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Let $A\subseteq \mathbb{R}$ be nonempty and bounded above. An upper bound $s$ equals $\sup A$ exactly when, for every $\epsilon >0$, some $a\in A$ satisfies $s-\epsilon <a\le s$. If no such member existed, $s-\epsilon$ would be a smaller upper bound. Conversely, the condition defeats every proposed upper bound $u<s$ by taking $\epsilon =s-u$. ^[ref].

For $(0,1)$ and $\epsilon =0.2$, $a=0.9$ is one witness: $0.8<0.9<1$. Choosing $a_n\in A$ with $s-1/n<a_n\le s$ produces a sequence converging to $s$ by the squeeze theorem. The choices need not form a monotone sequence.

A monotone sequence of real numbers has a finite limit of a sequence exactly when it is a bounded sequence. In the increasing case, only an upper bound needs to be supplied, because the first term already gives a lower bound; in the decreasing case, only a lower bound needs to be supplied. The limit of a sequence is the exact edge of the set of terms:The upper bound $M$ or lower bound $m$ need not itself be the limit of a sequence. A supremum or infimum identifies the sharp bound. ^[ref]

For example, $a_n=1-1/n$ is increasing and bounded above by one, and $b_n=1+1/n$ is decreasing and bounded below by one. Both have limit of a sequence one, although neither ever equals one. Their absolute errors are exactly $1/n$. They also bracket any sequence $x_n$ satisfying $a_n\le x_n\le b_n$: the squeeze theorem gives $x_n\to 1$, even if $(x_n)$ itself is not a monotone sequence.

The theorem is useful when a recurrence relation gives successive improvements and a bound, before the answer has been identified. It proves existence of a limit of a sequence; further algebra, often using continuity, identifies the answer. A finite plot illustrates the formulas but does not establish the infinite claim. Error bounds for bounded monotone sequences require additional quantitative information.

A real increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum. For the increasing case, the definition of supremum provides a term above $L-\epsilon$, and every later term remains between that term and $L$. Negation reduces the decreasing case to the increasing case. This elementary result is distinct from the measure-theoretic monotone convergence theorem for integrals.

The existence of the real supremum in this short proof uses the least-upper-bound property, equivalently completeness of the real numbers. The approximation property of a supremum supplies $a_N>L-\epsilon$, and monotonicity keeps every later term within the same tolerance:The corresponding decreasing statement follows from the approximation property of an infimum. The converse uses the fact that every convergent sequence is a bounded sequence. ^[ref]

For every real $x>0$,For a short proof, put $I(x)={\int }_x^{\infty }{e}^{-t^2}dt$. Integration by parts givesThe last integral is positive and strictly smaller than $I(x)/x^2$. Using these two inequalities gives the upper and lower bounds after multiplication by $2/\sqrt{\pi }$. Dividing both bounds by $e^{-x^2}/(\sqrt{\pi }x)$ also proves the leading asymptotic equivalence by the squeeze theorem. For small positive $x$, the bounds remain true but can be loose; they are especially useful when the Gaussian integral tail is already small.

Start at one and apply $n$ updates, each multiplying the current value by $1+1/n$. The final value is $(1+1/n{)}^n$: one update gives $2$, two give $2.25$, ten give approximately $2.59374$, and one hundred give approximately $2.70481$. Increasing the number of steps, while reducing each increment proportionally, approaches Euler's number rather than producing unbounded accumulation.

The natural logarithm makes the limit transparent. For $u>0$, its integral definition gives $u/(1+u)\le \ln (1+u)\le u$, by bounding $1/t$ on $[1,1+u]$. With $u=1/n$,The squeeze theorem gives a logarithmic limit of one; continuity of the exponential function gives $(1+1/n{)}^n\to e$. This links a discrete multiplicative rule with continuous exponential growth, without assuming that a finite number of steps equals the limiting value.

For an increasing sequence converging to $L$, its absolute error is $e_n=L-a_n\ge 0$ and is nonincreasing. For a decreasing sequence, the absolute error is $e_n=a_n-L\ge 0$ and is also nonincreasing. This tells us that later terms cannot be worse, but does not supply a computable stopping index for a chosen tolerance.

Suppose an increasing lower approximation $a_n$ and a decreasing upper approximation $b_n$ enclose the same limit of a sequence $L$. Their observable gap $w_n=b_n-a_n$ givesIf $w_n\to 0$, the squeeze theorem certifies a common limit of a sequence; the midpoint has a certified absolute error at most half the gap. This is the principle behind the bisection method. In the symmetric example $a_n=1-1/n$, $b_n=1+1/n$, the gap is $2/n$, each endpoint's exact absolute error is $1/n$, and the midpoint happens to equal one exactly. The bound is valid even when such symmetry is absent.

A fixed upper bound by itself supplies less. If $a_n\uparrow L$ and $a_n\le M$ for all $n$, then $L\le M$ and $0\le L-a_n\le M-a_n$, but the last gap need not vanish. Taking $a_n=1-1/n$ and $M=2$ makes $M-a_n\to 1$. A shrinking bracket or an explicit remainder formula adds information that qualitative convergence does not provide.

On the geometric progression $x_i=(1+ϵ{)}^i$, with $ϵ>0$, the interval widths are $ϵ(1+ϵ{)}^i$. The left-endpoint Riemann sum for the decreasing integrand $x^{-1-\alpha }$ is the geometric seriesThe right-endpoint sum is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$. Integral comparison on each interval gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds tend to $1/\alpha$ as $ϵ\to 0^{+}$, proving the improper integral by the squeeze theorem. Although the largest interval width is unbounded on the infinite domain, these explicit upper and lower bounds justify the limit without incorrectly applying a finite-interval mesh argument to the entire half-line.

The limit inferior and limit superior retain the lower and upper long-run behaviour of a real sequence, even when it has no ordinary limit of a sequence. For each tail of a sequence beginning at $N$, defineDiscarding more initial terms raises or preserves the infimum and lowers or preserves the supremum. Hence $(l_N)$ is nondecreasing and $(u_N)$ is nonincreasing. Their limits exist in the extended real number line:For a bounded sequence, both values are finite and $\mathrm{liminf}{a}_n\le \mathrm{limsup}{a}_n$. These bounds use the entire infinite tail of a sequence, rather than the last few observed terms. ^[ref].

Take the original example $a_n=(-1{)}^n(1+1/n)$, beginning $-2,3/2,-4/3,5/4,\dots$. The subsequence at even integers decreases toward one, while the subsequence at odd integers increases toward minus one. If $e_N$ is the first even integer at least $N$ and $o_N$ is the first odd integer at least $N$, thenThus $\mathrm{liminf}{a}_n=-1$ and $\mathrm{limsup}{a}_n=1$. The finite initial extreme $a_1=-2$ disappears from later lower bounds; neither long-run value is attained by any term of this example.

For a bounded sequence, the limit inferior is the smallest subsequential limit and the limit superior is the largest subsequential limit; both are reached by suitable convergent subsequences. Every other subsequential limit lies between them. In the example there are only two subsequential limits, $-1$ and $1$: a convergent subsequence cannot retain infinitely many terms of both parities, because those two branches stay separated. The whole interval between the two bounds need not consist of subsequential limits. ^[ref].

A finite ordinary limit of a sequence exists exactly when the limit inferior and limit superior agree at the same finite value $L$. The bounds $l_N\le a_N\le u_N$ and the squeeze theorem prove the sufficient direction. For a bounded sequence, its tail diameter of a sequence is $D_N=u_N-l_N$, so the limiting gap is zero exactly when the Cauchy sequence condition holds. In the example the gap approaches two, not zero. The same framework allows infinite bounds: $a_n=n$ has both values $+\infty$, whereas $a_n=(-1{)}^{n}n$ has limit inferior $-\infty$ and limit superior $+\infty$. Equality at infinity describes escape rather than convergence to a real value. ^[ref].

If $|h(x)|\le C$ near $c$ and $a(x)\to 0$ as $x\to c$, then $a(x)h(x)\to 0$ by the squeeze theorem, even if $h$ has no limit. Defining the product to be zero at $c$ gives continuity there. For example, $\sqrt{|x|}\sin (1/\sin x)$ extends continuously at zero; the same oscillation near a nonzero multiple of $\pi$ has nonvanishing amplitude and does not have a continuous zero extension.

For a partition $a=x_0<\cdots <x_k=b$ and tags ${\xi }_j\in [x_{j-1},x_j]$, form the Riemann sum ${\sum }_{j=1}^{k}f({\xi }_j)(x_j-x_{j-1})$. The Riemann integral is the unique real number $I$ such that for every $\epsilon >0$ there is $\delta >0$ for which every tagged partition with mesh ${\max }_j(x_j-x_{j-1})<\delta$ has its sum within $\epsilon$ of $I$. Then $I={\int }_a^{b}f(x)dx$. The stated continuity ensures existence, which need not be proved here.

The properties required are linearity, the integral of a constant, and monotonicity of the Riemann integral. Monotonicity follows directly because a nonnegative function has nonnegative Riemann sums, and therefore a nonnegative integral. Integrating $f-m\ge 0$ and $M-f\ge 0$ givesSince $g\ge 0$, multiplication preserves the pointwise inequalities: $mg\le fg\le Mg$. The continuous products are integrable, so the same monotonicity and linearity giveNeither $m$ nor $f$ needs to be nonnegative for this argument.

For the first limit, set $r_n=n^{-1/2}$, $g_n(x)=n{e}^{-nx}$, and ${\epsilon }_n={\max }_{0\le x\le r_n}|f(x)-f(0)|$. By continuity at zero, ${\epsilon }_n\to 0$. On this shrinking interval $f(0)-{\epsilon }_n\le f(x)\le f(0)+{\epsilon }_n$, andApplying the weighted bounds just proved givesBoth bounding expressions tend to $f(0)$, so the squeeze theorem proves the first limit.

For the full interval, continuity on $[0,1]$ gives a finite bound $B={\max }_{[0,1]}|f|$ by the extreme value theorem. The remaining tail has absolute value at mostAdding this to the already evaluated near-end integral provesThese are instances of a one-sided exponential approximate identity: the positive weight has total mass tending to one, concentrated in a shrinking neighborhood of zero. No interchange of limit and integral has been assumed.

A real function is differentiable at an interior point $x$ when the finite limitexists, with $h$ restricted so $x+h$ lies in its domain. In quantified form, for every $\epsilon >0$ there is $\delta >0$ such that $0<|h|<\delta$ implies that the difference quotient differs from $f^{\prime }(x)$ by less than $\epsilon$.

To prove differentiability implies continuity, the convergent difference quotient is bounded by $|f^{\prime }(x)|+1$ for sufficiently small nonzero $h$. Thus $|f(x+h)-f(x)|\le (|f^{\prime }(x)|+1)|h|\to 0$. This is exactly continuity at $x$.

At zero, the particular function has difference quotient $h\sin (1/h)$. The bound $|h\sin (1/h)|\le |h|$ and the squeeze theorem giveFor $x\ne 0$, the product rule and chain rule giveAt $x_k=1/(2\pi k)$ this equals $-1$, although $x_k\to 0$ and $f^{\prime }(0)=0$. At $y_k=1/((2k+1)\pi )$ it equals $+1$. Hence the derivative has no limit at zero and is not continuous there. This demonstrates that differentiability does not imply continuity of the derivative, even though the function itself is continuous.

Use integration by parts on ${\sin }^{n-1}x\sin x$, for $n\ge 2$:Both boundary terms are zero. Therefore the Wallis integrals satisfyStarting with $I_0=\pi /2$ and $I_1=1$, repeated application givesThese formulas also hold at $n=0$ by empty products.

For $0<x<\pi /2$, $0<\sin x<1$, so $0<{\sin }^{n}x<{\sin }^{n-1}x$ when $n\ge 1$. Integrating the strict inequalities proves $0<I_n<I_{n-1}$. In particular $I_{2n+1}/I_{2n}<1$. For $n\ge 1$, the recurrence and $I_{2n-1}>I_{2n}$ giveThusAt $n=0$ the same displayed bounds are $0<2/\pi <1$, also true directly from $I_1<I_0$.

The squeeze theorem now gives $I_{2n+1}/I_{2n}\to 1$. Substituting the evaluated integrals showsand henceThis is twice the finite-product form of the Wallis product; the proof used only the integral recurrence and strict comparison of its integrands.

Write $Q(x)=1-\Phi (x)={\int }_x^{\infty }\varphi (t)dt$, where $\varphi (t)=(2\pi {)}^{-1/2}{e}^{-t^2/2}$ is the standard normal density. For $x>0$, ${\varphi }^{\prime }(t)=-t\varphi (t)$ and integration by parts yieldThe boundary term at infinity vanishes. The remainder is nonnegative and is at most $Q(x)/x^2$, soDividing by the positive quantity $\varphi (x)/x$ gives lower and upper bounds $x^2/(x^2+1)$ and $1$. The squeeze theorem proves the Mills ratio asymptotic

For a bounded real function on $[a,\infty )$, convergence of its improper integral means that its restriction to every $[a,b]$, $b>a$, is Riemann integrable and that the finite limitexists. Boundedness by itself does not guarantee either local Riemann integrability or existence of this limit.

A monotone function on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower Darboux sums differ byFor nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to 0$. For every $\epsilon >0$ one can choose $m$ to make this gap smaller than $\epsilon$; the Riemann integrability criterion proves that $f$ is Riemann integrable.

Now let $f$ be decreasing on $[1,\infty )$ with limit zero. For each $x$, $f(x)\ge {\lim }_{y\to \infty }f(y)=0$. On $[n,n+1]$ its monotonicity givesSumming these bounds, with $S_N={\sum }_{n=1}^{N}f(n)$, yieldsBoth the partial sums $S_N$ and the truncated integrals are nondecreasing because $f\ge 0$. The inequalities show that one family is bounded above if and only if the other is. The bounded monotone sequence theorem then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the squeeze theorem supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the integral test for convergence in the required form.

For $\alpha <0$, use $f(x)=x^{\alpha }$, which is decreasing and tends to zero. Its truncated improper integral isThis has a finite limit exactly when $\alpha <-1$. If $\alpha \ge 0$, the terms $n^{\alpha }$ do not tend to zero, so the term test for divergence applies. Thus the full p-series criterion isFor every $\alpha \ge -1$ its nonnegative partial sums tend to $+\infty$.

Since $f(0)=0$, its difference quotient isThe absolute value of cosine is at most $1$, so this quotient lies between $0$ and $|h|$. The squeeze theorem proves

Since $0<a\le b$, we have $b^n\le a^n+b^n\le 2b^n$. Taking positive $n$th roots yieldsFor every $\epsilon >0$, the binomial theorem gives $(1+\epsilon {)}^n\ge 1+n\epsilon >2$ eventually, so $2^{1/n}\to 1$. The squeeze theorem therefore gives

Away from integer multiples of $\pi$, the function is a composition of continuous functions. At zero, the bound $|g(x)|\le \sqrt{|x|}$ gives $g(x)\to 0=g(0)$ by the squeeze theorem. This is oscillation multiplied by a vanishing amplitude.

Fix $n\ne 0$ and set $t_k=\pi /2+2\pi k$ and $x_k=n\pi +(-1{)}^n\arcsin (1/t_k)$. Then $x_k\to n\pi$, $\sin x_k=1/t_k$, andSo $g$ is discontinuous at every nonzero integer multiple of $\pi$. Its set of continuity points is

On $[0,\pi ]$, $g(x)\le \sqrt{x}<\sqrt{\pi }$ for $x<\pi$, while $g(\pi )=0$. With $x_k=\pi -\arcsin (1/t_k)$ as above, $g(x_k)=\sqrt{x_k}\to \sqrt{\pi }$. Consequently the supremum is $\sqrt{\pi }$ and is not attained. This illustrates that supremum can fail to be attained at an oscillatory endpoint.

On $[\pi ,3\pi /2]$, set $x_0=\pi +\arcsin (2/(3\pi ))$. It lies strictly inside the interval, and $1/\sin x_0=-3\pi /2$, so $g(x_0)=\sqrt{x_0}>\sqrt{\pi }$. Choose $0<\delta <x_0-\pi$. On $[\pi ,\pi +\delta ]$,On $[\pi +\delta ,3\pi /2]$, $g$ is a continuous function, so the extreme value theorem gives a maximum there, at least $g(x_0)$. It exceeds every value on the discarded interval. Therefore $g$ does attain its supremum on $[\pi ,3\pi /2]$. This is the interior-value criterion for an attained maximum.

For $0<x<\pi /2$, the usual unit-circle area comparison, with angles measured in radians, gives $\sin x\le x\le \tan x$. HenceThe cosine is a continuous function at zero, so the squeeze theorem yields the right-hand limit of a function $1$. The quotient $\sin x/x$ is even, so the left-hand limit of a function is the same. Thus the limit is $1$.

For $x>0$, the exponent $x/(1+x)$ lies in $(0,1)$, and $0\le {\cos }^{4}x\le 1$. ConsequentlyThe last inequality uses the positive-term power series $e^x={\sum }_{k\ge 0}{x}^k/k!\ge x^2/2$ for $x>0$. The final bound tends to zero, so the squeeze theorem gives limit $0$.

We first prove by induction thatIt holds for $n=1$, and if it holds at $n$, thenConsequentlyPart (c) and the squeeze theorem now give

Because $(a_n)$ is a monotone sequence, every one of its first $n$ terms is at most $a_n$, soOn the other hand, every term from $a_n$ through $a_{2n}$ is at least $a_n$, and henceBoth the lower bound and upper bound tend to $x$ because the Cesàro means converge to $x$. The squeeze theorem gives