# Math for Non-Geeks/Limit theorems

Limit proofs using the $\epsilon$-definition are quite laborious. In this chapter we will study some limit theorems that simplifies matters.

## Limit theorems

The limit theorems are the following:
Theorem (Limit theorems)
Let $\left(a_{n}\right)_{n\in \mathbb {N} }$ and $\left(b_{n}\right)_{n\in \mathbb {N} }$ be two convergent sequences with $\lim _{n\rightarrow \infty }a_{n}=a$ and $\lim _{n\rightarrow \infty }b_{n}=b$. Let also $k\in \mathbb {N}$ be arbitrary. Then we have
Template:List
If furthermore $b\neq 0$ and $b_{n}\neq 0$ for all $n\in \mathbb {N}$, then we also have
* $\lim _{n\rightarrow \infty }{\frac {1}{b_{n}}}={\frac {1}{b}}$
* $\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}={\frac {a}{b}}$
For $k\in \mathbb {N}$ and $a_{n}\geq 0$ for all $n\in \mathbb {N}$:
$\lim _{n\rightarrow \infty }{\sqrt[{k}]{a_{n}}}={\sqrt[{k}]{a}}$
Warning
These rules can only be applied if the respective sequences converge. As soon as one of the involved sequences diverge, we cannot use the rules.
Also note that $\infty$ and $-\infty$ are not real numbers and therefore no valid limits. If, for instance, $\lim _{n\rightarrow \infty }a_{n}=\infty$, then $\left(a_{n}\right)_{n\in \mathbb {N} }$ is divergent and we cannot use the limit theorems.

## Monotonicity rule: limit estimates

We also have the following monotonicity rule, which can be used to estimate limits:
Theorem (Monotonicity rule)
Le $\left(a_{n}\right)_{n\in \mathbb {N} }$ and $\left(b_{n}\right)_{n\in \mathbb {N} }$ be two convergent sequences. If $a_{n}\leq b_{n}$ for almost all $n\in \mathbb {N}$, then the limits satisfy the inequality
$\lim _{n\rightarrow \infty }a_{n}\leq \lim _{n\rightarrow \infty }b_{n}$

## Example: Computing the limit of a sequence

Consider the following sequence
$x_{n}={\frac {5-{\sqrt[{n}]{23}}}{{\sqrt[{n}]{n}}+{\tfrac {1}{n^{2}}}}}$
This sequence is convergent. A proof using the $\epsilon$-definition would be rather complicated. Fortunately we can break the whole sequence apart into individual sequences where the convergence and limits are known. For example $\lim _{n\rightarrow \infty }{\sqrt[{n}]{23}}=1$. Using the limit theorems we can compute the limit step by step:
${\begin{aligned}\lim _{n\rightarrow \infty }x_{n}&=\lim _{n\rightarrow \infty }{\frac {5-{\sqrt[{n}]{23}}}{{\sqrt[{n}]{n}}+{\tfrac {1}{n^{2}}}}}\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{quotient rule}}\right.}\\[0.3em]&={\frac {\lim _{n\rightarrow \infty }5-{\sqrt[{n}]{23}}}{\lim _{n\rightarrow \infty }{\sqrt[{n}]{n}}+{\tfrac {1}{n^{2}}}}}\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{sum rule}}\right.}\\[0.3em]&={\frac {\left(\lim _{n\rightarrow \infty }5\right)+\left(\lim _{n\rightarrow \infty }-{\sqrt[{n}]{23}}\right)}{\left(\lim _{n\rightarrow \infty }{\sqrt[{n}]{n}}\right)+\left(\lim _{n\rightarrow \infty }{\tfrac {1}{n^{2}}}\right)}}\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{factor rule}}\right.}\\[0.3em]&={\frac {\left(\lim _{n\rightarrow \infty }5\right)-\left(\lim _{n\rightarrow \infty }{\sqrt[{n}]{23}}\right)}{\left(\lim _{n\rightarrow \infty }{\sqrt[{n}]{n}}\right)+\left(\lim _{n\rightarrow \infty }{\tfrac {1}{n^{2}}}\right)}}\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{calculate known limit}}\right.}\\[0.3em]&={\frac {5-1}{1+0}}=4\end{aligned}}$
In this manner we can show that $(x_{n})_{n\in \mathbb {N} }$ is convergent and the limit is $4$. Unfortunately this derivation is flawed: We are applying the limit theorems before we showed that the individual sequences are convergent. That those sequences do indeed converges becomes clear only after we have already employed the limit theorems. Therefore the above is not a valid proof. Instead we could proceed as follows:
${\begin{array}{ll}&{\begin{aligned}\lim _{n\rightarrow \infty }5=5&\land \lim _{n\rightarrow \infty }{\sqrt[{n}]{23}}=1\land \lim _{n\rightarrow \infty }{\sqrt[{n}]{n}}=1\\[0.3em]&\land \lim _{n\rightarrow \infty }{\tfrac {1}{n^{2}}}=0\end{aligned}}\\[0.7em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{factor rule}}\right.}\\[0.7em]\implies &{\begin{aligned}\lim _{n\rightarrow \infty }5=5&\land \lim _{n\rightarrow \infty }-{\sqrt[{n}]{23}}=-1\land \lim _{n\rightarrow \infty }{\sqrt[{n}]{n}}=1\\[0.3em]&\land \lim _{n\rightarrow \infty }{\tfrac {1}{n^{2}}}=0\end{aligned}}\\[0.7em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{sum rule}}\right.}\\[0.7em]\implies &\lim _{n\rightarrow \infty }5-{\sqrt[{n}]{23}}=5-1=4\land \lim _{n\rightarrow \infty }{\sqrt[{n}]{n}}+{\tfrac {1}{n^{2}}}=1+0=1\\[0.7em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{quotient rule}}\right.}\\[0.7em]\implies &\lim _{n\rightarrow \infty }{\frac {5-{\sqrt[{n}]{23}}}{{\sqrt[{n}]{n}}+{\tfrac {1}{n^{2}}}}}={\frac {4}{1}}=4\end{array}}$
We start with the sequences of which we know that they converge. By applying the limit theorems in reverse order we can derive the convergence and limit of the original sequence $(x_{n})_{n\in \mathbb {N} }$. The symbol $\land$ is the logical conjunction, which should be read as "and".
Writing the proof in the above way is time-consuming and no fun. Most of the time we prefer writing down the first version. We use the limit theorems even though we don't know if the sequences converge. We must argue afterwards, that it was okay to use the limit theorems in the first place. But this is the case, because at the end, everything converges. Since the last steps worked, we were allowed to do the steps before. So if we want to write the proof like in the first version, we need to make sure at the end to give a justification why applying the limit theorems was a valid thing to do.

## Problems with divergent sequences

As stated many times, we cannot use the limit theorems if one of the partial sequences diverges. If we forget this, we can quickly get nonsensical results:
${\begin{aligned}1&=\lim _{n\rightarrow \infty }1\\&=\lim _{n\rightarrow \infty }n\cdot {\tfrac {1}{n}}\\&=\left(\lim _{n\rightarrow \infty }n\right)\cdot \left(\lim _{n\rightarrow \infty }{\tfrac {1}{n}}\right)\\&=\infty \cdot 0\\&=0\end{aligned}}$
Question: What is wrong in the above argumentation?
$\infty$ is not a real number and thus $(n)_{n\in \mathbb {N} }$ is a divergent sequence with $\lim _{n\rightarrow \infty }n=\infty$. A sequence is convergent only if the limit is a real number. The product rule $\lim _{n\rightarrow \infty }a_{n}\cdot b_{n}=\left(\lim _{n\rightarrow \infty }a_{n}\right)\cdot \left(\lim _{n\rightarrow \infty }b_{n}\right)$ cannot be applied.

## Proof of limit theorems
### Absolute value rule

Theorem (Absolute value limit rule)
Let $(a_{n})_{n\in \mathbb {N} }$ be a convergent sequence with limit $a$. Then $\lim _{n\rightarrow \infty }|a_{n}|=|a|$.
How to get to the proof? (Absolute value limit rule)
If $\lim _{n\to \infty }a_{n}=a$ then the distance $|a_{n}-a|$ becomes arbitrary small. We need to show that $||a_{n}|-|a||$ also becomes arbitrary small. In the chapter Absolute value we proved the following inequality:
$||x|-|y||\leq |x-y|$
Thus we have
$||a_{n}|-|a||\leq |a_{n}-a|$
This means that if $|a_{n}-a|$ is smaller than $\epsilon$, by transitivity this is true also for $||a_{n}|-|a||$. We can use this in our convergence proof. Let $\epsilon >0$. We need to find $N\in \mathbb {N}$, so that $||a_{n}|-|a||<\epsilon$ for all $n\geq N$. Because of $\lim _{n\to \infty }a_{n}=a$ we know that there is ${\tilde {N}}\in \mathbb {N}$, so that $|a_{n}-a|<\epsilon$ for all $n\geq {\tilde {N}}$.
But as we have seen, if $|a_{n}-a|<\epsilon$ then also $||a_{n}|-|a||<\epsilon$. So we can set $N={\tilde {N}}$. Since $|a_{n}-a|<\epsilon$ is true for all $n\geq N$, it follows that $||a_{n}|-|a||<\epsilon$ is true for all $n\geq N$.
Proof (Absolute value limit rule)
Let $\epsilon >0$ be arbitrary. Because $(a_{n})_{n\in \mathbb {N} }$ converges to $a$, there is $N\in \mathbb {N}$ with $|a_{n}-a|<\epsilon$ for all $n\geq N$. From the inverted triangle inequality $||x|-|y||\leq |x-y|$ it follows
$||a_{n}|-|a||\leq |a_{n}-a|<\epsilon$
for all $n\geq N$.

### Inversion of the absolute value rule

If $(|a_{n}|)_{n\in \mathbb {N} }$ is a null sequence, then the inversion of the absolute value rule holds, as well. From $\lim _{n\to \infty }|a_{n}|=0$ , we have $\lim _{n\to \infty }a_{n}=0$:
Theorem
Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence. If $\lim _{n\to \infty }|a_{n}|=0$ , then $(a_{n})_{n\in \mathbb {N} }$ converges to zero, i.e. $\lim _{n\to \infty }a_{n}=0$.
Proof
Since $\lim _{n\to \infty }|a_{n}|=0$ , we have that
Now, $||a_{n}|-0|=||a_{n}||=|a_{n}|=|a_{n}-0|$. Therefore, we also have that
Or in other words, $\lim _{n\to \infty }a_{n}=0$.
This inversion only holds for null sequences. For a general sequence, it may not hold. An example is the divergent sequence $a_{n}=(-1)^{n}$. The absolute value sequence $\lim _{n\to \infty }|a_{n}|=1$ converges. So $\lim _{n\to \infty }|a_{n}|=1$ but $\lim _{n\to \infty }a_{n}\neq 1$. The latter limit even does not exist.

### The sum rule

Theorem (Limit theorems for sums)
Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence converging to $a$ and $(b_{n})_{n\in \mathbb {N} }$ a sequence converging to $b$. Then, the sequence $(a_{n}+b_{n})_{n\in \mathbb {N} }$ also converges with $\lim _{n\rightarrow \infty }a_{n}+b_{n}=a+b$.
How to get to the proof? (Limit theorems for sums)
We need to prove that $|(a_{n}+b_{n})-(a+b)|$ gets arbitrarily small. What we can use is that $|a_{n}-a|$ and $|b_{n}-b|$ get arbitrarily small. That means, we need to construct an upper bound for $|(a_{n}+b_{n})-(a+b)|$ which uses $|a_{n}-a|$ and/ or $|b_{n}-b|$ . The trick is to use the triangle inequality in $(a_{n}+b_{n})-(a+b)$:
${\begin{aligned}|(a_{n}+b_{n})-(a+b)|&=|(a_{n}-a)+(b_{n}-b)|\\&\leq |a_{n}-a|+|b_{n}-b|\end{aligned}}$
Since $|a_{n}-a|$ and $|b_{n}-b|$ get arbitrarily small, this estimate should suffice. It remains to establish the bounds in terms of $\epsilon$. This is done by bounding each of the two summands against ${\tfrac {\epsilon }{2}}$ : When we get $|a_{n}-a|<{\tfrac {\epsilon }{2}}$ and $|b_{n}-b|<{\tfrac {\epsilon }{2}}$ , then
$|a_{n}-a|+|b_{n}-b|<{\tfrac {\epsilon }{2}}+{\tfrac {\epsilon }{2}}=\epsilon$
We know that for some $N_{1}$ there is $|a_{n}-a|<{\tfrac {\epsilon }{2}}$ for all $n\geq N_{1}$ . Analogously, there is some $N_{2}$ with $|b_{n}-b|<{\tfrac {\epsilon }{2}}$ for all $n\geq N_{2}$. For the proof, we need $|a_{n}-a|<{\tfrac {\epsilon }{2}}$ and $|b_{n}-b|<{\tfrac {\epsilon }{2}}$ simultaneously. So both $n\geq N_{1}$ and $n\geq N_{2}$ should hold for all $n\geq N$ with some suitable $N\in \mathbb {N}$ . The smallest suitable $N$ is then given by $N=\max\{N_{1},\,N_{2}\}$ : From $n\geq \max\{N_{1},\,N_{2}\}$ , we get $n\geq N_{1}$ and $n\geq N_{2}$.
Proof (Limit theorems for sums)
Let $\epsilon >0$ be arbitrary. Then, there is an $N_{1}$ with $|a_{n}-a|<{\tfrac {\epsilon }{2}}$ for all $n\geq N_{1}$, since $\lim _{n\rightarrow \infty }a_{n}=a$ . Analogously, since $\lim _{n\to \infty }b_{n}=b$ there is an $N_{2}$ with $|b_{n}-b|<{\tfrac {\epsilon }{2}}$ for all $n\geq N_{2}$. We choose $N=\max\{N_{1},\,N_{2}\}$. Let $n\geq N$ be arbitrary. Then,
${\begin{aligned}|(a_{n}+b_{n})-(a+b)|&=|(a_{n}-a)+(b_{n}-b)|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{triangle inequality}}\right.}\\[0.5em]&\leq |a_{n}-a|+|b_{n}-b|\\&<{\tfrac {\epsilon }{2}}+{\tfrac {\epsilon }{2}}=\epsilon \end{aligned}}$

### The factor rule

Theorem (Factor rule for limits)
Let $\lambda \in \mathbb {R}$ be arbitrary and $(a_{n})_{n\in \mathbb {N} }$ a converging sequence with limit $a$. Then, the sequence $(\lambda a_{n})_{n\in \mathbb {N} }$ converges, as well with $\lim _{n\rightarrow \infty }\lambda a_{n}=\lambda a$.
How to get to the proof? (Factor rule for limits)
In order to prove $\lim _{n\rightarrow \infty }\lambda a_{n}=\lambda a$ , we need to establish a bound $|\lambda a_{n}-\lambda a|<\epsilon$ for almost all $n\in \mathbb {N}$ . This "to-be-achieved"-inequality can be equivalently reformulated:
${\begin{array}{lrl}&|\lambda a_{n}-\lambda a|&<\epsilon \\\iff&|\lambda \cdot (a_{n}-a)|&<\epsilon \\\iff&|\lambda |\cdot |a_{n}-a|&<\epsilon \end{array}}$
It is not allowed to directly divide by $|\lambda |$ , since there might be $\lambda =0$ . However, we can easily treat the case $\lambda =0$ : there, we need to show $\lim _{n\to \infty }\lambda a_{n}=\lambda a=0\cdot a=0$ . And since $\lambda a_{n}=0\cdot a_{n}=0$ , we trivially have $\lim _{n\to \infty }\lambda a_{n}=\lim _{n\to \infty }0=0$. So the assumption is established within case $\lambda =0$ . In the case $\lambda \neq 0$ , we can divide by $|\lambda |$:
${\begin{array}{lrl}&|\lambda |\cdot |a_{n}-a|&<\epsilon \\\iff&|a_{n}-a|&<{\frac {\epsilon }{|\lambda |}}\end{array}}$
Since $|a_{n}-a|$ converges to $0$ , there is an $N\in \mathbb {N}$, such that $|a_{n}-a|<{\tfrac {\epsilon }{|\lambda |}}$ for all $n\geq N$.
Proof (Factor rule for limits)
Let $\epsilon >0$ be arbitrary. in addition, let $\lambda \in \mathbb {R}$ and $(a_{n})_{n\in \mathbb {N} }$ a sequence converging to $a$.
Case 1: $\lambda =0$
There is $\lambda a_{n}=0\cdot a_{n}=0$ so we have convergence
$\lim _{n\rightarrow \infty }\lambda a_{n}=\lim _{n\rightarrow \infty }0=0=0\cdot a=\lambda \cdot a$
Case 2: $\lambda \neq 0$
Choose $N$ such that $|a_{n}-a|<{\tfrac {\epsilon }{|\lambda |}}$ for all $n\geq N$ . Such an $N$ can be found, since $(a_{n})_{n\in \mathbb {N} }$ converges to $a$ . Then,
${\begin{array}{lrl}&|a_{n}-a|&<{\frac {\epsilon }{|\lambda |}}\\\implies&|\lambda |\cdot |a_{n}-a|&<\epsilon \\\implies&|\lambda \cdot (a_{n}-a)|&<\epsilon \\\implies&|\lambda a_{n}-\lambda a|&<\epsilon \end{array}}$
this shown the assumption $\lim _{n\to \infty }\lambda a_{n}=\lambda a$ .

### The product rule

Theorem (Product rule for limits)
Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence converging to $a$ and $(b_{n})_{n\in \mathbb {N} }$ a sequence converging to $b$. Then, the product sequence $(a_{n}\cdot b_{n})_{n\in \mathbb {N} }$ converges as well with $\lim _{n\rightarrow \infty }a_{n}\cdot b_{n}=a\cdot b$.
Proof (Product rule for limits)
Let $\epsilon >0$ be arbitrary.
We need to show that $|a_{n}b_{n}-ab|<\epsilon$ for all $n\geq N$ , where we need to choose $N$ depending on $\epsilon$ . What we can use is that $|a_{n}-a|$ and $|b_{n}-b|$ get arbitrarily small, since $(a_{n})_{n\in \mathbb {N} }$ converges to $a$ and $(b_{n})_{n\in \mathbb {N} }$ converges to $b$ .This requires a deliberate reformulation of $|a_{n}b_{n}-ab|$ , such that we get an upper bound including $|a_{n}-a|$ and $|b_{n}-b|$ . The trick is to add a "smart zero" in the form of $ab_{n}-ab_{n}$:
${\begin{aligned}&|a_{n}b_{n}-ab|\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ ab_{n}-ab_{n}=0\right.}\\[0.3em]=\ &|a_{n}b_{n}-ab+ab_{n}-ab_{n}|\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{reformulate}}\right.}\\[0.3em]=\ &|a_{n}b_{n}-ab_{n}+ab_{n}-ab|\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{factor out}}\right.}\\[0.3em]=\ &|b_{n}(a_{n}-a)+a(b_{n}-b)|\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{triangle inequality}}\right.}\\[0.3em]\leq \ &|b_{n}(a_{n}-a)|+|a(b_{n}-b)|\\[0.3em]\leq \ &|b_{n}|\cdot |a_{n}-a|+|a|\cdot |b_{n}-b|\end{aligned}}$
If we can get the two summands below ${\tfrac {\epsilon }{2}}$ for all$n\geq N$ , then we are done.
Bounding the second summand
The sequence $(b_{n})_{n\in \mathbb {N} }$ converges to $b$ and by the factor rule with $\lambda =a$ we have $\lim _{n\to \infty }a\cdot b_{n}=a\cdot b$. Hence, $\lim _{n\to \infty }a\cdot (b_{n}-b)=0$ by the sum rule, so there is an $N_{1}\in \mathbb {N}$ such that for all $n\geq N_{1}$ we have $|a|\cdot |b_{n}-b|=|a(b_{n}-b)|<{\tfrac {\epsilon }{2}}$.
Bounding the first summand
This term is a bit more complicated, since the factor $|b_{n}|$ is not constant. However, we can bound it from above by a constant:
In the previous chapter, we have seen that convergent sequences are bounded. So there is an $M\in \mathbb {R} _{0}^{+}$ with $|b_{n}|\leq M$ for all $n\in \mathbb {N}$.
Now we can replace $|b_{n}|$ within the bound by the constant $M$ and we have $|b_{n}|\cdot |a_{n}-a|\leq M\cdot |a_{n}-a|$ . Now, we proceed as for the second summand. We use the factor rule with $\lambda =M$ and get some $N_{2}\in \mathbb {N}$ with $M\cdot |a_{n}-a|<{\tfrac {\epsilon }{2}}$ for all $n\geq N_{2}$, which in turn implies $|b_{n}|\cdot |a_{n}-a|<{\tfrac {\epsilon }{2}}$.
Conclusion
It remains to choose a suitable $N$. For any $n\geq N$ we need to have $n\geq N_{1}$ and $n\geq N_{2}$ . This can be done by choosing $N:=\max\{N_{1},N_{2}\}$, which only depends on $\epsilon$, as $N_{1}$ and $N_{2}$ also only depend on $\epsilon$.
Now, for all $n\geq N$ there is $|a_{n}b_{n}-ab|\leq \ |b_{n}|\cdot |a_{n}-a|+|a|\cdot |b_{n}-b|<{\tfrac {\epsilon }{2}}+{\tfrac {\epsilon }{2}}=\epsilon$

### The power rule

Theorem (Limit theorems for powers)
Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence converging to $a$. Let $k\in \mathbb {N}$ be any natural number . Then, the sequence $\left(a_{n}^{k}\right)_{n\in \mathbb {N} }$ also converges with $\lim _{n\to \infty }a_{n}^{k}=a^{k}$.
How to get to the proof? (Limit theorems for powers)
The power rule is a consequence of the product rule- For the sequence $\left(a_{n}^{2}\right)_{n\in \mathbb {N} }$ we have:
${\begin{aligned}\lim _{n\to \infty }a_{n}^{2}&=\lim _{n\to \infty }a_{n}\cdot a_{n}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{product rule}}\right.}\\[0.5em]&=\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }a_{n}\\&=a\cdot a=a^{2}\end{aligned}}$
Repeating this step $k$ times yields:
${\begin{aligned}\lim _{n\to \infty }a_{n}^{k}&=\lim _{n\to \infty }\underbrace {a_{n}\cdot a_{n}\cdot \ldots \cdot a_{n}} _{k{\text{ times}}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{product rule}}\right.}\\[0.5em]&=\underbrace {\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }a_{n}\cdot \ldots \cdot \lim _{n\to \infty }a_{n}} _{k{\text{ times}}}\\[0.5em]&=\underbrace {a\cdot a\cdot \ldots \cdot a} _{k{\text{ times}}}=a^{k}\end{aligned}}$
Now, the above argument with dots is not accepted as a proper proof in mathematics. The "clean" formulation is via induction in $k$.
Proof (Limit theorems for powers)
The proof is by induction, using the product rule.
Theorem whose validity shall be proven for $k\in \mathbb {N}$ bewiesen werden soll:
$\lim _{n\to \infty }a_{n}^{k}=a^{k}$
1. Base case:
$\lim _{n\to \infty }a_{n}^{1}=\lim _{n\to \infty }a_{n}=a=a^{1}$
2. Inductive step:
2a. Inductive hypothesis:
$\lim _{n\to \infty }a_{n}^{k}=a^{k}$
2b. Induction theorem:
$\lim _{n\to \infty }a_{n}^{k+1}=a^{k+1}$
2c. Proof of induction step:
${\begin{aligned}\lim _{n\to \infty }a_{n}^{k+1}&=\lim _{n\to \infty }a_{n}^{k}\cdot a_{n}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{product rule}}\right.}\\[0.5em]&=\lim _{n\to \infty }a_{n}^{k}\cdot \lim _{n\to \infty }a_{n}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{induction assumption}}\right.}\\[0.5em]&=a^{k}\cdot a=a^{k+1}\end{aligned}}$
Example (An example for the power rule)
We can show for all $k\in \mathbb {N}$ that the sequence $\left({\sqrt[{n}]{n^{k}}}\right)_{n\in \mathbb {N} }$ converges to $1$:
${\begin{aligned}\lim _{n\to \infty }{\sqrt[{n}]{n^{k}}}&=\lim _{n\to \infty }({\sqrt[{n}]{n}})^{k}\\&=\left(\lim _{n\to \infty }{\sqrt[{n}]{n}}\right)^{k}\\&=1^{k}=1\end{aligned}}$

### The quotient rule

Theorem (Quotient rule for limits)
Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence converging to $a$ and $(b_{n})_{n\in \mathbb {N} }$ a sequence converging to $b\neq 0$ where $b_{n}\neq 0$ for all $n\in \mathbb {N}$. Then, the quotient sequence $({\tfrac {a_{n}}{b_{n}}})_{n\in \mathbb {N} }$ converges with $\lim _{n\rightarrow \infty }{\tfrac {a_{n}}{b_{n}}}={\tfrac {a}{b}}$.
How to get to the proof? (Quotient rule for limits)
It suffices to show $\lim _{n\to \infty }{\tfrac {1}{b_{n}}}={\tfrac {1}{b}}$ . Then, by the product rule, we get
$\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=\lim _{n\to \infty }a_{n}\cdot {\tfrac {1}{b_{n}}}=\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }{\tfrac {1}{b_{n}}}=a\cdot {\tfrac {1}{b}}={\tfrac {a}{b}}$
In order to show $\lim _{n\to \infty }{\tfrac {1}{b_{n}}}={\tfrac {1}{b}}$, we need to show that $|{\tfrac {1}{b_{n}}}-{\tfrac {1}{b}}|$ gets arbitrarily small. We may use that $|b_{n}-b|$ gets arbitrarily small, as $(b_{n})_{n\in \mathbb {N} }$ converges to $b$ . So we need to re-formulate $|{\tfrac {1}{b_{n}}}-{\tfrac {1}{b}}|$ in a suitable way:
${\begin{aligned}\left|{\frac {1}{b_{n}}}-{\frac {1}{b}}\right|&=\left|{\frac {b}{b_{n}b}}-{\frac {b_{n}}{bb_{n}}}\right|\\[0.3em]&=\left|{\frac {b-b_{n}}{b_{n}b}}\right|\\[0.3em]&={\frac {|b-b_{n}|}{|b_{n}||b|}}\end{aligned}}$
The term $|b_{n}-b|$ can be controlled, meaning we can make it arbitrarily small. The $|b|$ in the denominator is just a constant and will not affect the convergence. The term $|b_{n}|$ needs some more work, since it depends on $n$. More precisely, we need to bound ${\tfrac {1}{|b_{n}|}}$ from above by a constant. That means, we require a lower bound for $|b_{n}|$.
By assumption, $\lim _{n\to \infty }b_{n}=b\neq 0$, so $b$ keeps a positive distance from 0. Due to convergence, we have an $N_{1}\in \mathbb {N}$, such that all sequence elements $(b_{n})_{n\in \mathbb {N} }$ with $n\geq N_{1}$ satisfy $|b_{n}-b|<{\tfrac {|b|}{2}}$ , i.e. they are not further away from $b$ than half the distance from $b$ to 0. Hence, $|b_{n}|\geq |b|-|b_{n}-b|\geq |b|-{\tfrac {|b|}{2}}={\tfrac {|b|}{2}}$ for all $n\geq N_{1}$. For the entire expression, that means
${\begin{aligned}\left|{\frac {1}{b_{n}}}-{\frac {1}{b}}\right|&={\frac {|b-b_{n}|}{|b_{n}||b|}}\\[0.5em]&\leq {\frac {|b-b_{n}|}{{\tfrac {|b|}{2}}|b|}}\\[0.5em]&={\frac {2}{|b|^{2}}}|b-b_{n}|\end{aligned}}$
This expression can be made arbitrarily small, as $|b-b_{n}|$ can be chosen arbitrarily small. For any given $\epsilon >0$ we choose an $N_{2}\in \mathbb {N}$ such that for all $n\geq N_{2}$ there is
${\frac {2}{|b|^{2}}}|b-b_{n}|<\epsilon \iff |b_{n}-b|<{\frac {\epsilon |b|^{2}}{2}}$
For all $n\geq \max\{N_{1},N_{2}\}$ it then follows that:
${\begin{aligned}\left|{\frac {1}{b_{n}}}-{\frac {1}{b}}\right|&\leq {\frac {2}{|b|^{2}}}|b-b_{n}|\\[0.5em]&<{\frac {2}{|b|^{2}}}\cdot {\frac {\epsilon |b|^{2}}{2}}\\[0.5em]&=\epsilon \end{aligned}}$
and we will be done with the proof. Now, let's write down the proof for $\lim _{n\to \infty }{\tfrac {1}{b_{n}}}={\tfrac {1}{b}}$ in a formal way.
Proof (Quotient rule for limits)
Let $\epsilon >0$ be given. Since $\lim _{n\to \infty }b_{n}=b\neq 0$ there is an $N_{1}\in \mathbb {N}$, such that $|b_{n}|\geq {\tfrac {|b|}{2}}$ for all $n\geq N_{1}$ . In addition, there is an $N_{2}\in \mathbb {N}$ with $|b_{n}-b|<{\frac {\epsilon |b|^{2}}{2}}$ for all $n\geq N_{2}$. Then, for $n\geq \max\{N_{1},N_{2}\}$, we have:
${\begin{aligned}\left|{\frac {1}{b_{n}}}-{\frac {1}{b}}\right|&=\left|{\frac {b}{b_{n}b}}-{\frac {b_{n}}{bb_{n}}}\right|\\[0.3em]&=\left|{\frac {b-b_{n}}{b_{n}b}}\right|\\[0.3em]&={\frac {|b-b_{n}|}{|b_{n}||b|}}\\[0.5em]&\leq {\frac {|b-b_{n}|}{{\tfrac {|b|}{2}}|b|}}\\[0.5em]&={\frac {2}{|b|^{2}}}|b-b_{n}|\\[0.5em]&<{\frac {2}{|b|^{2}}}\cdot {\frac {\epsilon |b|^{2}}{2}}\\[0.5em]&=\epsilon \end{aligned}}$
Therefore, $\lim _{n\to \infty }{\tfrac {1}{b}}_{n}={\tfrac {1}{b}}$. The product rule now implies
$\lim _{n\to \infty }{\tfrac {a_{n}}{b_{n}}}=\lim _{n\to \infty }a_{n}\cdot {\tfrac {1}{b_{n}}}=\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }{\tfrac {1}{b_{n}}}=a\cdot {\tfrac {1}{b}}={\tfrac {a}{b}}$

### The root rule

The following rule is in fact a generalization to the power rule above. It extends its validity from integer powers (like $a^{m},m\in \mathbb {N}$) to powers of the form $a^{1/k}={\sqrt[{k}]{a}}$. Combining both rules, we get a limit theorem for all positive rational powers (like $a^{m/k}$).
Theorem (Limit theorems for roots)
Let $(a_{n})_{n\in \mathbb {N} }$ be a non-negative sequence converging to $a$. In addition, let $k\in \mathbb {N}$. Then, the sequence $({\sqrt[{k}]{a_{n}}})_{n\in \mathbb {N} }$ converges to $\lim _{n\to \infty }{\sqrt[{k}]{a_{n}}}={\sqrt[{k}]{a}}$.
How to get to the proof? (Limit theorems for roots)
We need to find an upper bound for the absolute value $|{\sqrt[{k}]{a_{n}}}-{\sqrt[{k}]{a}}|$ . What we can use is that $|a_{n}-a|$ can be made arbitrarily small. Both expressions can be related by an auxiliary formula: For $x\geq y\geq 0$ and $k\in \mathbb {N}$ there is
$0<{\sqrt[{k}]{x}}-{\sqrt[{k}]{y}}\leq {\sqrt[{k}]{x-y}}$
The reason why this holds is that the root function $f(x)={\sqrt[{k}]{x}}$ is concave (meaning it curves down). You may draw this function yourself and verify why this holds. Taking the absolute of this equation, we get:
$|{\sqrt[{k}]{x}}-{\sqrt[{k}]{y}}|={\sqrt[{k}]{x}}-{\sqrt[{k}]{y}}\leq {\sqrt[{k}]{x-y}}={\sqrt[{k}]{|x-y|}}$
For $y\geq x\geq 0$, we get
$|{\sqrt[{k}]{x}}-{\sqrt[{k}]{y}}|={\sqrt[{k}]{y}}-{\sqrt[{k}]{x}}\leq {\sqrt[{k}]{y-x}}={\sqrt[{k}]{|x-y|}}$
Hence, $|{\sqrt[{k}]{x}}-{\sqrt[{k}]{y}}|\leq {\sqrt[{k}]{|x-y|}}$. We apply this auxiliary formula with $x=a_{n}$ and $y=a$:
$|{\sqrt[{k}]{a_{n}}}-{\sqrt[{k}]{a}}|\leq {\sqrt[{k}]{|a_{n}-a|}}$
The expression ${\sqrt[{k}]{|a_{n}-a|}}$ can be made arbitrarily small by setting $|a_{n}-a|$ "small enough":
${\sqrt[{k}]{|a_{n}-a|}}<\epsilon \iff |a_{n}-a|<\epsilon ^{k}$
So $|a_{n}-a|<\epsilon ^{k}$ is "small enough" and we have proven the desired inequality $|{\sqrt[{k}]{a_{n}}}-{\sqrt[{k}]{a}}|<\epsilon$.
Proof (Limit theorems for roots)
Let $k\in \mathbb {N}$ and $(a_{n})_{n\in \mathbb {N} }$ be a non-negative sequence converging to $a$. In addition, let $\epsilon >0$ be given. Since $\lim _{n\rightarrow \infty }a_{n}=a$ there is an $N\in \mathbb {N}$ with $|a_{n}-a|<\epsilon ^{k}$ for all $n\geq N$. Hence, for $n\geq N$ there is
${\begin{aligned}|{\sqrt[{k}]{a_{n}}}-{\sqrt[{k}]{a}}|&{\underset {\text{formula}}{\overset {\text{auxiliary-}}{\leq }}}{\sqrt[{k}]{|a_{n}-a|}}\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ |a_{n}-a|<\epsilon \right.}\\[0.3em]&<{\sqrt[{k}]{\epsilon ^{k}}}=\epsilon \end{aligned}}$

### The monotonicity rule

Theorem (Monotonicity rule for limits)
Let $(a_{n})_{n\in \mathbb {N} }$ and $(b_{n})_{n\in \mathbb {N} }$ be sequences converging to $a$ and $b$. In addition, let $a_{n}\leq b_{n}$ for almost all $n\in \mathbb {N}$. Then, we also have $a\leq b$.
Summary of proof (Monotonicity rule for limits)
The proof is by contradiction: We assume $a>b$ and show that this cannot hold true.
Proof (Monotonicity rule for limits)
Assume $a>b$. Since $\lim _{n\to \infty }a_{n}=a$ and $\lim _{n\to \infty }b_{n}=b$, for $\epsilon ={\tfrac {a-b}{2}}>0$ (an interval width, which is half of the spacing between $a$ and $b$), there are thresholds $N_{1},N_{2}\in \mathbb {N}$ such that $|a_{n}-a|<{\tfrac {a-b}{2}}$ for all $n\geq N_{1}$ and $|b_{n}-b|<{\tfrac {a-b}{2}}$ for all $n\geq N_{2}$. Hence for $n\geq \max\{N_{1},N_{2}\}$ (beyond both thresholds):
${\begin{aligned}b_{n}-a_{n}&=b_{n}-a_{n}+0+0\\&=b_{n}-a_{n}+b-a+a-b\\&=b_{n}-b+(b-a)+a-a_{n}\\&\leq |b_{n}-b|+(b-a)+|a_{n}-a|\\&<{\tfrac {a-b}{2}}+(b-a)+{\tfrac {a-b}{2}}\\&=a-b+b-a=0\end{aligned}}$
Therefore, $b_{n}<a_{n}$ for all $n\geq \max\{N_{1},N_{2}\}$. This contradicts $a_{n}\leq b_{n}$ which must hold for almost all $n$. So the assumption $a>b$ was wrong and $a\leq b$ must hold.

### Remarks concerning the monotonicity rule

There is a special case: we consider a constant $b_{n}=c$ :
The above proposition implies:
Both cases can be connected: „$a_{n}\leq b_{n}$“ and „$a_{n}\geq b_{n}$“:
Warning
The monotonicity rule does not hold for "$<$ and $>$", since within the limit, strict inequalities "become $\leq$ and $\geq$". An example are the sequences $(a_{n})_{n\in \mathbb {N} }=(-{\tfrac {1}{n}})_{n\in \mathbb {N} }$ and $(b_{n})_{n\in \mathbb {N} }=({\tfrac {1}{n}})_{n\in \mathbb {N} }$. In this case $a_{n}<0$ for all $n\in \mathbb {N}$, but in the limit, there is $\lim _{n\to \infty }a_{n}=0$ . So the limit of $(a_{n})_{n\in \mathbb {N} }$ is not strictly smaller than $0$. Analogously, for all $n\in \mathbb {N}$ there is $b_{n}>0$ but in the limit,$\lim _{n\to \infty }b_{n}=0\ngtr 0$.