Limit of function

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4.1.1 Definition

$E \subset \mathbb{R}, f: E \rightarrow \mathbb{R}$: a function, $p \in E^\prime$. $f$ has a limit at p if there exists $L\in \mathbb{R}$ such that $\forall \epsilon >0, \exists \delta >0$ s.t. \(\begin{align} 0 <|x-p|<\delta \:(\forall x \in E) \Rightarrow |f(x)-L| < \epsilon. \end{align}\) We write $\displaystyle{\lim_{x\to p}f(x) = L}$.


  • The definition does not consider the case when $x=p$.
  • Why should $p$ be a limit point? We want to guarantee the existence of $x\in N^\prime_\delta (x) \cap E$. Since every deleted neighborhood of $p$ contains infinitely many points of $E$, there is such $x$ for every $\delta >0.$
  • $\displaystyle{\lim_{x\to p}f(x) \neq L} \iff \forall L \in \mathbb{R}, \exists \epsilon >0$ such that $\forall \delta >0, \exists x\in E$ such that $0<|x-p|<\delta \text{ and } |f(x) - L| \geq \epsilon .$

Examples 4.1.2

(a) $\displaystyle{\lim_{x\to p}c = c}$

$\because \forall \epsilon>0, \forall \delta >0, $ $0<|x-p|<\delta \Rightarrow |c-c| <\epsilon$

(b) $\displaystyle{\lim_{x\to p}x = p}$

$\because \forall \epsilon >0, \exists \delta >0$ such that $0<|x-p|<\delta \Rightarrow |x-p|<\epsilon$. Take $\delta :=\epsilon$. Then, $0<|x-p|<\delta \Rightarrow |x-p|<\delta=\epsilon$.

(c) $\displaystyle{\lim_{x\to p}x^2 = p^2}$

<proof> Let $\epsilon >0$ be given. We want to show that the existence of $\delta >0$ such that $0<|x-p|<\delta \Rightarrow |x^2-p^2| <\epsilon.$ Now we factorize $|x^2-p^2| = |x+p|\cdot |x-p|$. If $0<\delta \leq 1$,

\(\begin{align} \begin{split} |x|-|p| \leq |x-p| &\leq \delta \\ |x| &< |p| + \delta \\ &\leq |p| + 1 \\ \end{split} \end{align}\) Similarly, we can bound $|x+p|$ with triangular inequality, \(\begin{align} \begin{split} |x+p| &\leq |x| + |p| \\ &< 2|p| + \delta \\ &\leq 2|p| + 1 \end{split} \end{align}\) Now, we take $\delta := \min (\frac{\epsilon}{2|p|+1}, 1)$. Then, \(\begin{align} |x+p||x-p| < (2|p|+1)\frac{\epsilon}{2|p|+1} = \epsilon \end{align}\) $\therefore \displaystyle{\lim_{x\to p}}x^2 = p^2.$ \(\tag*{$\square$}\)

(e) $f(x) = 1 \text{ if } x \in \mathbb{Q}, f(x) = 0 \text{ if } x \not\in \mathbb{Q}.$

$\displaystyle{\lim_{x\to p}f(x)}$ does not exist for every $p$.

<proof> Let $L \in \mathbb{R}$ be given. Take $\epsilon := \max (|L-1|, |L|)$. Then for any $\delta >0$, we can take $x \in \mathbb{Q}$ such that $0<|x-p| <\delta$ by the density of rational number. Thus $|f(x) - L| = |L-1| \leq \epsilon$.

Similarly, $\forall \delta >0$, we can take $x\in \mathbb{R}\setminus \mathbb{Q}$ such that $0<|x-p| <\delta$ by the density of irrational number. Thus, $|f(x) - L| = |L| \leq \epsilon$. \(\tag*{$\square$}\)

(f) $f(x) = 0 \text{ if } x \in \mathbb{Q}$ or $f(x) = x \text{ if } x \in \mathbb{R}\setminus \mathbb{Q}$. $\displaystyle{\lim_{x\to 0} f(x)=0}$ and $\displaystyle{\lim_{x\to p} f(x)}$ does not exist for $\forall p\neq0.$

Let $\epsilon >0$ be given. Since $|f(x)| \leq |x| <\delta$, take $\delta := \epsilon$. Then $|f(x) < \epsilon$. $\therefore \displaystyle{\lim_{x\to 0} f(x) = 0}$. \(\tag*{$\square$}\)

Theorem 4.1.3

$E \subset \mathbb{R}, f: E\rightarrow \mathbb{R}, p \in E^\prime$. \(\begin{align} \begin{split} \lim_{x\to p}f(x) = L \iff &\lim_{n\to\infty} f(p_n) = L \\ &\text{for all seq } (p_n) \text{ with } p_n \rightarrow p \text{ as } n\rightarrow \infty \\ &\text{and } p_n \neq p \end{split} \end{align}\)

<proof> $\Rightarrow$ Suppose that $\displaystyle{\lim_{x\to p}f(x)=L}$. Let $\epsilon >0 $ be given. Since $\displaystyle{\lim_{x\to p}f(x)=L}$, there is $\delta >0$, such that $0<|x-p| <\delta \Rightarrow |f(x)-L| <\epsilon.$ Since $p$ is a limit point, there exists a sequence $(p_n)$ with $p_n \rightarrow p$ as $n\rightarrow \infty$ and $p_n \neq p$. Let $(p_n)$ be any sequence satisfying such condition. Since $p_n \rightarrow p$ as $n\rightarrow \infty$, there exists $N\in \mathbb{N}$ such that $n\geq N \Rightarrow 0<|p_n -p| <\delta (\because p_n \neq p)$.
$\therefore |f(p_n) - L | <\epsilon$, i.e. $\displaystyle{\lim_{n\to\infty}f(p_n) = L}.$

$\Leftarrow$ Suppose that $\displaystyle{\lim_{x\to p} f(x) \neq L}.$ $\exists \epsilon >0$ such that $\forall \delta >0, \exists x \in E$, $0<|x-p|<\delta \text{ and } |f(x)-L| \geq \epsilon$. Take $\delta := \frac{1}{n}$. Then there exists $p_n \in E$ such that \(\begin{align} 0< |p_n-p|<\frac{1}{n} \text{ but } |f(p_n) - L | \geq \epsilon \end{align}\) which contradicts to the assumption that $\displaystyle{\lim_{n\to\infty}f(p_n)=L}.$

$\therefore \displaystyle{\lim_{x\to p}f(x)=L}.$ \(\tag*{$\square$}\)

Corollary 4.1.4

$[f(x) \rightarrow L_1$ and $f(x) \rightarrow L_2$ as $x\rightarrow p] \Rightarrow [L_1 = L_2]$.

<proof> By the previous theorem, for any seq. $(p_n)$ with $p_n \rightarrow p$ as $n\rightarrow \infty$ and $p_n \neq p$ $\Rightarrow \displaystyle{\lim_{n\to\infty}f(p_n)} = L_1 \text{ and } \displaystyle{\lim_{n\to\infty}f(p_n) = L_2}.$ Since the limit of sequence is unique, $L_1 = L_2$. \(\tag*{$\square$}\)

Examples 4.1.5

(a) $E = (0, \infty), f(x) = \sin \frac{1}{x}$ If we take a sequence $p_n = \frac{(2n+1)\pi}{2}$, then $\sin \frac{1}{p_n} = (-1)^n$.

$\therefore \displaystyle{\lim_{x\to p}f(x)}$ does not exist for every $p \in \mathbb{R}$.

(b) $f(x) = 0 \text{ if } x \in \mathbb{Q}$ or $f(x) = x \text{ if } x \not\in \mathbb{Q}$.

Let $p \in \mathbb{R}$ and $p\neq 0$. $\exists (p_n) \subset \mathbb{Q}$ s.t. $p_n \to p$ as $n\to\infty$ and $p_n \neq p$. $\exists (q_n) \subset \mathbb{R}\setminus\mathbb{Q}$ s.t. $q_n \to p$ as $n\to\infty$ and $q_n \neq p$. $(\because$ For every $n\in \mathbb{N}$, take $p_n \in \mathbb{Q}$ or $q_n \in \mathbb{R}\setminus \mathbb{Q}$ $p_n, q_n \in (p, p+\frac{1}{n})$. By the density of rational/irrational numbers, we can construct such sequences.) \(\begin{align} \begin{split} 0<|p_n -p| <\delta &\Rightarrow f(p_n) = 0 \to 0 \\ 0< |q_n -p| <\delta &\Rightarrow f(q_n) =q_n \to 0 \end{split} \end{align}\) Since $q_n \to p$ and $q_n \to 0$, $p=0$. But it is contradiction. \(\tag*{$\square$}\)

Theorem 4.1.6

$E\subset \mathbb{R}, f,g: E \rightarrow \mathbb{R}, p\in E^\prime$ $\displaystyle{\lim_{x\to p}f(x)=A, \lim_{x\to p}g(x) = B}$.

(a) $\displaystyle{\lim_{x\to p}(f(x) + g(x))} = A+B$

(b) $\displaystyle{\lim_{x\to p}(f(x) \cdot g(x))} = A\cdot B$

(c) $\displaystyle{\lim_{x\to p}\frac{f(x)}{ g(x)}} = \frac{A}{B}$

<proof> (a) Since $f(x) \to A$ as $x\to p$ and $g(x)\to B$ as $x\to p$. Let $(p_n)$ be a sequence with $p_n \to p$ as $n\to \infty$ and $p_n \neq p$. Thus, \(\begin{align} \lim_{n\to\infty}f(p_n)=A, \lim_{n\to\infty}g(p_n)=B \end{align}\) $\therefore \displaystyle{\lim_{n\to\infty}(f(p_n) + g(p_n)) = A+B}$

$\therefore \displaystyle{\lim_{x\to p}(f(x) + g(x))= A + B}$ \(\tag*{$\square$}\)

(b) Similarly, we can prove it.

(c) It is enough to show that $\displaystyle{\lim_{x\to p}\frac{1}{g(x)}} = \frac{1}{B}$

<proof> Since $\displaystyle{\lim_{x\to p}g(x) = B}$, \(\begin{align} \exists \delta_1 >0 \text{ s.t. } 0<|x-p|<\delta_1 \Rightarrow |g(x) - B| <\epsilon \end{align}\) Take $\epsilon := \frac{|B|}{2}$. Then, \(\begin{align} |B| - |g(x)| \leq |g(x) - B| < \frac{|B|}{2} \end{align}\) $\therefore |g(x)| > \frac{|B|}{2}$ for $0<|x-p|<\delta_1$.

Let $(p_n)$ be a sequence with $p_n \to p$ as $n\to \infty$ and $p_n \neq p$. Then there is $N \in \mathbb{N}$ such that \(\begin{align}n\geq N \Rightarrow 0<|p_n -p| <\delta_1. \end{align}\)

So, $g(p_n)\to B$ as $n\to \infty$. i.e. $\frac{1}{g(p_n)} \to \frac{1}{B}$ as $n\to \infty$.

$\therefore \displaystyle{\lim_{n\to\infty}\frac{1}{g(p_n)}=\frac{1}{B}}.$

$\therefore \displaystyle{\lim_{x\to p}\frac{1}{g(x)}=\frac{1}{B}}.$

Definition 4.1.7

$f:E\to \mathbb{R}$ is bounded if $\exists M\in \mathbb{R}$ such that $|f(x)| \leq M$ for all $x\in E$.

Theorem 4.1.8

$f:E\to\mathbb{R}, p\in E^\prime, g:E\to\mathbb{R}$ is bounded and $\displaystyle{\lim_{x\to p}f(x)} = 0$.

$\Rightarrow \displaystyle{\lim_{x\to p}f(x)g(x)} = 0$

<proof> Let $\epsilon >0$ be given. Since $f(x)\to 0$ as $x\to p$, there exist $\delta >0$ such that $|f(x)|<\epsilon$, whenever $0<|x-p|<\delta, x\in E$. Since $g$ is bounded, there is a constant $M$ such that $|g(x)|\leq M$ for all $x\in E$.

If $0<|x-p|<\delta, x\in E$, then $|f(x)g(x)| = |f(x)| |g(x)| < \epsilon$.
$\therefore \displaystyle{\lim_{x\to p}f(x)g(x)}$ \(\tag*{$\square$}\)

Theorem 4.1.9

$f,g,h: E\to \mathbb{R}, p\in E^\prime$

Suppose that $g(x)\leq f(x)\leq h(x)$ for all $x\in E$.

If $\displaystyle{\lim_{x\to p}g(x)=L=\lim_{x\to p}h(x)}$, then $\displaystyle{\lim_{x\to p}f(x)=L}$.

<proof> Let $\epsilon >0$ be given. Since $\displaystyle{\lim_{x\to p}g(x) = L = \lim_{x\to p}h(x)}$, there exists $\delta >0$ such that $0<|x-p|<\delta, x\in E \Rightarrow L-\epsilon < g(x), h(x)<L +\epsilon$.

Since $g(x)\leq f(x) \leq h(x)$ for all $x\in E$, we have \(\begin{align} L-\epsilon < g(x) \leq <f(x) \leq h(x) <L+\epsilon \end{align}\) if $0<|x-p|<\delta, x\in E$, i.e. $\displaystyle{\lim_{x\to p}f(x) =L}.$ \(\tag*{$\square$}\)

Examples 4.1.10

$p(x):= a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$, where $a_n \neq 0$. We call it polynomial function of degree $n$.

\[\lim_{x\to c} p(x) = a_n c^n + \cdots + a_0 = p(c)\]

Definition 4.1.11

$f$ is real-valued function such that $\text{Dom}f \cap (a,\infty)\neq \emptyset$ for all $a\in \mathbb{R}$.
The function $f$ has a limit at $\infty$ if $\exists L\in \mathbb{R}$ such that \(\forall \epsilon >0, \exists M\in \mathbb{R} \text{ s.t. } x\in \text{Dom}f \cap (M,\infty)\neq \emptyset \Rightarrow |f(x)-L|<\epsilon.\)
We write $\displaystyle{\lim_{x\to\infty}f(x)=L}.$

Examples 4.1.12

(a) $f(x)= \frac{\sin x}{x}$ for all $x\in (0,\infty)$.

Since, $|\sin x|\leq 1$, $|f(x)|\leq \frac{1}{x}$ for all $x\in (0,\infty)$. Let $\epsilon >0$ be given. Take $M :=\frac{1}{\epsilon}$. If $x>\frac{1}{\epsilon}$, then $\frac{1}{x} <\epsilon$. \(|f(x)| \leq \frac{1}{x} < \epsilon, \text{ for all }x > \frac{1}{\epsilon}=M\)
$\therefore \displaystyle{\lim_{x\to\infty}\frac{\sin x}{x}=0}$

(b) $f(x) = x\sin \pi x$

Put $p_n :=(n+\frac{1}{2})$. Then $\sin \pi (p_n) = \sin(\frac{2n+1}{2})\pi = (-1)^n,$ i.e. $f(p_n) = (n+\frac{1}{2})\cdot (-1)^n.$
Since $\displaystyle{\lim_{x\to\infty}f(p_n)}$ does not exist, $\displaystyle{\lim_{x\to\infty}f(x)}$ does not exist.


  • Manfred Stoll, Introduction to Real Analysis, Pearson