Lesson 1.12 · Unit 1 · CED topic 1.15
Connecting limits at infinity and horizontal asymptotes
Lesson 1.14 asked what happens when the output grows without bound. This lesson asks a different question: what happens to the output when the input grows without bound? If you let \(x\) become huge, positive or negative, does \(f(x)\) settle down? A limit at infinity describes the end behavior of a function, and the number it settles on gives a horizontal asymptote.
Learning objective.
- LIM-2.D. Interpret the behavior of functions using limits involving infinity.
Essential knowledge.
- LIM-2.D.3. The concept of a limit can be extended to include limits at infinity.
- LIM-2.D.4. Limits at infinity describe end behavior.
- LIM-2.D.5. Relative magnitudes of functions and their rates of change can be compared using limits.
Limits at infinity
We write \(\displaystyle\lim_{x\to\infty}f(x)=L\) if the values \(f(x)\) can be made as close to \(L\) as we like by taking \(x\) sufficiently large, and \(\displaystyle\lim_{x\to-\infty}f(x)=L\) if the same is true when \(x\) is large and negative. If such a limit is a real number \(L\), the line \(y=L\) is a horizontal asymptote of the graph. A function can have a different horizontal asymptote on each end, or one, or none.
The basic facts: \(\displaystyle\lim_{x\to\pm\infty}\frac{1}{x^n}=0\) for any positive \(n\), and \(\displaystyle\lim_{x\to\infty}e^{-x}=0\) (as does \(\displaystyle\lim_{x\to-\infty}e^{x}=0\)). Everything else comes from rewriting an expression so these facts apply.
Rational functions
To find \(\displaystyle\lim_{x\to\pm\infty}\frac{p(x)}{q(x)}\), divide the numerator and denominator by the highest power of \(x\) in the denominator. The result depends on the degrees:
| Degrees | Limit as \(x\to\pm\infty\) | Horizontal asymptote |
|---|---|---|
| degree of \(p\) less than degree of \(q\) | 0 | \(y=0\) |
| degrees equal | ratio of leading coefficients | \(y=\) that ratio |
| degree of \(p\) greater than degree of \(q\) | \(\pm\infty\) (unbounded) | none |
Find \(\displaystyle\lim_{x\to\infty}\frac{5x^2-3x+1}{2x^2+7}\).
Divide top and bottom by \(x^2\): \(\dfrac{5-\frac{3}{x}+\frac{1}{x^2}}{2+\frac{7}{x^2}}\). As \(x\to\infty\), \(\frac3x\to0\), \(\frac1{x^2}\to0\), \(\frac7{x^2}\to0\), so the limit is \(\dfrac{5}{2}\). The line \(y=\dfrac52\) is a horizontal asymptote.
Find \(\displaystyle\lim_{x\to\infty}\frac{4x+1}{x^2+3}\) and \(\displaystyle\lim_{x\to\infty}\frac{x^3+1}{2x^2-5}\).
The first has a lower degree on top. Dividing by \(x^2\): \(\dfrac{\frac4x+\frac1{x^2}}{1+\frac3{x^2}}\to\dfrac{0}{1}=0\). The second has a higher degree on top; dividing by \(x^2\) gives \(\dfrac{x+\frac1{x^2}}{2-\frac5{x^2}}\), whose numerator grows without bound while the denominator approaches \(2\), so the limit is \(+\infty\) (the function is unbounded, with no horizontal asymptote).
Square roots and the sign of \(x\)
When a square root of \(x^2\) appears, remember that \(\sqrt{x^2}=|x|\), which equals \(x\) for positive \(x\) and \(-x\) for negative \(x\). The two ends can then give different answers.
Find \(\displaystyle\lim_{x\to\infty}\frac{3x+2}{\sqrt{x^2+1}}\) and \(\displaystyle\lim_{x\to-\infty}\frac{3x+2}{\sqrt{x^2+1}}\).
Divide numerator and denominator by \(|x|\), since \(\sqrt{x^2+1}\approx|x|\) for large \(|x|\).
As \(x\to\infty\), \(|x|=x\): \(\dfrac{3+\frac2x}{\sqrt{1+\frac{1}{x^2}}}\to\dfrac{3}{1}=3\).
As \(x\to-\infty\), \(|x|=-x\), so dividing by \(|x|\) means dividing the numerator by \(-x\): \(\dfrac{-3-\frac2x}{\sqrt{1+\frac1{x^2}}}\to-3\).
The function has two different horizontal asymptotes, \(y=3\) on the right and \(y=-3\) on the left.
Comparing growth rates
Limits at infinity let us compare how fast functions grow: the relative magnitudes of functions and their rates of change. A short ranking, from slowest to fastest growth as \(x\to\infty\), is: logarithms, then powers (\(x^p\) with \(p\gt0\)), then exponentials (\(b^x\) with \(b\gt1\)). In limit form, for any positive \(n\) and \(p\),
\[\lim_{x\to\infty}\frac{\ln x}{x^p}=0\qquad\text{and}\qquad\lim_{x\to\infty}\frac{x^n}{e^x}=0.\]
So \(e^x\) eventually dwarfs every power of \(x\), and every power of \(x\) dwarfs \(\ln x\). When one function in a quotient dominates the other, the limit is \(0\) or \(\pm\infty\) accordingly.
Evaluate \(\displaystyle\lim_{x\to\infty}\frac{x^5}{e^x}\) and \(\displaystyle\lim_{x\to\infty}\frac{e^x}{x^5}\).
An exponential grows faster than any polynomial, so the denominator dominates in the first quotient: the limit is \(0\). The second is the reciprocal and is unbounded: the limit is \(+\infty\). (L'Hospital's Rule in Lesson 4.7 can prove these.)
- \(\displaystyle\lim_{x\to\infty}f(x)=L\) means the end behavior of \(f\) approaches \(L\); \(y=L\) is then a horizontal asymptote. Check both ends.
- Rational functions: compare degrees (lower: \(0\); equal: ratio of leading coefficients; higher: unbounded).
- Treat \(\sqrt{x^2}\) as \(|x|\); the left and right ends may give opposite signs.
- Growth order: \(\ln x\ll x^p\ll b^x\). Use it to decide limits of quotients at infinity.
- A function can cross a horizontal asymptote; the asymptote describes the far ends, not every point.
- Evaluate \(\displaystyle\lim_{x\to\infty}\dfrac{7x-2}{3x+5}\).
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The degrees are equal; the ratio of leading coefficients is \(\dfrac73\).
- Evaluate \(\displaystyle\lim_{x\to-\infty}\dfrac{2x^3-x}{x^3+4x^2}\).
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Divide by \(x^3\): \(\dfrac{2-\frac1{x^2}}{1+\frac4x}\to\dfrac21=2\).
- Evaluate \(\displaystyle\lim_{x\to\infty}\dfrac{x^2+1}{3x-1}\). Does the graph have a horizontal asymptote on the right?
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The degree on top is higher, so the quotient is unbounded: the limit is \(+\infty\). There is no horizontal asymptote on the right.
- Evaluate \(\displaystyle\lim_{x\to\infty}\left(3+e^{-2x}\right)\) and state the horizontal asymptote.
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\(e^{-2x}\to0\), so the limit is 3. The line \(y=3\) is a horizontal asymptote.
- Evaluate \(\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{\sqrt x}\) and explain in a sentence.
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The limit is 0. A power function \(x^{1/2}\) grows faster than \(\ln x\), so the denominator dominates.
- Find all horizontal asymptotes of \(f(x)=\dfrac{5x}{\sqrt{x^2+4}}\).
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As \(x\to\infty\), \(\sqrt{x^2+4}\approx x\) and the limit is 5. As \(x\to-\infty\), \(\sqrt{x^2+4}\approx|x|=-x\) and the limit is \(-5\). The asymptotes are \(y=5\) and \(y=-5\).