Limit

analysis

Definition

Grenzwert

Limit of a Sequence

A real number $a$ is called limit of sequence $(a_n{)}_{n\ge 0}$ if almost all sequence elements $a_n$ lie within every epsilon neighbourhood of $a$:

$$\forall \epsilon >0:\exists N(\epsilon )\in \mathbb{N}:\forall n>N(\epsilon ):|a_n-a|<\epsilon$$

A number $a$ is thus the limit of a sequence if the following holds: Given any distance $\epsilon$, an index $N$ can be found such that from this index $N$ onward, all sequence elements have a from/to $a$ that is smaller than $\epsilon$. Such an $N$ must exist for every $\epsilon$, no matter how small $\epsilon$ is chosen.

If there exists an epsilon neighbourhood of $a$ that does not contain infinitely many sequence elements, then $a$ cannot be the limit, then $a$ cannot be the limit of that sequence.

Every limit is a limit point of that sequence.

Limit of a Function

A real number $c$ is called limit of function $f:D\to \mathbb{R}$ iff:

$$\underset{x\to a}{\lim }f(x)=c:⟺\forall (x_n{)}_{n\ge 0}:\underset{n\to \infty }{\lim }{x}_n=a⟹\underset{n\to \infty }{\lim }f(x_n)=c$$

Not all functions converge, i.e. signum function $\mathrm{sgn}(x)$. For instance, the signum function has two different limits, depending on the side:

$$\begin{array}{rl}\underset{x\to 0-}{\lim }\mathrm{sgn}(x)& =-1\\ \underset{x\to 0+}{\lim }\mathrm{sgn}(x)& =1\end{array}$$

Note that $x\to 0-$ means $x$ approaches $0$ from the left side and $x\to 0+$ means $x$ approaches $0$ from the right side.

Uniqueness

Different Limit Points

Let $a$ and $b$ be two different limit points of sequence $(a_n{)}_{n{\ge }_0}$, then for $\epsilon <\frac{1}{2}|a-b|$ it holds that $U_{\epsilon }(a)\cap U_{\epsilon }(b)=\varnothing$. Hence, it is impossible for almost all $a_n$ to lie in both $U_{\epsilon }(a)$ and $U_{\epsilon }(b)$.

Uniqueness of Limits

From this consideration it follows the uniqueness of the limit: since almost all terms of the sequence lie in every neighbourhood of the limit, a convergent sequence can have only one limit point. In the case of convergence, it therefore holds that:

$$\underset{n\to \infty }{\lim }{a}_n=\lim \underset{n\to \infty }{\sup }{a}_n=\lim \underset{n\to \infty }{\inf }{a}_n$$ Link to original

Rules

Sum

Let $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ be two sequences.

$$\underset{n\to \infty }{\lim }\left(a_n+b_n\right)=\underset{n\to \infty }{\lim }{a}_n+\underset{n\to \infty }{\lim }{b}_n$$

Product

Let $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ be two sequences.

$$\underset{n\to \infty }{\lim }\left(a_n\cdot b_n\right)=\{\begin{array}{ll}\infty & \text{if }{\lim }_{n\to \infty }{a}_n>0\\ -\infty & \text{if }{\lim }_{n\to \infty }{a}_n<0\end{array}$$

Product with Constant

Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence and $c\in \mathbb{R}$ be a constant.

$$\underset{n\to \infty }{\lim }\left(c\cdot a_n\right)=c\cdot \underset{n\to \infty }{\lim }{a}_n$$

Quotient

Let $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ be two sequences.

$$\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)=\{\begin{array}{ll}\frac{{\lim }_{n\to \infty }{a}_n}{{\lim }_{n\to \infty }{b}_n}& \text{if }{b}_n\ne 0\wedge {\lim }_{n\to \infty }{b}_n\ne 0\\ \infty & \text{if }{\lim }_{n\to \infty }{b}_n=0\wedge {\lim }_{n\to \infty }{a}_n>0\end{array}$$

Examples

Find Limit of $(\frac{1}{n^2}{)}_{n\ge 1}$

We want to prove that ${\lim }_{n\to \infty }\frac{1}{n^2}=0$, which, by definition, means that for every $\epsilon >0$, there exists an $N(\epsilon )$ such that for all $n>N(\epsilon )$:

$$\left|\frac{1}{n^2}-0\right|<\epsilon$$

Since $\frac{1}{n^2}$ is always non-negative, we can simplify the above to:

$$\frac{1}{n^2}<\epsilon$$

To solve the inequality for $n$, we take the reciprocal (nothing that all quantities are positive) and get:

$$n^2>\frac{1}{\epsilon }$$

Taking the square root on both sides (again, all values are positive) gives:

$$n>\frac{1}{\sqrt{\epsilon }}$$

Now, choose:

$$N(\epsilon )=\lfloor \frac{1}{\sqrt{\epsilon }}\rfloor$$

with $N(\epsilon )$. we can establish that for an arbitrary $\epsilon >0$, there exists an $N(\epsilon )$ such that $\forall n>N(\epsilon ):|a_n-0|<\epsilon$, which is the definition of a limit.

Find Limit of “Jumping” Sequence

Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence with:

$$\begin{array}{rl}{a}_n& =\{\begin{array}{ll}\frac{1}{n}& n\cancel{\equiv }0(\mathrm{mod}3)\\ \frac{10}{n}& \text{else}\end{array}\\ & =\left(1,\frac{1}{2},\frac{10}{3},\frac{1}{4},\frac{1}{5},\frac{10}{6},\dots \right)\end{array}$$

We assume that ${\lim }_{n\to \infty }{a}_n=0$. Let $\epsilon >0$ be given.

$$\begin{array}{rl}& |a_n-0|=|a_n|=a_n\\ & \forall n>N(\epsilon ):a_n<\epsilon \\ ⟹& a_n\le \frac{10}{n}<\epsilon \\ ⟺& 10>\epsilon \cdot n\\ ⟺& \frac{10}{\epsilon }<n\\ ⟹& N(\epsilon )=\lfloor \frac{10}{\epsilon }\rfloor \\ ⟹& \forall n>N(\epsilon ):|a_n-0|<\epsilon \\ ⟹& \underset{n\to \infty }{\lim }{a}_n=0\end{array}$$