limit

Limit

The limit of a convergent sequence $(a_n{)}_{n\in \mathbb{N}}$ or a function $f$ describes the value that the sequence/function approaches as the index/argument grows (or approaches some point).

Limit arithmetic

Let $(a_n),(b_n)$ be convergent sequences with $a:={\lim }_{n\to \infty }{a}_n$ and $b:={\lim }_{n\to \infty }{b}_n$, and let $\lambda \in \mathbb{C}$. Then:

$$\begin{array}{rl}\underset{n\to \infty }{\lim }(a_n+b_n)& =a+b\\ \underset{n\to \infty }{\lim }(\lambda \cdot a_n)& =\lambda \cdot a\\ \underset{n\to \infty }{\lim }(a_n\cdot b_n)& =a\cdot b\\ \underset{n\to \infty }{\lim }\frac{a_n}{b_n}& =\frac{a}{b}\text{if }b\ne 0\end{array}$$

The same rules hold for function limits ${\lim }_{x\to c}f(x)$.

Proofs

Sum rule: Use the triangle inequality and “epsilon budgeting”:

$$|a_n+b_n-(a+b)|\le |a_n-a|+|b_n-b|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$$

for $n$ large enough. Each term gets half the epsilon budget.

Product rule: Add and subtract $a{b}_n$:

$$|a_n{b}_n-ab|=|a_n{b}_n-a{b}_n+a{b}_n-ab|\le |b_n||a_n-a|+|a||b_n-b|$$

Since convergent sequences are bounded, $|b_n|\le C$ for some $C$. Both terms can be made arbitrarily small.

Scalar rule: Follows directly from $|\lambda a_n-\lambda a|=|\lambda ||a_n-a|$.

Quotient rule: Show $\frac{1}{b_n}\to \frac{1}{b}$ first (using $|b_n|>\frac{|b|}{2}$ for large $n$), then apply the product rule.

Arithmetic for definitely divergent sequences

If $a_n\to \infty$:

$a_n+b_n\to \infty$ when $b_n\to \infty$ or $b_n\to b\in \mathbb{R}$
$\frac{\alpha }{a_n}\to 0$ for any $\alpha \in \mathbb{R}$
$\alpha \cdot a_n\to \infty$ if $\alpha >0$, and $\to -\infty$ if $\alpha <0$

Indeterminate forms (no general rule):
$\infty -\infty$
$\frac{\infty }{\infty }$
$0\cdot \infty$

Example: $a_n=(-1{)}^n/n\to 0$ and $b_n=n\to \infty$, but $a_n\cdot b_n=(-1{)}^n$ diverges, so since $0\cdot \infty$, it is indeterminate.

See sequence, null sequence, convergence for some examples.
See also: sandwich rule

Limes superior and limes inferior

Let $(a_n{)}_{n\in \mathbb{N}}\subset \mathbb{R}$. Define

$$\underset{n\to \infty }{\mathrm{limsup}}{a}_n:=\underset{n\to \infty }{\lim }\left(\underset{k\ge n}{\sup }{a}_k\right),\underset{n\to \infty }{\mathrm{liminf}}{a}_n:=\underset{n\to \infty }{\lim }\left(\underset{k\ge n}{\inf }{a}_k\right)$$

These always exist (in $\mathbb{R}\cup \{-\infty ,+\infty \}$) for any real-valued sequence, even divergent ones.
$\mathrm{liminf}$ / $\mathrm{limsup}{a}_n$ are the smallest / largest accumulation points.
They generalize the ordinary limit: every sequence has a $\mathrm{limsup}$ and $\mathrm{liminf}$, but only convergent sequences have a limit. When the limit exists, all three are equal.

Unpacking the definition step by step, for $\mathrm{limsup}$

1. Fix an index $n$. Look at all terms from $n$ onward: $a_n,a_{n+1},a_{n+2},\dots$
2. Take the supremum of those terms: $s_n:={\sup }_{k\ge n}{a}_k$. This is the largest value the sequence reaches from position $n$ onward.
3. As $n$ increases, we’re taking the sup over a smaller set (we dropped $a_n$), so $s_n$ can only decrease or stay: $s_{n+1}\le s_n$. The sequence $(s_n)$ is non-increasing.
4. A non-increasing sequence always has a limit (by the monotonicity principle: it either converges if bounded below, or $\to -\infty$). That limit is $\mathrm{limsup}{a}_n$.

Same construction flipped for $\mathrm{liminf}$: $i_n:={\inf }_{k\ge n}{a}_k$ is non-decreasing (dropping terms can only raise the inf), so its limit always exists too.

Compare with convergence (figure): Limsup/inf give the eventual bounds of the sequence, whereas convergence, if it exists, is the single value a sequence approaches.

Equivalent formulation

Since $s_n={\sup }_{k\ge n}{a}_k$ is non-increasing, the monotonicity principle gives $\lim s_n=\inf \{s_n\}$:

$$\underset{n\to \infty }{\mathrm{limsup}}{a}_n=\underset{n\in \mathbb{N}}{\inf }\underset{k\ge n}{\sup }{a}_k,\underset{n\to \infty }{\mathrm{liminf}}{a}_n=\underset{n\in \mathbb{N}}{\sup }\underset{k\ge n}{\inf }{a}_k$$

Eventually, all terms are near $[\mathrm{liminf}{a}_n,\mathrm{limsup}{a}_n]$

For any $\epsilon >0$, eventually:

$$\mathrm{liminf}{a}_n-\epsilon \le a_n\le \mathrm{limsup}{a}_n+\epsilon$$

Individual terms can still lie outside the interval itself: $a_n=(-1{)}^n(1+\frac{1}{n})$ always overshoots $[-1,1]$ slightly. But the overshoot gets arbitrarily small.

Convergence criterion

$$\underset{n\to \infty }{\lim }{a}_n\text{ exists}⟺\underset{n\to \infty }{\mathrm{liminf}}{a}_n=\underset{n\to \infty }{\mathrm{limsup}}{a}_n=\underset{n\to \infty }{\lim }{a}_n$$

Equivalently: A bounded sequence converges iff it has exactly one accumulation point.

$a_n=(-1{)}^n$

${\sup }_{k\ge n}{a}_k=1$ for all $n$ (there’s always an even index ahead), so $\mathrm{limsup}{a}_n=1$.
${\inf }_{k\ge n}{a}_k=-1$ for all $n$ (there’s always an odd index ahead), so $\mathrm{liminf}{a}_n=-1$.

$\mathrm{limsup}\ne \mathrm{liminf}$, so $(a_n)$ diverges. The two values are the accumulation points.

$a_n=(-1{)}^n+\frac{1}{n}$

Terms: $0,\frac{3}{2},-\frac{2}{3},\frac{5}{4},-\frac{4}{5},\dots$

${\sup }_{k\ge n}{a}_k$: the even-indexed terms are largest, and $a_{2k}=1+\frac{1}{2k}$, so ${\sup }_{k\ge n}{a}_k\to 1$. Hence $\mathrm{limsup}{a}_n=1$.

${\inf }_{k\ge n}{a}_k$: the odd-indexed terms are smallest, and $a_{2k-1}=-1+\frac{1}{2k-1}$, so ${\inf }_{k\ge n}{a}_k\to -1$. Hence $\mathrm{liminf}{a}_n=-1$.

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What is $\mathrm{limsup}{a}_n$ conceptually?

The largest value a sequence revisits infinitely many times, the smallest asymptotic (eventual) upper bound on the sequence (can be violated finitely many times).
The largest accumulatiion point.

What property do sequences for which $\mathrm{limsup}$ and $\mathrm{liminf}$ coincide have?

They are convergent.

Explain why we take the sup of the tail of a sequence and then the limit: $$

\limsup_{n \to \infty} a_n := \lim_{n \to \infty} \left(\sup_{k \ge n} a_k\right), \qquad \liminf_{n \to \infty} a_n := \lim_{n \to \infty} \left(\inf_{k \ge n} a_k\right)

$\sup_{k\ge n} a_k$ is the highest the sequence reaches from index $n$ onward. Letting $n\to\infty$ discards ever-larger *finite* prefixes, so only *eventual* behaviour survives.

What question does $\mathrm{limsup}$ answer that $\lim$ can't?

$\lim$ asks “does the whole sequence settle on one value?”, which might not have an answer.
$\mathrm{limsup}$ asks “what’s the highest value it keeps returning to?”, which always has an answer.

Give the definition of $\mathrm{limsup}$, $\mathrm{liminf}$

$$\underset{n\to \infty }{\mathrm{limsup}}{a}_n:=\underset{n\to \infty }{\lim }\left(\underset{k\ge n}{\sup }{a}_k\right),\underset{n\to \infty }{\mathrm{liminf}}{a}_n:=\underset{n\to \infty }{\lim }\left(\underset{k\ge n}{\inf }{a}_k\right)$$

Give an intuitive explanation for why $s_n:={\sup }_{k\ge n}{a}_k$ is non-increasing.

Removing an element from a set can not increase its upper bound (nor can it decrease the lower bound).

When are bounded sequences convergent? (3 different characterizations)

When they have only a single accumulation point.
Equivalenty:

$$\underset{n\to \infty }{\lim }{a}_n\text{ exists}⟺\underset{n\to \infty }{\mathrm{liminf}}{a}_n=\underset{n\to \infty }{\mathrm{limsup}}{a}_n$$

Equivalently: When they are monotone.

$\mathrm{limsup}$ and $\mathrm{liminf}$ of $a_n=(-1{)}^n(1+\frac{1}{n})$

$\mathrm{limsup}{a}_n=1$, $\mathrm{liminf}{a}_n=-1$. Note the terms sit outside $[-1,1]$ (each overshoots), yet the overshoot $\to 0$, so the limsup/liminf still land on $\pm 1$; since they differ, $(a_n)$ diverges, and $\pm 1$ are its accumulation points.

$\mathrm{limsup}$ and $\mathrm{liminf}$ of $a_n=(-1{)}^{n}n$

$\mathrm{limsup}{a}_n=+\infty$, $\mathrm{liminf}{a}_n=-\infty$.
Reminder: these live in the extended reals $\mathbb{R}\cup \{\pm \infty \}$, unlike an ordinary limit.