Overview
Horizontal and vertical asymptotes are quick clues to a function’s behavior. A vertical asymptote marks an \(x\)-value where the function’s outputs blow up near that point—often signaling a domain break (like dividing by zero).
A horizontal asymptote describes what the function settles toward far out to the left or right—the long-run baseline of the graph.
Together, they let us sketch and reason about limits: vertical for “what happens near a trouble \(x\),” horizontal for “what happens as \(x\) goes to \(\pm \infty\).”
Vertical asymptotes
Definition. A vertical line \(x=a\) is a vertical asymptote of \(f\) if
as \(x\) approaches \(a\) from the left or right, the \(y\)-values grow without bound:
\(\displaystyle \lim_{x\to a^-} f(x)=\pm\infty\) or \(\displaystyle \lim_{x\to a^+} f(x)=\pm\infty\).
Let \(f(x)=\dfrac{x+5}{x+6}\). Find any vertical asymptote(s) and compute the corresponding one-sided limits. Examine the one-sided behavior:
\[
\lim_{x\to-6^-}\frac{x+5}{x+6}
= \frac{\text{negative}}{\text{negative small}}=+\infty,\qquad
\lim_{x\to-6^+}\frac{x+5}{x+6}
= \frac{\text{negative}}{\text{positive small}}=-\infty.
\]
There is a vertical asymptote at \(x=-6\); from the left, \(f(x)\to+\infty\), and from the right \(f(x) \to-\infty\). Let \(g(x)=\dfrac{3}{(x-2)^2}\). Describe the asymptote. Since the numerator is finite and the denominator is \(0\) at \(x=2\), there is a vertical asymptote at \(x=2\). Because of the square,
\((x-2)^2>0\) on both sides, so \(\displaystyle \lim_{x\to2^-}g(x)=\lim_{x\to2^+}g(x)=+\infty.\) Why doesn't \(\dfrac{x^2-1}{x-1}\) have a vertical asymptote at \(x=1\)? Factor: \(\dfrac{(x-1)(x+1)}{x-1} = x+1\) for \(x\ne1\). The common factor cancels, so \(x=1\) is a hole, not a vertical asymptote. The simplified function is the line \(y=x+1\) with a missing point at \(x=1\). \(\displaystyle \lim_{x\to1} \dfrac{x^2-1}{x-1} = 2\) exists and is finite. Let \(h(x)=\ln(x-3)\). Where is the vertical asymptote? Domain requires \(x-3>0\Rightarrow x>3\). As \(x\to3^+\), \(\ln(x-3)\to -\infty\), so there’s a vertical asymptote at \(x=3\) (approached from the right only). Let \(p(x)=\tan x\). What happens near \(x=\tfrac{\pi}{2}\)? \(\tan x=\dfrac{\sin x}{\cos x}\). Therefore, the zeros of \(\cos x\) give vertical asymptotes:
\(x=\tfrac{\pi}{2}+k\pi\). The one-sided behavior is:
\(\displaystyle \lim_{x\to(\pi/2)^-}\tan x=+\infty\),
\(\displaystyle \lim_{x\to(\pi/2)^+}\tan x=-\infty\).Example: simple rational function
Solution
Example: squared denominator
Solution
Example: is there an asymptote?
Solution
Example: logarithm
Example: tangent
Solution
Horizontal asymptotes
Definition. A line \(y=L\) is a horizontal asymptote if \(\displaystyle\lim_{x\to\infty}f(x)=L\) or \(\displaystyle\lim_{x\to-\infty}f(x)=L\). A function may have different horizontal asymptotes on the two ends and may cross them.
Example
\(\displaystyle f(x)=\frac{6x^2-1}{3x^2+2} \Rightarrow \lim_{x\to\pm\infty}f(x)=\frac{6}{3}=2\). Horizontal asymptote: \(y=2\) (both ends).
Rational functions
Example: Denominator degree bigger
\(\displaystyle \lim_{x\to\infty}\frac{3x^2-7}{5x^3+2} =\lim_{x\to\infty}\frac{3-7/x^2}{5x+2/x^3}=\lim_{x\to\infty}\frac{3-0}{5x+0}=0.\)
Example: Equal degrees
\(\displaystyle \lim_{x\to-\infty}\frac{4x^3-2x+9}{-2x^3+7x} =\frac{4}{-2}=-2.\)
Example: Numerator degree bigger
\(\displaystyle \lim_{x\to\infty}\frac{x^4-1}{2x^2+5}=\infty.\) (Here \(n=4>2=m\). If you long-divide you’ll see a quadratic “asymptote,” not horizontal.)
Example: Different horizontal asymptotes (left vs. right)
Let \[ f(x)=\frac{\sqrt{x^2+1}-x}{x} \;=\;\frac{\sqrt{x^2+1}}{x}-1. \] Note that \(\sqrt{x^2}=|x|\), so \[ \frac{\sqrt{x^2+1}}{x} =\frac{|x|}{x}\,\sqrt{\,1+\tfrac{1}{x^2}\,}. \]
As \(x\to+\infty\): \(\dfrac{|x|}{x}=1\) and \(\sqrt{1+\tfrac{1}{x^2}}\to1\), hence \[ \frac{\sqrt{x^2+1}}{x}\to 1 \quad\Rightarrow\quad \lim_{x\to+\infty}f(x)=1-1=0. \] Horizontal asymptote on the right: \(y=0\).
As \(x\to-\infty\): \(\dfrac{|x|}{x}=-1\) and \(\sqrt{1+\tfrac{1}{x^2}}\to1\), hence \[ \frac{\sqrt{x^2+1}}{x}\to -1 \quad\Rightarrow\quad \lim_{x\to-\infty}f(x)=-1-1=-2. \] Horizontal asymptote on the left: \(y=-2\).
Conclusion: \[ \boxed{\lim_{x\to+\infty} f(x)=0\;\Rightarrow\;y=0},\qquad \boxed{\lim_{x\to-\infty} f(x)=-2\;\Rightarrow\;y=-2}. \]
Radicals
Example: Classic “\(\sqrt{\cdot}-x\)” form
\(\displaystyle \lim_{x\to\infty}\big(\sqrt{x^2+5x}-x\big) =\lim_{x\to\infty}\frac{(x^2+5x)-x^2}{\sqrt{x^2+5x}+x} =\lim_{x\to\infty}\frac{5x}{\sqrt{x^2+5x}+x}=\frac{5}{2}.\) So \(y=\tfrac{5}{2}\) is a horizontal asymptote on the right.
Example: Ratio with a radical in the denominator
\(\displaystyle \lim_{x\to-\infty}\frac{x}{\sqrt{x^2+9}} =\lim_{x\to-\infty}\frac{x}{|x|\sqrt{1+9/x^2}} =\lim_{x\to-\infty}\frac{x}{(-x)\sqrt{1+0}}=-1.\)
Trig & exponential functions
Example: Bounded numerator over growing denominator
\(\displaystyle \lim_{x\to\infty}\frac{\sin(5x)}{x}=0\) since \(0\leq |\sin(5x)|\le 1\).
Example: Exponentials beat polynomials
\(\displaystyle \lim_{x\to\infty}\frac{e^{x}}{x^7}=\infty,\qquad \lim_{x\to\infty}x^5e^{-x}=0.\)