I need some help with figuring out the end behavior of a Rational Function. I really do not understand how you figure it out. I looked at this question:How do you determine the end behavior of a rational function? but it made me even more confused on how to figure out the end behavior. So I was wondering if anybody could help me out. I am in Highschool Honors Precalculus, so I am doing Algebra based calculus. For example, what is the end behavior of: $$f(x)=\frac{2x^2+2}{x^2+9} $$
Thanks in advance!
2 Answers
\begin{align}\frac{2x^2+2}{x^2+9}&=\frac{2(x^2+9)+2-2(9)}{x^2+9}\\&=\frac{2(x^2+9)-16}{x^2+9} \\ &=2 - \color{blue}{\frac{16}{x^2+9}} \end{align}
As $x$ is large, the term in blue will vanish, as the denominator have far bigger magnitude compared to the numerator. Hence the limit would be $2$.
Of course, not every rational number will have their numerator and denominator to have the same degree.
Try to write $$\frac{p(x)}{q(x)} = t(x) + \color{blue}{\frac{p_1(x)}{q_1(x)}}$$
where degree of $p_1(x)$ is smaller than degree of $q_1(x)$. As $x$ is huge, $\color{blue}{\frac{p_1(x)}{q_1(x)}}$ becomes close to $0$ and it will behave like $t(x)$.
The usual trick to find asymptotes as $x\to\infty$ or $x\to-\infty$ is to divide the numerator and denominator by the highest power of $x$ that appears in the denominator. In your case, this is $x^2$:
$$f(x)=\frac{2x^2+2}{x^2+9}=\frac{2+\frac2{x^2}}{1+\frac9{x^2}}.$$
Now as $x\to\pm\infty$, you can see that the terms $\frac2{x^2}$ and $\frac1{x^2}$ disappear, so we have
$$\lim_{x\to\pm\infty}f(x) = \lim_{x\to\pm\infty}\frac{2+\frac2{x^2}}{1+\frac9{x^2}} = \frac{2+0}{1+0} = 2.$$
In general, by the same reasoning, we see that if the numerator and denominator have the same highest power, then the function will approach a constant:
$$\lim_{x\to\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0} = \lim_{x\to\infty}\frac{a_n+\frac{a_{n-1}}x+\cdots+\frac{a_0}{x^n}}{b_n+\frac{b_{n-1}}x+\cdots+\frac{b_0}{x^n}} = \frac{a_0}{b_0}.$$
The intuition is that $x$ gets bigger, most of the terms in the expression "don't matter". In your example, when $x=1000$, say, the fraction is $\frac{2000002}{1000009}$, which is very close to $2$.
Edit: We have established above that $f(x)\to2$ as $x\to\infty$ or $x\to-\infty$. We can now look for other asymptotic behaviour. For a function $f(x) = \frac{g(x)}{h(x)}$, there will typically be a vertical asymptote near roots of $h(x)$, i.e., any point $x_0$ where $h(x_0)=0$.
In your case, the denominator of $f(x)$ is $x^2+9$. Solving for $x^2+9=0$, we see that there are no real solutions. (Observe that since $x^2 \geq 0$ for all $x$, it follows that $x^2+9 \geq 0+9 = 9$ for all $x$, so we can't have $x^2+9=0$.) Therefore, your function has no vertical asymptotes.
