Epsilon-Delta Argument

$$\forall ϵ>0,\exists \delta >0,0<|x-a|<\delta \Rightarrow |f(x)-L|<ϵ$$

모든 $ϵ$에 대하여 어떤 $\delta$가 존재하여 $0<|x-a|<\delta \Rightarrow |f(x)-L|<ϵ$을 만족시킬 수 있다 극한값 ${\lim }_{x\to a}f(x)=L$이라고 할 수 있다. ($ϵ$이 아무리 작을지라도 이를 만족시키는 어떤 $\delta$를 항상 제시할 수 있다면, ${\lim }_{x\to a}f(x)$는 $L$에 한없이 가까워진다.)

1 연산

1.1 덧셈

$$\begin{array}{r}\underset{x\to c}{\lim }f(x)=L,\underset{x\to c}{\lim }g(x)=M⟹\underset{x\to c}{\lim }\{f(x)+g(x)\}=L+M\end{array}$$ $$\begin{array}{rl}& \underset{x\to c}{\lim }f(x)=L,\underset{x\to c}{\lim }g(x)=M\text{일 때}\\ & \forall ϵ>0\\ & \exists {\delta }_1>0\left(s.t.0<|x-c|<{\delta }_1⟹|f(x)-L|<\frac{ϵ}{2}\right)\\ & \exists {\delta }_2>0\left(s.t.0<|x-c|<{\delta }_2⟹|g(x)-M|<\frac{ϵ}{2}\right)\\ & let{\delta }_0=min({\delta }_1,{\delta }_2)\\ & 0<|x-c|<{\delta }_0⟹|f(x)+g(x)-L-M|\le |f(x)-L|+|g(x)-M|<ϵ\end{array}$$

1.2 곱셈

$$\begin{array}{r}\underset{x\to c}{\lim }f(x)=L,\underset{x\to c}{\lim }g(x)=M⟹\underset{x\to c}{\lim }\{f(x)g(x)\}=LM\end{array}$$ $$\begin{array}{rl}& \underset{x\to c}{\lim }f(x)=L,\underset{x\to c}{\lim }g(x)=M\text{일 때}\\ & \forall ϵ>0\\ & \exists {\delta }_1>0\left(s.t.0<|x-c|<{\delta }_1⟹|f(x)-L|<\frac{ϵ}{2|M|+2}\right)\\ & \exists {\delta }_2>0\left(s.t.0<|x-c|<{\delta }_2⟹|g(x)-M|<\frac{ϵ}{2|L|+2}\right)\\ & \exists {\delta }_3>0\left(s.t.0<|x-c|<{\delta }_3⟹|f(x)|<|L|+1\right)\\ & let\delta =min({\delta }_1,{\delta }_2,{\delta }_3)\\ & 0<|x-c|<{\delta }_0\\ & ⟹|f(x)g(x)-LM|=|f(x)g(x)-f(x)M+f(x)M-LM|\\ & \le |f(x)||g(x)-M|+|M||f(x)-L|\le (|L|+1)|g(x)-M|+|M||f(x)-L|<\frac{|L|+1}{2|L|+2}ϵ+\frac{|M|}{2|M|+1}ϵ<ϵ\\ & \therefore \underset{x\to c}{\lim }f(x)g(x)=LM\end{array}$$

1.3 나눗셈

$$\begin{array}{r}\underset{x\to c}{\lim }f(x)=L,\underset{x\to c}{\lim }g(x)=M⟹\underset{x\to c}{\lim }\left\{\frac{f(x)}{g(x)}\right\}=\frac{L}{M}\end{array}$$ $$\begin{array}{rl}& \underset{x\to c}{\lim }f(x)=L,\underset{x\to c}{\lim }g(x)=M\text{일 때}\\ & \forall ϵ>0\\ & \exists {\delta }_1>0(0<|x-c|<{\delta }_1⟹|f(x)-L|<1)\\ & ⟹|f(x)|<|L|+1\\ & \exists {\delta }_2>0(0<|x-c|<{\delta }_1⟹|g(x)-M|<1)\\ & ⟹|g(x)|>|M|-1\\ & \exists {\delta }_3>0\left(0<|x-c|<{\delta }_2⟹|f(x)-L|<\frac{|M|-1}{2}ϵ\right)\\ & \exists {\delta }_4>0\left(0<|x-c|<{\delta }_3⟹|g(x)-M|<\frac{|M|(|M|-1)}{2(|L|+1)}ϵ\right)\\ & \mathrm{let}{\delta }_0=\min ({\delta }_1,{\delta }_2,{\delta }_3,{\delta }_4)\\ & 0<|x-c|<{\delta }_0\\ & ⟹\left|\frac{f(x)}{g(x)}-\frac{L}{M}\right|=\frac{|f(x)M-g(x)L|}{g(x)M}\le \frac{|f(x)-L|}{|g(x)|}+\frac{|L||M-g(x)|}{|g(x)||M|}\\ & <\frac{|f(x)-L|}{|M|-1}+\frac{(|L|+1)|M-g(x)|}{(|M|-1)|M|}<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ\end{array}$$

1.4 거듭제곱

$$\underset{x\to a}{\lim }[f(x){]}^n=\underset{x\to a}{\lim }[f(x{)}^n]$$

2 연속

$$\underset{x\to c^{+}}{\lim }f(x)=L,\underset{x\to c^{-}}{\lim }f(x)=L\Rightarrow \underset{x\to c}{\lim }=L$$ $$\begin{array}{rl}& \forall ϵ>0\\ & \exists {\delta }_1>0\left(c<x<c+\delta ⟹|f(x)-L|<\frac{ϵ}{2}\right)(\because \underset{x\to c^{+}}{\lim }=L)\\ & \exists {\delta }_2>0\left(c-\delta <x<c⟹|f(x)-L|<\frac{ϵ}{2}\right)(\because \underset{x\to c^{-}}{\lim }=L)\\ & 0<|x-c|<\delta ⟹|f(x)-L|<ϵ\end{array}$$

극한은 하나의 값으로 수렴하는가?

$$\underset{x\to c}{\lim }f(x)=L,\underset{x\to c}{\lim }f(x)=m\Rightarrow l=M$$ $$\begin{array}{rl}& \text{suppose}L\ne M\\ & \text{let}0<|L-M|<ϵ\\ & \forall ϵ>0\\ & \exists {\delta }_1>0\left(0<|x-c|<{\delta }_1⟹|f(x)-L|<\frac{ϵ}{2}\right)(\because \underset{x\to c^{+}}{\lim }=L)\\ & \exists {\delta }_2>0\left(0<|x-c|<{\delta }_2⟹|f(x)-L|<\frac{ϵ}{2}\right)(\because \underset{x\to c^{-}}{\lim }=L)\\ & \text{let}{x}_0=c+\frac{\min ({\delta }_1,{\delta }_2)}{2}\\ & ϵ=|L-M|\le |f(x)-L|+|g(x)-M|<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ(\to \leftarrow )\end{array}$$