Connect end behavior with asymptotes
A cup of tea cooling
The curve shows the temperature of a cup of tea, in degrees Celsius, t minutes after it is poured. The formula is \(T(t) = 20 + 60e^{-0.3t}\).
Drag the orange point to the right. The tea cools quickly at first and more slowly later.
At \(t = 5\) the temperature is about 33.4, at \(t = 10\) it is about 23.0, and at \(t = 20\) it is about 20.15.
The temperature comes close to 20, the temperature of the room. The toggle marks the height 20 with a dashed orange line.
The curve never reaches 20, but it comes as close to 20 as you like. A horizontal line that a curve approaches in this way is a horizontal asymptote.
Three ways to approach a line
The dashed orange line is at the height L, and the curve comes close to it as x grows.
Press the three buttons. The curve can approach the line from above, from below, or with waves that cross the line again and again.
In every case the heights of the curve come close to L, so their limit as x grows is L.
A horizontal asymptote depends only on this limit, so the curve may cross the line.
Notation
The line \(y = L\) is a horizontal asymptote when at least one of these limits equals L. The first limit describes the right end, and the second describes the left end.
\[\begin{aligned} \lim_{x \to \infty} f(x) &= \cy{L} \\ \lim_{x \to -\infty} f(x) &= \cy{L} \end{aligned}\]Check each end separately. Each end has at most one limit, so a curve has at most two horizontal asymptotes, and it can have none.
The limit must be a finite number. If the limit is infinite, that end gives no asymptote.
\[\lim_{x \to \infty} x^2 = \infty\]
When it fails
Take \(f(x) = \sin x\). Its heights move between \(-1\) and 1 forever and do not approach one number. The limit does not exist, so there is no horizontal asymptote.
Take \(f(x) = x^2\). Its heights grow without bound, so the limit is infinite and no horizontal line fits the curve.
Example 1: a different line on each end
Find the horizontal asymptotes of \(f(x) = \dfrac{5e^x}{e^x + 2}\).
Start with the right end. Divide the top and the bottom by \(e^x\), so that x appears only in the faint \(2e^{-x}\).
\[\frac{5e^x}{e^x + 2} = \frac{5}{1 + \cq{2e^{-x}}}\]As x grows, \(e^{-x}\) approaches 0, so the fraction approaches \(5/(1 + 0) = 5\).
\[\lim_{x \to \infty} f(x) = \cy{5}\]At the left end, x decreases without bound and \(e^x\) approaches 0, so the top approaches 0 and the bottom approaches 2.
\[\lim_{x \to -\infty} f(x) = \frac{0}{0 + 2} = \cy{0}\]Both limits are finite, so the horizontal asymptotes are y = 5 on the right and y = 0 on the left.
Example 2: waves that shrink
Find the horizontal asymptotes of \(g(x) = 3 + \dfrac{\cos x}{x}\).
The 3 is a constant, so the question is what \(\cos x / x\) does at each end. Near \(x = 0\) the curve leaves the figure at a vertical asymptote, which does not affect the ends.
\[g(x) = 3 + \cq{\frac{\cos x}{x}}\]A tempting wrong move: The function cos x has no limit, so g has no limit and no horizontal asymptote.
\[\cr{\text{no asymptote}}\]At \(x = 60\) the curve is within 0.02 of the height 3. Dividing by the growing x shrinks the waves of \(\cos x\), so g has a limit even though \(\cos x\) has none.
The size of \(\cos x\) is at most 1, so \(\cos x / x\) lies between the two bounds below. Both bounds approach 0, so \(\cos x / x\) approaches 0 too, by the squeeze theorem.
\[-\frac{1}{|x|} \le \frac{\cos x}{x} \le \frac{1}{|x|}\]The limit is 3 at both ends, so y = 3 is a horizontal asymptote. The curve crosses this line where \(\cos x = 0\), which the definition allows.
\[\lim_{x \to \pm\infty} g(x) = \cy{3}\]
Practice
Find all horizontal asymptotes of \(f(x) = 1 + 6e^x\).
Check each end of the picture. On which end does the curve approach a height?
Solution
Start with the left end. As x decreases without bound, \(e^x\) approaches 0, so f approaches 1 and the line y = 1 is a horizontal asymptote.
\[\lim_{x \to -\infty} (1 + 6e^x) = \cy{1}\]At the right end, \(e^x\) grows without bound, so f does too. This limit is not a finite number, so the right end gives no asymptote.
\[\lim_{x \to \infty} (1 + 6e^x) = \infty\]The only horizontal asymptote is y = 1.
Questions
The graph shows \(u(x) = 2 + \dfrac{5 \sin x}{x}\) for x from \(-30\) to 30.
Which line is a horizontal asymptote of the graph?
- y = 2
- y = 7
- y = 0
Answer
y = 2
The curve crosses the line y = 2 again and again. What does this show about y = 2?
- It is still a horizontal asymptote, because the heights approach 2
- It is not an asymptote, because a curve cannot cross its asymptote
- It is an asymptote on the right end only
Answer
It is still a horizontal asymptote, because the heights approach 2
How many horizontal asymptotes does \(f(x) = \dfrac{5x^3 + 1}{x^2 + 4}\) have?
- None
- One: y = 5
- One: y = 0
- Two: y = 5 and y = −5
Answer
None
Summary
You find the limit at each end, and each finite limit L gives the asymptote y = L.
This works because the curve comes as close to the line as you want, even if it crosses the line on the way.
Use it to say what a function does far to the right and far to the left.
This is one of four skills in topic 1.15 of Which & Why: AP Calculus.