Limits of Real-Valued Functions

Accumulation Point

Definition 30.1. Let $X \subset \mathbb{R}$ and let \( a \in \mathbb{R} \). We say that \( a \) is an accumulation point of \( X \) if for every \( \delta > 0 \), there exists a number \( x \in X \) such that \( 0 < |x - a| < \delta \).
We say that $a$ is a left (right) accumulation point of $X$ if for every $\delta > 0$, there exists a number $x \in X$ such that $0 < a - x < \delta (0 < x - a < \delta)$.

다른말로, $a$ 근방에 $a$와는 다른 $x \in X$가 항상 존재하면 $a$를 $X$의 accumulation point라고 부른다. 이때 upper and lower bound와 같이 $a$는 $X$에 속할 필요는 없다. 

Limits of Real-Valued Functions

Definition 30.2. Let \( f \) be a function from \( X \subset \mathbb{R} \) into \( \mathbb{R} \), and let \( a \) be an accumulation point of \( X \). We say that the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \) and write \[ \lim_{x \to a} f(x) = L \] if for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \) and \( x \in X \), then \( |f(x) - L| < \varepsilon \).

수열의 극한과 유사하게, 함수 $f(x)$가 $x \to a$일 때 극한값은 유일함이 보장된다.

Theorem 31.2

Theorem 31.2. Let \( f \colon X \to \mathbb{R} \), \( X \subset \mathbb{R} \) and suppose \( a \) is an accumulation point of \( X \). Then \( \lim_{x \to a} f(x) = L \) if and only if for every sequence \( \{ a_n \} \) in \( X \) such that \( \lim_{n \to \infty} a_n = a \) and \( a_n \ne a \) for every positive integer \( n \), we have \( \lim_{n \to \infty} f(a_n) = L \).

Proof. ($\Longrightarrow$) Suppose that $\lim_{x \to a} f(x) = L$. Then $\forall \varepsilon > 0, \exists \delta > 0$ such that $0 < |x - a| < \delta \Longrightarrow |f(x) - L| < \varepsilon$. Let $\{ a_n \}$ be a sequence such that $\lim_{n \to \infty} a_n = a$. Then $\exists N$ such that $0 < |a_n - a| < \delta, \forall n \geq N$. Thus we have that for $n \geq N$, $0 < |a_n - a| < \delta \Longrightarrow |f(a_n) - L| < \varepsilon$. This means that $\lim_{x \to a} f(x) = L$. 
($\Longleftarrow$) Suppose that $\lim_{n \to \infty} f(a_n) = L$ for every sequence $\{ a_n \}$ such that $\lim_{n \to \infty} a_n = a$ and $a_n \neq a, \forall n \in \mathbb{P}$. Assume that $\lim_{x \to a} f(x) \neq L$. Then $\exists \varepsilon > 0$ such that $\forall \delta > 0, \exists x \in X$ with $0 < |x - a| < \delta$ such that $|f(x) - L| \geq \varepsilon$. 
Take $\delta = \frac{1}{n}$. Since $\{ a_n \}$ converges to $a$, there is $a_n$ such that $$0 < |a_n - a| < \frac{1}{n} \quad \text{ and } \quad |f(a_n) - L| \geq \varepsilon,$$ which contradicts that $\lim_{n \to \infty} f(a_n) = L$. Thus $\lim_{x \to a} f(x) = L$. $\blacksquare$

Theorem 31.3

Theorem 31.3. Let \( f \) and \( g \) be two functions from \( X \) into \( \mathbb{R} \) with \( X \subset \mathbb{R} \). If \( \lim_{x \to a} f(x) = L \) and \( \lim_{x \to a} g(x) = M \), then
(i) \( \displaystyle \lim_{x \to a} |f(x)| = |L| \)
(ii) \( \displaystyle \lim_{x \to a} cf(x) = cL \), where \( c \in \mathbb{R} \)
(iii) \( \displaystyle \lim_{x \to a} [f(x) + g(x)] = L + M \)
(iv) \( \displaystyle \lim_{x \to a} [f(x) - g(x)] = L - M \)
(v) \( \displaystyle \lim_{x \to a} f(x)g(x) = LM \)
(vi) \( \displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M} \), provided that \( M \ne 0 \)

Proof. 

Exercise

Exercise 30.1. $\lim_{x \to a} f(x) = c,$ where $f(x) = c$ for every real $x$. 
Exercise 31.3. Let \( f \) and \( g \) be functions such that for some \( \delta > 0 \), \( f(x) < g(x) \) if \( 0 < |x - a| < \delta \). Suppose that \( \lim_{x \to a} f(x) = L \) and \( \lim_{x \to a} g(x) = M \). Then \( L < M \).
Exercise 31.4 (Squeeze Theorem). Let \( f \), \( g \), and \( h \) be functions such that for some \( \delta > 0 \), \( f(x) < g(x) < h(x) \) if \( 0 < |x - a| < \delta \). Suppose that \( \lim_{x \to a} f(x) = L = \lim_{x \to a} h(x) \). Then \( \lim_{x \to a} g(x) = L \).
Exercise 31.5 (Cauchy Condition for Functions). \( f \) has a limit at \( a \) if and only if for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \) and \( 0 < |y - a| < \delta \), then \( |f(x) - f(y)| < \varepsilon \).

Solution. 

Remark

Remark. Let $f(x) = \frac{|x|}{x}$. Then $f(x)$ does not have a limit at $x = 0$. 
(($\because$) Suppose that $f(x)$ has the limit $L$ at $x = 0$. Take $\varepsilon = 1$. Then $\exists \delta > 0$ such that $0 < |x| < \delta \Longrightarrow |f(x) - L| < 1$. If $x > 0$, then $0 < x < \delta \Longrightarrow |1 - L| < 1$ and if $x < 0$, then $- \delta < x < 0 \Longrightarrow |-1 - L| < 1$. Then $$x \in ( -\delta, 0) \cup (0, \delta) \Longrightarrow 2 = |1 + L + 1 - L| \leq |1+L| + |1-L| < 2 \bigotimes.$$ Thus $f(x)$ does not have the limit at $x = 0.$)
But $f(x)$ has the right-hand limit and left-hand limit at $x = 0$.

One-Sided Limits

Definition 32.1. Let \( f \) be a function from \( X \subset \mathbb{R} \) into \( \mathbb{R} \), and let \( a \) be a left (right) accumulation point of \( X \). We say that the limit of \( f(x) \) as \( x \) approaches \( a \) from the left (right) is \( L \) and write \[ \lim_{x \to a^-} f(x) = L \quad (\lim_{x \to a^+} f(x) = L) \] if for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that if \( 0 < a - x < \delta \) (respectively, \( 0 < x - a < \delta \)), then \( |f(x) - L| < \varepsilon \).

Theorem 32.2

Theorem 32.2. If \( a \) is a right and left accumulation point of \( X \), then \( \lim_{x \to a} f(x) = L \) if and only if \( \lim_{x \to a^+} f(x) = L = \lim_{x \to a^-} f(x) \).

Proof. ($\Longrightarrow$) Assume that $\lim_{x \to a} f(x) = L$. Then $\forall \varepsilon > 0, \exists \delta > 0$ such that $0 < |x-a| < \delta \Longrightarrow |f(x) - L| < \varepsilon$. If $x \geq a$, then $0 < x - a < \delta \Longrightarrow |f(x) - L| < \varepsilon$, which means that $\lim_{x \to a^{+}} f(x) = L$. If $x < a$, then $0 < a - x < \delta \Longrightarrow |f(x) - L| < \varepsilon$, which means that $\lim_{x \to a^{-}} f(x) = L$. 
($\Longleftarrow$) Assume that $\lim_{x \to a^+} f(x) = L = \lim_{x \to a^-} f(x)$. Then $\forall \varepsilon > 0, \exists \delta_1, \delta_2 > 0$ such that $0 < x - a < \delta_1 \Longrightarrow |f(x) - L| < \varepsilon$ and $0 < a - x < \delta_2 \Longrightarrow |f(x) - L| < \varepsilon$. Put $\delta = \min \{ \delta_1, \delta_2 \}$. Then $$0 < |x-a| < \delta \Longrightarrow 0 < x - a < \delta_1 \quad \text{or} \quad 0 < a-x < \delta_2 \\ \Longrightarrow |f(x) - L| < \varepsilon \Longrightarrow \lim_{x \to a} f(x) = L. \blacksquare$$

Infinite Limits

Definition 32.3. Let \( f \colon X \to \mathbb{R} \) with \( X \subset \mathbb{R} \). We say that the limit of \( f(x) \) as \( x \) approaches infinity (minus infinity) is \( L \) and write \[ \lim_{x \to \infty} f(x) = L \quad \text{[} \lim_{x \to -\infty} f(x) = L \text{]} \] if for every \( \varepsilon > 0 \), there exists a number \( M \) such that if \( x > M \) (respectively, \( x < M \)), then \( |f(x) - L| < \varepsilon \). However, if \( X \) is bounded above (below), we say \( \lim_{x \to \infty} f(x) \) \([ \lim_{x \to -\infty} f(x) ]\) is undefined.