Rational Functions and Vertical Asymptotes

End behavior describes a rational function far from the origin. This topic looks at the places where it blows up: inputs where the denominator is $0$ and the outputs grow without bound. Those vertical lines are vertical asymptotes, and the way to describe them precisely is with one-sided limits.

Where vertical asymptotes come from

Near an input $a$ where the denominator of $r(x) = \frac{p(x)}{q(x)}$ is $0$ but the numerator is not, the outputs are a fixed number divided by a tiny number: they grow without bound in size. The vertical line $x = a$ is then a vertical asymptote.

When the numerator is $0$ at $a$ too, compare multiplicities. If $(x - a)$ divides the denominator more times than the numerator, there is still a vertical asymptote at $a$. If not, the factor cancels away and the outputs approach a number instead. The next topic studies that case.

One-sided limits

A one-sided limit describes the outputs as the input approaches $a$ from one side only. $\lim_{x \to a^-} f(x)$ means "as $x$ approaches $a$ from the left" (inputs less than $a$) and $\lim_{x \to a^+} f(x)$ "from the right". At a vertical asymptote each one-sided limit is $\infty$ or $-\infty$.

To find the sign, look at each factor just beside $a$.

When the leftover power of $(x - a)$ is even, it is positive on both sides, so both one-sided limits match. For $f(x)$$= \frac{x + 1}{(x - 2)^2}$: $\lim_{x \to 2^-} f(x) = \infty$ and $\lim_{x \to 2^+} f(x) = \infty$.

Writing it for a grader

Write both one-sided limits in full: "$\lim_{x \to 1^-} r(x) = -\infty$ and $\lim_{x \to 1^+} r(x) = \infty$." These lose the point: "$r(1) = \infty$" ($r(1)$ does not exist), "$r \to \infty$ at $1$" (no limit notation, no side), or a limit with the wrong side.

In context, translate both parts: $\lim_{V \to 0^+} P(V) = \infty$ for $P(V) = \frac{600}{V}$ says that as the volume shrinks toward $0$, the pressure grows without bound.

From a table

A table of inputs closing in on $a$ from each side shows the same thing. For $g(x) = \frac{x^2 + 1}{x - 1}$: $g(0.99) = -198.01$ and $g(0.999) = -1998.00$ on the left; $g(1.01) = 202.01$ and $g(1.001) = 2002.00$ on the right. So $\lim_{x \to 1^-} g(x) = -\infty$ and $\lim_{x \to 1^+} g(x) = \infty$.