Limits at infinity

What a limit at infinity means

Lessons 2.1.1 and 2.1.2 let $x$ close in on a number. Here $x$ runs off without end, and the question is where $f(x)$ heads. The symbol $\infty$ (infinity) is not a number: $x \to \infty$ is short for "$x$ grows without end", and $x \to -\infty$ for "$x$ falls without end".

Start with $\frac{1}{x}$. For $x = 10$, 100, 1000 and $10^6$ it is 0.1, 0.01, 0.001 and 0.000001. A fixed top over a bottom that keeps growing heads for 0.

For $x = -10$ and $-1000$ it is $-0.1$ and $-0.001$, so it heads for 0 from below. Both directions give the same limit, 0.

The same holds for any fixed number over a growing power of $x$. $\frac{7}{x^2}$ and $\frac{4}{\sqrt{x}}$ both tend to 0 as $x \to \infty$, because their bottoms grow without end.

If $f(x)$ itself grows without end, write $\lim_{x \to \infty} f(x) = \infty$, as for a blow-up in lesson 2.1.2. That says how the limit fails; it is not a number. For example $x^2 \to \infty$ in both directions.

Ratios of polynomials

For large $x$, only the highest power on top and on the bottom matters. Take $\dfrac{3x + 1}{x + 2}$. Both top and bottom grow without end, so neither part alone settles it:

The values head for 3: from below as $x$ grows, from above as $x$ falls. Algebra explains why. Divide top and bottom by $x$, the highest power on the bottom, and every leftover term has the form $\frac{\text{number}}{x}$, which tends to 0:

The same algebra gives the size of the miss: $3 - \frac{3x + 1}{x + 2} = \frac{5}{x + 2}$. At $x = 48$ the function is $\frac{5}{50} = 0.1$ below 3.

Equal, smaller and bigger top degree

The degree of a polynomial is its highest power (lesson 1.3.1), and the leading coefficient is the number in front of it. Three ratios with the same bottom, $2x^2 + 3$, show the three cases:

Divide each top and bottom by $x^2$. The bottom becomes $2 + \frac{3}{x^2}$, which tends to 2. The tops become $\frac{4}{x} + \frac{1}{x^2}$, then $4 + \frac{1}{x^2}$, then $4x + \frac{1}{x^2}$.

• Smaller top degree: the top tends to 0, so the ratio tends to $\frac{0}{2} = 0$.
• Equal degrees: the top tends to 4, so the ratio tends to $\frac{4}{2} = 2$, the ratio of the leading coefficients.
• Bigger top degree: the top grows without end, so the ratio does too. It behaves like $\frac{4x}{2} = 2x$, and at $x = 1000$ it is 1999.997001.

When the top degree is bigger, the sign depends on the direction. $\frac{x^2}{x + 1}$ behaves like $\frac{x^2}{x} = x$, so it tends to $\infty$ as $x \to \infty$ and to $-\infty$ as $x \to -\infty$.

A polynomial on its own is the case with bottom 1, and its leading term decides. $x^3 \to \infty$ as $x \to \infty$ and $x^3 \to -\infty$ as $x \to -\infty$: the two ends point opposite ways, as lesson 1.3.1 said of every cubic. $-2x^3 + 5x$ does the reverse.

Exponentials against powers

$e^x$ grows faster than any power of $x$. Lesson 1.4.2 showed $2^x$ overtaking $x^2$; the same happens with any base above 1 and any power. So a power over $e^x$ tends to 0, and $e^x$ over a power grows without end.

The race can take a while to decide. Here is $\frac{x^3}{e^x}$:

It rises at first: $x^3$ is ahead of $e^x$ between about $x = 1.857$ and $x = 4.536$. After that $e^x$ pulls away and the ratio heads for 0. A bigger power holds out longer: $\frac{x^{10}}{e^x}$ is about 454,000 at $x = 10$ but 0.00001884 at $x = 50$.

The same race settles exponential decay. $e^{-x} = \frac{1}{e^x}$, so $e^{-x} \to 0$ as $x \to \infty$; at $x = 10$ it is already 0.0000454. Likewise $e^x \to 0$ as $x \to -\infty$.

Logarithms lose the race to every power. $\ln x$ grows without end, but slowly: $\ln(10^6)$ is only 13.815511. So $\frac{\ln x}{x} \to 0$, and it is 0.00001382 at $x = 10^6$.

Horizontal asymptotes

When $f(x) \to L$ as $x \to \infty$ or as $x \to -\infty$, the graph levels off along the line $y = L$. That line is a horizontal asymptote.

A ratio of polynomials with equal degrees, like $\frac{3x + 1}{x + 2}$, has the same one in both directions. A smaller top degree gives the asymptote $y = 0$, and a bigger one gives none.

The two directions can give different asymptotes. Take $\frac{e^x}{1 + e^x}$, which equals $\frac{1}{1 + e^{-x}}$ after dividing top and bottom by $e^x$.

As $x \to \infty$, $e^{-x} \to 0$, so the value tends to $\frac{1}{1 + 0} = 1$. As $x \to -\infty$, $e^x \to 0$, so the first form tends to $\frac{0}{1 + 0} = 0$. Its values at $-10$, $-5$, 0, 5 and 10 are 0.000045, 0.006693, 0.5, 0.993307 and 0.999955.

Two more cases to watch for:

• A graph can cross its asymptote. $\frac{x}{x^2 + 1}$ has asymptote $y = 0$ in both directions, yet at $x = 0$ the graph sits on it and crosses.
• Some functions have no limit at infinity. $\sin x$ keeps swinging between $-1$ and 1. At $x = 10$, 100, 1000 and $10^6$ it is $-0.544$, $-0.5064$, 0.8269 and $-0.35$.

Limits as $n \to \infty$

A sequence (lesson 1.5.1) is a function of a whole number $n$, and its limit as $n \to \infty$ works the same way. So $\frac{2n + 1}{n} \to 2$, by the equal-degree rule, and $0.5^n \to 0$, as lesson 1.5.1 showed.

One limit needs care. In $\left(1 + \frac{1}{n}\right)^n$ the base tends to 1, but the power grows without end. Guessing "1 to any power is 1" is wrong:

The base shrinking towards 1 and the power growing pull against each other, and they settle on $e$, as in lesson 1.4.2. Case study 2.1.5 uses the same limit with interest: $\left(1 + \frac{r}{n}\right)^n \to e^r$.

Common mistakes

1. $\frac{\infty}{\infty}$ is not 1. When top and bottom both grow, compare their leading terms: $\frac{3x + 1}{x + 2} \to 3$.
2. Compare powers, not the size of the numbers. $\frac{x + 1000}{2x} \to \frac{1}{2}$, even though at $x = 10$ it is 50.5.
3. Leading terms win only once $x$ is large. $x^3 - 100x^2$ is $-125{,}000$ at $x = 50$ and 0 at $x = 100$. Likewise $\frac{x^3}{e^x}$ rises until $x = 3$ before it falls.
4. Check both directions. $x^3$ tends to $-\infty$ on the left, and $\frac{e^x}{1 + e^x}$ has a different asymptote on each side.
5. A graph can cross its horizontal asymptote. The asymptote describes the far ends only.
6. $1^\infty$ is not 1. $\left(1 + \frac{1}{n}\right)^n \to e$, not 1.