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Limits of functions Support

Limits of functions

Cluster points

To start talking about limits of functions first we need to introduce the concept of cluster point.

Let $S\subset \R$ be a set. A numbwer $x\in \R$ is called an cluster point of $S$ if for every $\vre >0,$ the set $\left( x- \vre , x + \vre \right) \cap S \setminus \{x\}$ is not empty.

Intuitively, $x$ is an cluster point of $S$ if there are points of $S$ arbitrarily close to $x.$ Notice also that a cluster point of $S$ does not need to lie in $S.$

Consider the set
$$\left[0,1\right)\cup\left\{2\right\}=\{x\in\mathbb R: 0\leq x\lt 1\}\cup\left\{2\right\}.$$
$0$ is the only cluster point.
The set of cluster points for
$$(0,1) = \left\{x\in \R~:~ 0\lt x\lt 1 \right\}$$
are all points in the closed interval $\left[0,1\right].$
The set of cluster points for
$$\left[0,1\right)\cup\left\{2\right\}=\{x\in\mathbb R: 0\leq x\lt 1\}\cup\left\{2\right\}$$
is the interval $\left[0,1\right].$
A number $c \in \R$ is a cluster point of $S\subset $ if and only if there exists a convergent sequence of numbers $\{ x_n \}_{n=1}^\infty$ such that $x_n \not= c$ and $x_n \in S$ for all $n,$ and $\lim\limits_{n\to\infty} x_n = c.$
Suppose $c$ is a cluster point of $S.$ For every $n \in \N,$ pick $x_n$ to be an arbitrary point of $$\left(c-\frac{1}{n},c+\frac{1}{n}\right) \cap S \setminus \{c\}.$$ Note that this is a nonempty set because $c$ is a cluster point of $S.$ Then $x_n$ is within $\dfrac{1}{n}$ of $c.$ In other words, \begin{equation*} \abs{x_n-c} \lt \frac{1}{n} . \end{equation*} Since $\{ \frac{1}{n} \}_{n=1}^\infty$ converges to zero, $\{ x_n \}_{n=1}^\infty$ converges to $c.$ Now, if we start with a sequence of numbers $\{ x_n \}_{n=1}^\infty$ in $S$ converging to $c$ such that $x_n \not=c$ for all $n,$ then for every $\epsilon> 0$ there is an $N$ such that, in particular, $\abs{x_N -c} \lt \varepsilon.$ That is, $$x_N\in(c-\varepsilon,c+\varepsilon) \cap S \setminus \{c\}.$$

Limits of functions

If a function $f$ is defined on a set $S$ and $c$ is an cluster point of $S,$ then we define the limit of $f(x)$ as $x$ gets close to $c.$ It is irrelevant for the definition whether $f$ is defined at $c$ or not. Even if the function is defined at $c,$ the limit of the function as $x$ goes to $c$ can be totally different from $f(c).$

Let $f \colon S \to \R$ be a function and $c$ an cluster point of $S \subset \R.$ Suppose there exists an $L \in \R$ and for every $\vre > 0,$ there exists a $\delta > 0$ such that whenever $x \in S \setminus \{ c \}$ and $\abs{x - c} \lt \delta,$ we have \begin{equation*} \abs{f(x) - L} \lt \vre . \end{equation*} We then say $f(x)$ converges to $L$ as $x$ goes to $c$ and $L$ is the limit of $f(x)$ as $x$ goes to $c.$ We write \begin{equation*} \lim_{x \to c} f(x) :=L , \end{equation*} or \begin{equation*} f(x) \to L \quad\text{as}\quad x \to c . \end{equation*} If no such $L$ exists, then we say that the limit does not exist or that $f$ diverges at $c.$

Claim: $\displaystyle \lim_{x\ra 2} 3x+1 = 7.$

Discussion: For every $\vre >0 ,$ we need to find a $\delta>0$ so that \[ \abs{x-2}\lt\delta \quad \implies \quad \abs{(3x+1) - 7}\lt \vre. \] So we start from what we want to estimate: \[ \abs{(3x+1) - 7} = \abs{3x-6} = 3\abs{x-2}\lt 3 \delta \] If we choose $\delta := \vre/3,$ then $\abs{(3x+1) - 7} \lt \vre .$ Now we are ready to write a formal proof.

Proof. Let $\vre >0 .$ Put $\delta := \vre/3.$ Thus for all $x\in \R \setminus \{2\}$

\[ \abs{x-2}\lt \delta \quad \implies \quad \abs{(3x+1) - 7}\lt 3 \left(\frac{\vre}{3} \right) = \vre.\quad \bs \]

Claim: $\displaystyle \lim_{x\ra 2} x^2 = 4.$

Discussion: Given an arbitrary $\vre >0 ,$ our goal is to find a $\delta>0$ so that \[ \abs{x-2} \lt \delta \quad \implies \quad \abs{x^2 - 4}\lt \vre. \] Note that \begin{eqnarray}\label{eq:quadratic} \abs{x^2 - 4} = \abs{x+2}\abs{x-2} \end{eqnarray} Here we can make $\abs{x-2}$ as small as we like, but we need an upper bound on $\abs{x+2}$ in order to know how small to choose $\delta.$ In this case, the presence of the factor $\abs{x+2}$ in the expression (\ref{eq:quadratic}) may cause some initial confusion, but keep in mind that we are discussing the limit as $x$ approaches $2.$ So, we need to manipulate the inequality $\abs{x-2}\lt\delta$ in order to get an upper bound for $\abs{x+2},$ that is \begin{eqnarray*} &\abs{x-2}& \lt \delta \\ -\delta \lt & x-2 & \lt \delta \\ 4 -\delta \lt& x + 2 & \lt 4 + \delta\\ &\abs{x+2}& \lt \delta + 4 \end{eqnarray*} If $\delta =1,$ then $\abs{x+2} \lt 1+ 4 = 5 .$
Thus, if we choose $\delta := \min\{1, \vre / 5\},$ $\abs{x-2}\lt\delta$ implies $\abs{x^2 - 4} \lt \vre .$ Now let's write the formal proof.

Let $\vre >0 .$ Choose $\delta := \min\{1, \vre / 5\}.$ For all $x\in \R \setminus \{2\}$ and $\abs{x-2}\lt \delta,$ we have that \[ \abs{x^2-4} \lt \abs{x+2}\abs{x-2} \lt (5) \frac{\vre}{5} \lt \vre \] and the limit is proved.

Why do we need cluster points? 🤔

At this point, it is likely you are asking yourself:

Why do we need the concept of cluster point in the definition of limit?

To answer this question suppose \(c\) is not a cluster point of \(S.\) Then there exists some \(r\) such that the interval \((c-r,c+r)\) contains no points in \(S\) other than \(c\) itself.

Then, for any function \(f,\) choose any number \(L.\) Let \(\varepsilon\gt0\) be arbitrary and choose \(\delta = r.\) Since there are no points \(x\in S\setminus\{c\}\) such that \(|x-c|\lt\delta=r,\) then the statement:

For every \(\varepsilon \gt 0\) there exists a \(\delta \gt 0\) such that if for every \(x\in S\setminus\{c\}\) and \(|x-c|\lt \delta,\) then \(|f(x)-L|\lt \varepsilon.\)

is vacuously true!