

**Horizontal Asymptote**

In order to find the horizontal asymptotes of our function, we need to find the limit of \(f(x)\) as \(x\) gets large in either the positive or negative direction. We, therefore, find

\begin{align*}

\lim_{x \to \infty}f(x)=\lim_{x \to \infty}\frac{x^{2}+3x-4}{2x^{2}+x-3}.

\end{align*}

While \(\infty\) is not a number, we still begin by thinking about what happens to \(f(x)\) if we “plug in” \(\infty\). That is, the top would look like \(\infty^{2}+3\infty-3\) which would look like \(\infty+\infty-3\) which would look like \(\infty\). If we do the same for the bottom, we will also arrive at \(\infty\). We therefore get that the function will look like \(\frac{\infty}{\infty}\) as \(x\) goes toward \(\infty\). Since this is another one of our indeterminate forms, we will know that we have to do more work.

In this case, the “trick” (or technique) that we will use is that we will multiply the top and the bottom by \(1\) over the largest power of \(x\) in the function. In this case, that would be \(\frac{1}{x^{2}}\). By doing this, we get

\begin{align*}

\lim_{x \to \infty}\frac{x^{2}+3x-4}{2x^{2}+x-3}&=\lim_{x \to \infty}\frac{x^{2}+3x-4}{2x^{2}+x-3}\frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}} \\

&=\lim_{x \to \infty}\frac{\frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}-\frac{4}{x^{2}}}{\frac{2x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{3}{x^{2}}} \\

&=\lim_{x \to \infty}\frac{1+\frac{3}{x}-\frac{4}{x^{2}}}{2+\frac{1}{x}-\frac{3}{x^{2}}}.

\end{align*}

Now that we have simplified the problem after multiplying, we are now in the position to see what happens \(x\) goeas to \(\infty\). Note that each of the pieces of the from some number over \(\infty\) will go to \(0\) since making the bottom arbitrarily large makes the fraction arbitrarily small. We, therefore, get that

\begin{align*}

\lim_{x \to \infty}\frac{1+\frac{3}{x}-\frac{4}{x^{2}}}{2+\frac{1}{x}-\frac{3}{x^{2}}}&=\frac{1+0-0}{2+0-0} \\

&=\frac{1}{2}.

\end{align*}

Because \(\lim_{x \to \infty}f(x)=\frac{1}{2}\) and \(\frac{1}{2}\) is a number, we know that \(f(x)\) will have a horizontal asymptote at \(y=\frac{1}{2}\).

We would now also have to check the limit as \(x \to -\infty\), but I will leave it up to you to show that you will again get a limit of \(\frac{1}{2}\). Therefore, \(y=\frac{1}{2}\) is the only horizontal asymptote of \(f(x)\).

As a quick note here, if you are looking at this when reviewing for the final for Calculus, there is another way to find the limit as \(x\) goes to \(\infty\). In particular, we could make use of L’Hopital’s rule and make this shorter, but I will refrain from doing so in this post. I will look at that in another post.

**Conclusion**

By looking at the limits of the function at points of discontinuity and as \(x\) get arbitrarily large in the positive or negative direction, we were able to find all the asymptotes of the function. When the function is undefined, we potentially but not necessarily have vertical asymptotes. On the other, if the limit of the function as \(x\) gets large is a number, this will give us horizontal asymptotes.

As always, I hope this helped you with your studying and that you learned something about asymptotes. If you learned something or enjoyed the post, please take the time to like the post or share it on Social Media.