Limits at Infinity

Definition The notation $x \to \infty$ means that $x$ increases without bound, taking values larger than any given number. The notation $x \to -\infty$ means $x$ decreases without bound. Neither $\infty$ nor $-\infty$ is a real number, and $x$ never equals them. To talk about $x \to \infty$, the function must be defined on some interval $(c, \infty)$, and for $x \to -\infty$ on some interval $(-\infty, c)$.

Definition Let $f$ be defined on some interval $(c, \infty)$ and let $L$ be a real number. Then $\lim_{x \to \infty} f(x) = L$ means that for every $\varepsilon > 0$ there exists a number $M$ such that

For $f$ defined on some interval $(-\infty, c)$, the statement $\lim_{x \to -\infty} f(x) = L$ means that for every $\varepsilon > 0$ there exists $M$ such that $x < M$ implies $\lvert f(x) - L \rvert < \varepsilon$.

Definition When $\lim_{x \to \infty} f(x)$ or $\lim_{x \to -\infty} f(x)$ is a real number $L$, it is called a finite limit at infinity. Two basic ones are

The limit laws hold for limits at infinity as well. The limits at $\infty$ and at $-\infty$ are separate questions, and their answers can differ.

Definition Let $p(x) = a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0$ with $n \ge 1$ and $a_n \neq 0$. For $x \neq 0$,

and the bracket tends to $1$ as $x \to \pm\infty$. Hence

The term $a_n x^n$ is called the dominant (or leading) term.

Definition Let $r = p/q$, where $p$ has degree $m$ and leading coefficient $a \neq 0$, and $q$ has degree $n$ and leading coefficient $b \neq 0$. As $x \to \infty$ and as $x \to -\infty$,

1. if $m < n$, then $r(x) \to 0$,
2. if $m = n$, then $r(x) \to a/b$,
3. if $m > n$, then $r(x)$ has no finite limit, because $\lvert r(x) \rvert$ grows without bound.

To see this, divide the numerator and the denominator by $x^n$.

Definition The limit

exists, and the same limit holds as $x \to -\infty$. Here is why, starting from the fact that $\ln(1 + t)/t \to 1$ as $t \to 0$, from the lesson on special limits. For $x > 0$ put $t = 1/x$. As $x \to \infty$, $t \to 0$ through positive values, so

Now

and the exponent tends to $1$. To conclude that the power tends to $e^1 = e$ we need one more fact, that $e^u \to e^1$ whenever $u \to 1$. This is the continuity of the exponential function at $1$, which the continuity unit proves. As $x \to -\infty$ the same steps work with $x < -1$, where $1 + 1/x > 0$ and $t \to 0$ through negative values. Some texts instead define $e$ as this limit and prove its existence another way. Then the logarithm limit is derived from it, and the argument above runs in the opposite direction.