Remark 1.3.1. $f(x)$ can be interpreted in two different ways; one is to see $f(x)$ just as a formal expression and the other is, after introducing a function $f$, to see $f(x)$ as the output value of $f$ when input is $x$. Whether it is advantageous to introduce function structures depends on the situation.
Definition 1.3.2. We define a limit as follows:
\[\lim_{x\to a}f(x)=L:\Longleftrightarrow \forall \varepsilon >0\ \exists \delta >0\ \left( 0<\vert x-a\vert <\ \delta \Rightarrow \vert f(x)-L\vert <\varepsilon\right).\]This definition is called epsilon-delta definition, which is the most proof-friendly definition of limit. There can be a number of different definitions of limit, which are semantically different but equivalent to each other. For example, we can also define a limit as follows: “Let $W$ be an arbitrary neighborhood of $L$. Then there exists $U$, a punctured neighborhood of $a$, such that $f(U)\subseteq V$.”
Remark 1.3.3. Note that $\lim_{x\to a}f(x)$ is not a mathematical object itself until it is proved that $\lim_{x\to a}f(x)=L.$
Example 1.3.4. Prove that $\lim_{x\to 3}(4x-5)=7$.
Proof. Let $\varepsilon>0$ be given, and choose $\delta = {\varepsilon}/{4}$. Suppose that $0 < \vert x - 3\vert < \delta$. Then we have:
Since $\vert x - 3\vert < \delta$, it follows that:
\[4\vert x - 3\vert < 4\delta = \varepsilon.\]Thus, whenever $0 < \vert x - 3\vert < \delta$, we have $\vert (4x - 5) - 7\vert < \varepsilon$. Since $\varepsilon > 0$ was chosen arbitrarily, the result holds for all $\varepsilon > 0$, and, by Definition 1.3.2, $\lim_{x \to 3} (4x-5) = 7$.$\square$
Definition 1.3.5. We define a left-hand limit as follows:
\[\lim_{x\to a^-}f(x)=L:\Longleftrightarrow \forall \varepsilon >0\ \exists \delta >0\ \left( a-\delta < x < a \Rightarrow \vert f(x)-L\vert <\varepsilon\right).\]And, we define a right-hand limit as follows:
\[\lim_{x\to a^+}f(x)=L:\Longleftrightarrow \forall \varepsilon >0\ \exists \delta >0\ \left( a < x < a+\delta \Rightarrow \vert f(x)-L\vert <\varepsilon\right).\]Theorem 1.3.6 (Limit Laws). Let $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$. And, let $c\in \mathbb R$ be a constant.
- $\lim_{x\to a}[f(x)+g(x)]=L+M$
- $\lim_{x\to a}[f(x)-g(x)]=L-M$
- $\lim_{x\to a}[c\cdot f(x)]=c\cdot L$
- $\lim_{x\to a}[f(x)\cdot g(x)]=L\cdot M$
- If $M\neq 0$, then $\lim_{x\to a}\left[f(x)/g(x)\right] = L/M$
Theorem 1.3.7. Let $c\in \mathbb R$ be a constant and $n\in \mathbb N_{\geq 2}$.
- $\lim_{x\to a}c=c$.
- $\lim_{x\to a}x=a$.
- $\lim_{x\to a}x^n=a^n$.
- $\lim_{x\to a}\sqrt [n]x=\sqrt [n]a$.
Definition 1.3.8.
\[\lim_{x\to a}f(x)=\infty:\Longleftrightarrow \forall M >0\ \exists \delta >0\ \left( 0<\vert x-a\vert <\ \delta \Rightarrow f(x) > M\right).\]Example 1.3.9. Prove that $\lim_{x\to 0}1/x^2=\infty$.
Proof. Let $M>0$ be given, and choose $\delta = \sqrt{1/M}$. Suppose that $\vert x \vert < \delta$. Then we have:
Thus whenever $\vert x \vert < \delta$, we have ${1}/{x^2} > M.$ Since $M>0$ was chosen arbitrarily, the result holds for all $M>0$, and, by Definition 1.3.8, $\lim_{x\to 0}1/x^2=\infty$.$\square$
Definition 1.3.10.
\[\lim_{x\to a}f(x)=-\infty:\Longleftrightarrow \forall N <0\ \exists \delta >0\ \left( 0<\vert x-a\vert <\ \delta \Rightarrow f(x) < N\right).\]Theorem 1.3.11. Let $f$ and $g$ be real-valued functions defined on a domain $D\subseteq\mathbb{R}$, and let $a$ be an accumulation point of $D$. Suppose that there exists a punctured neighborhood $U$ of $a$ such that $x\in U \Rightarrow f(x)=g(x)$. Then, for every $L\in\mathbb{R}$,
\[\lim_{x\to a}f(x)=L \Longleftrightarrow \lim_{x\to a}g(x)=L.\]Exercises
Exercise 1.3.1. For given $\varepsilon > 0$, find $\delta$ such that $0<\vert x-2\vert <\delta \Rightarrow \vert 1 / x - 1 / 2\vert <\varepsilon$.
Solution. 일단 출력 오차가 입력 오차로 완전히 표현되는지를 봐야할 것 같다:
$\delta$를 제한하게 되면 input error의 boundary가 제한되어 $1/2\vert x \vert$의 boundary에도 제한이 생기긴 할 것이다. $1/2\vert x \vert$의 boundary의 경계값 $C$가 존재한다면 $C\cdot \delta\leq \varepsilon$, 즉 $\delta\leq \varepsilon / C$를 만족하기만 하면 될 것이다. $\delta < 2$로 제한하면 적어도 $C$가 존재하게 되고, $\delta \leq 1.9$, 또는 $\delta \leq 1$로 더 제한하면 boundary-value가 constant로 생김을 알 수 있다. 계산 편의를 위해 $\delta \leq 1$로 한다면 $1/2$로 잡히게 된다. 그러면 $\delta \cdot 1/2 \leq \varepsilon$, 즉, $\delta \leq 2\varepsilon$을 만족하면 되는 것이다. 따라서 boundary-value-inequality가 다음처럼 된다:
\[(\delta \leq 1) \wedge (\delta \leq 2\varepsilon).\]이로부터 $\delta \leq \min \lbrace 1, 2\varepsilon \rbrace$이면 됨을 알 수 있고, $\delta = \min \lbrace 1, 2\varepsilon \rbrace$로 하나로 정할 수도 있겠다.
Exercise 1.3.2. For given $\varepsilon > 0$, find $\delta$ such that $0<\vert x-a\vert <\delta \Rightarrow \vert \sqrt x - \sqrt a \vert <\varepsilon$.
Solution. $\delta$를 제한함에 따라 input error의 boundary가 제한되어 output error의 boundary도 줄긴 할 것이다. 이를 확인하기 위해 algebraic manipulation을 하면
가 되겠다. $\vert x-a\vert$과 $1/(\sqrt x + \sqrt a)$는 당연히 $\delta$로 bound되어 있기는 하겠다. variable $\delta$를 제한하기 전에 그 boundarie들은 정해지지 않는다. 그런데 $1/(\sqrt x + \sqrt a)$는 이미 constant boundary가 있다:
\[0< \frac{1}{\sqrt x + \sqrt a} \leq \frac {1}{\sqrt a}.\]그러면 boundary-value-inequality가 다음처럼 된다:
\[\delta \cdot \frac {1}{\sqrt a}\leq \varepsilon.\]따라서 $\delta \leq \sqrt a \varepsilon$이면 됨을 알 수 있고, $\delta = \sqrt a \varepsilon$으로 하나로 정할 수도 있겠다.
Exercise 1.3.3. Find the value of
\[\lim_{x\to 2}\frac{x^2-x-2}{x-2}.\]Solution. 어떤 $x$에 대해 출력값은 의미론적으로 ${x^2-x-2}$를 ${x-2}$로 나눈 값이다. $2$ 근방에서 각각을 보면 둘 다 $0$에 가깝다. 하지만 둘은 같은 수가 아니다. 이를 확인하기 위해 numerator를 denominator로 표현해봐야겠다:
\[x^2-x-2 = (x-2)^2 + 3(x-2)=(x-2)(x+1).\]어떤 $x$에 대해, quotient가 $1$이 아니라 정확히 $x+1$임을 확인하였다. Theorem 1.3.11에 의하여
\[\lim_{x\to 2}\frac{x^2-x-2}{x-2}=\lim_{x\to 2}{[x+1]}=3.\]Exercise 1.3.4. Prove that
\[\lim_{x\to 2}\frac{x^2-x-2}{x-2}=3.\]using epsilon-delta argument.
Proof. Let $\varepsilon>0$ be given, and choose $\delta = \varepsilon$. Suppose that $0<\vert x-2\vert <\delta$. Then we have
Since $\varepsilon > 0$ was chosen arbitrarily, the result holds for all $\varepsilon > 0$.$\square$
Exercise 1.3.5. Let $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$. Prove that $\lim_{x\to a}[f(x)+g(x)]=L+M$.
Proof. Let $W$ be an arbitrary punctured neighborhood of $L+M$. There exists $\varepsilon>0$ such that $(L+M-\varepsilon, L+M+\varepsilon)\subseteq W$. Let $V=(L-\varepsilon /2, L+\varepsilon /2)$ and $(M-\varepsilon /2, M+\varepsilon /2)$. Since $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$, there exists $U_f$ such that $f(U_f)\subseteq V$, and there exists $U_g$ such that $f(U_g)\subseteq T$. Let $U=U_f\cap U_g$. Then
Therefore, by the definition of limit, $\lim_{x\to a}[f(x)+g(x)]=L+M$. $\square$
Exercise 1.3.6. Let $a, L\in \mathbb R$ be constants. Prove that
\[\lim_{x\to a}f(x)=L \Longrightarrow \lim_{x\to a^-}f(x)= \lim_{x\to a^+}f(x)=L\]Proof. Let $\varepsilon$ be given. There exists $\delta$ such that $0<\vert x-a\vert <\ \delta \Rightarrow \vert f(x)-L\vert <\varepsilon$. We have $a-\delta < x < a \Rightarrow 0<\vert x-a\vert <\ \delta$. Thus, $a-\delta < x < a \Rightarrow \vert f(x)-L\vert <\varepsilon$. Since $\varepsilon > 0$ was chosen arbitrarily, the result holds for all $\varepsilon > 0$. Therefore, by Definition 1.3.5, $\lim_{x\to a^-}f(x)=L$. In the same way, $\lim_{x\to a^+}f(x)=L$.$\square$