# Monotonicity of Limits and Sandwich Theorem

Taking limits respects weak inequalities and sandwiching a sequence by two converging sequences reveals its limit. The technique of sandwiching can be used to determine limits via nested intervals.

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Concept | Content |
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Logical Statements and Operations | Logic is the foundation to formulate proofs and to understand the language of mathematics. |

Convergence | Convergent sequences have a well-defined limit. |

Bounded Sequences | Sequences can be bounded from above and from below. The values of a bounded sequence cannot become arbitrarily large. |

Limit Theorems | Combining limits is a useful tool to deduce convergence and the limit of a more complicated sequence from the convergence of simpler building blocks. |

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Intermediate Value Theorem | This theorem tells us that continuous functions don't jump. They have to attain every value between two values in their image. |

Uncountability of the Reals | The real numbers cannot be enumerated. |

## Study *Monotonicity of Limits and Sandwich Theorem* #

**Theorem 1** (Monotonicity of limits). *Let \((a_n)_{n\in\mathbb{N}}\) and \((b_n)_{n\in\mathbb{N}}\) be two convergent
real sequences with \[\lim_{n\to\infty}a_n=a,\qquad
\lim_{n\to\infty}b_n=b.\] Further, assume that for all \(n\in\mathbb{N}\) holds \(a_n\leq b_n\). Then the following holds
true:*

*\(a\leq b\);**Sandwich-Theorem*: If \(a=b\) and \((c_n)_{n\in\mathbb{N}}\) is another sequence with \(a_n\leq c_n\leq b_n\) for all \(n \in \mathbb{N}\), then \((c_n)_{n\in\mathbb{N}}\) is convergent with \[\lim_{n\to\infty}c_n=a=b.\]

*Proof.*

Consider the sequence of differences between \(b_n\) and \(a_n\), i.e., \((b_n-a_n)_{n\in\mathbb{N}}\). By the calculation rules for converging sequences, it suffices to show that \[b-a=\lim_{n\to\infty}(b_n-a_n)\geq0.\] Assume the converse statement, i.e., \(b-a<0\). Then, we have that both numbers \(a-b\) and \(b_n-a_n\) are positive and thus \[|a-b-(a_n-b_n)|=a-b+(b_n-a_n)>a-b.\] In particular, there exists no \(n\in\mathbb{N}\) such that \(|a-b-(a_n-b_n)|<\varepsilon\) for \(\varepsilon=a-b>0\). This is a contradiction to \(\lim_{n\to\infty}(b_n-a_n)=b-a\).

Again consider the sequence \((b_n-a_n)_{n\in\mathbb{N}}\) which is tending to zero according to the formulae for converging sequences. Further, consider the sequence \((c_n-a_n)_{n\in\mathbb{N}}\). Then we have for all \(n\in\mathbb{N}\) that \(0\leq c_n-a_n\leq b_n-a_n\). Let \(\varepsilon>0\). Since \(b_n-a_n\) is tending to zero, there exists some \(N\) such that for all \(n\geq N\) holds \(|b_n-a_n-0|<\varepsilon\). Due to \(0\leq c_n-a_n\leq b_n-a_n\), we can conclude that for \(n\geq N\) holds \[|c_n-a_n-0|=c_n-a_n\leq b_n-a_n= |b_n-a_n-0|<\varepsilon.\] This implies that \((c_n-a_n)_{n\in\mathbb{N}}\) is convergent with \(\lim_{n\to\infty}(c_n-a_n)=0\). Hence \(a=0+a=\lim_{n\rightarrow\infty} (c_n-a_n) + \lim_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty} c_n\).

◻

**Remark 2**. Since the modification of finitely many
sequence elements does not change the limits (take a closer look at the
definition of convergence, the statements of Theorem 1 can be slightly generalised by only
claiming that there exists some \(n_0\)
such that for all \(n\geq n_0\) holds
\(a_n\leq b_n\) (resp. for all \(n\geq n_0\) holds \(a_n\leq c_n\leq b_n\) in (ii)). In the
proof of (i), one has to replace the words “there exists no \(n\in\mathbb{N}\) such that” by “there
exists no \(n\geq n_0\) such that” and
in the proof of (ii) the number \(N\)
has to be replaced by \(\max\{N,n_0\}\).

**Attention 3**. From the fact that we have the strict
inequality \(a_n<b_n\), we cannot
conclude that the limits satisfy \(a<b\). To see this, consider the
sequences\((a_n)_{n\in\mathbb{N}}=(0,0,0,\ldots)\) and
\((b_n)_{n\in\mathbb{N}}=(\frac1n)_{n\in\mathbb{N}}\).
In this case, we have \(a=b=0\) though
the strict inequality \(a_n=0<\frac1n=b_n\) holds true for all
\(n\in\mathbb{N}\).

**Example 4**.

Consider \((\frac1{n^k})_{n\in\mathbb{N}}\) for some \(k\in\mathbb{N}\). We state two alternative ways to show that this sequence tends to zero. The first possibility is, of course, an argumentation as in statement (b) of the remarks on limit theorems. The second way to treat this problem is by making use of the inequality \[\frac1n\geq\frac1{n^k}>0.\] Since we know from example \(q^n\), the sequence \((\frac1n)_{n\in\mathbb{N}}\) tends to zero, statement (ii) of Theorem 1 directly leads to the fact that \((\frac1{n^k})_{n\in\mathbb{N}}\) also tends to zero.

Consider \((a_n)_{n\in\mathbb{N}}\) with \[a_n=\frac{2n^2+5n-1}{-5n^2+n+1}.\] Rewriting \[a_n=\frac{2+\frac5n-\frac1{n^2}}{-5+\frac1n+\frac1{n^2}},\] and using that both \((\frac1{n})_{n\in\mathbb{N}}\) and \((\frac1{n^2})_{n\in\mathbb{N}}\) tend to zero, we can apply the limit theorems to obtain that \[\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{2n^2+5n-1}{-5n^2+n+1}=\lim_{n\to\infty}\frac{2+\frac5n-\frac1{n^2}}{-5+\frac1n+\frac1{n^2}}=-\frac25.\]

Consider \((a_n)_{n\in\mathbb{N}}\) with \(a_n=\sqrt{n^2+1}-n\). At first glance, none of the so far presented results seem to help to analyse convergence of this sequence. However, we can compute \[\begin{aligned} a_n\,=&\sqrt{n^2+1}-n =\frac{(\sqrt{n^2+1}-n)(\sqrt{n^2+1}+n)}{\sqrt{n^2+1}+n}\\ =&\frac{n^2+1-n^2}{\sqrt{n^2+1}+n} =\frac{1}{\sqrt{n^2+1}+n}<\frac1n. \end{aligned}\] By Theorem 1, we now get that \(\lim_{n\to\infty}a_n=0\).

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