Difference between r1.12 and the current
@@ -34,6 +34,24 @@
i.e.
$|\frac1{x}-0|<\epsilon$
= Delta Epsilon Proofs.pdf 1 =
$\lim_{x\to 2}(3x-1)=5$
$|\frac1{x}-0|<\epsilon$
= 김도형 3 =
1:52분
$\lim_{x\to 0}\frac1{x^2}=\infty$
Goal: $\forall M>0,\exists\delta>0$ such that
$0<|x-0|<\delta\Rightarrow \frac1{x^2}>M$
Guess:
$|x|<\delta$
$|x|^2<\delta^2$
$\frac1{x^2}>\frac1{\delta^2}$
Sol: 주어진 M>0에 대하여
$\delta=\frac1{\sqrt{M}}$ 이라 하면,
$0<|x|<\delta\Rightarrow |x|<\frac1{\sqrt{M}}$
$\Rightarrow x^2<\frac1{M}$
$\Rightarrow \frac1{x^2}>M$
= Delta Epsilon Proofs.pdf 1 =
$\lim_{x\to 2}(3x-1)=5$
Up: 극한(limit)
1. 김도형 1 ¶
$\displaystyle \lim_{x\to3}(4x-5)=7$ 을 증명하라.
pf.
We want: $\displaystyle \forall \epsilon>0,\,\exists \delta >0$
pf.
We want: $\displaystyle \forall \epsilon>0,\,\exists \delta >0$
s.t. $\displaystyle 0<|x-3|<\delta\Rightarrow |(4x-5)-7|<\epsilon$
우변은$\displaystyle 4|x-3|<\epsilon$
$\displaystyle |x-3|<\frac{\epsilon}4$
주어진 양수 $\displaystyle \epsilon$ 에 대하여, $\displaystyle \delta=\frac{\epsilon}4$ 이라 하면,$\displaystyle |x-3|<\frac{\epsilon}4$
$\displaystyle 0<|x-3|<\delta \Rightarrow |x-3|<\frac{\epsilon}4$
$\displaystyle \Rightarrow 4|x-3|<\epsilon$
$\displaystyle \Rightarrow |(4x-5)-7|<\epsilon$
$\displaystyle \Rightarrow |f(x)-L|<\epsilon$
$\displaystyle \Rightarrow |(4x-5)-7|<\epsilon$
$\displaystyle \Rightarrow |f(x)-L|<\epsilon$
2. 김도형 2 ¶
2강 1:50분
$\displaystyle \lim_{x\to\infty}\frac1{x}=0$
Goal: $\displaystyle \forall\epsilon>0,\exists N>0$ such that
$\displaystyle x>N\Rightarrow \frac1{|x|}<\epsilon$
추측: 어떠한 $\displaystyle \epsilon$ 이 주어지더라도 $\displaystyle N$ 을 $\displaystyle \epsilon^{-1}$ 로 하면 될 것 같다.Sol: 주어진 $\displaystyle \epsilon>0$ 에 대하여 $\displaystyle N=\frac1{\epsilon}$ 이라 하면,
$\displaystyle x>N\Rightarrow |x|>N$
$\displaystyle x>N\Rightarrow |x|>N$
$\displaystyle \Rightarrow \frac{1}{|x|}<\frac1{N}=\epsilon$
i.e.$\displaystyle |\frac1{x}-0|<\epsilon$
3. 김도형 3 ¶
1:52분
$\displaystyle \lim_{x\to 0}\frac1{x^2}=\infty$
Goal: $\displaystyle \forall M>0,\exists\delta>0$ such that
$\displaystyle 0<|x-0|<\delta\Rightarrow \frac1{x^2}>M$
Guess:$\displaystyle |x|<\delta$
$\displaystyle |x|^2<\delta^2$
$\displaystyle \frac1{x^2}>\frac1{\delta^2}$
Sol: 주어진 M>0에 대하여$\displaystyle |x|^2<\delta^2$
$\displaystyle \frac1{x^2}>\frac1{\delta^2}$
$\displaystyle \delta=\frac1{\sqrt{M}}$ 이라 하면,
$\displaystyle 0<|x|<\delta\Rightarrow |x|<\frac1{\sqrt{M}}$$\displaystyle \Rightarrow x^2<\frac1{M}$
$\displaystyle \Rightarrow \frac1{x^2}>M$
$\displaystyle \Rightarrow \frac1{x^2}>M$
4. Delta Epsilon Proofs.pdf 1 ¶
$\displaystyle \lim_{x\to 2}(3x-1)=5$
다시 말해
Find an $\displaystyle \epsilon>0$ such that if $\displaystyle 0<|x-2|<\delta$ then $\displaystyle |(3x-1)-5|<\epsilon$
다시 말해
Find an $\displaystyle \epsilon>0$ such that if $\displaystyle 0<|x-2|<\delta$ then $\displaystyle |(3x-1)-5|<\epsilon$
먼저 delta를 찾는다
$\displaystyle |3x-6|<\epsilon$
$\displaystyle |x-2|<\epsilon/3$
$\displaystyle |3x-6|<\epsilon$
$\displaystyle |x-2|<\epsilon/3$
그래서 $\displaystyle \delta=\frac{\epsilon}3$ 으로 pick, 그러면 $\displaystyle |x-2|<\frac{\epsilon}{3}$ 이 되는지?
Yes.
5. Delta Epsilon Proofs.pdf 2 ¶
$\displaystyle \lim_{x\to 2}(3x^2-4x+1)=5$ 를 증명하라.
다시 말해 양수 epsilon을 찾아라. such that
if $\displaystyle 0<|x-2|<\delta$ then $\displaystyle |(3x^2-4x+1)-5|<\epsilon$
if $\displaystyle 0<|x-2|<\delta$ then $\displaystyle |(3x^2-4x+1)-5|<\epsilon$
$\displaystyle 3x^2-4x-4=(x-2)(3x+2)$
TODO
이상 http://mathnmath.tistory.com/36 의
Delta Epsilon Proofs.pdf 참조했음
"This handout is available at: www.uvu.edu/mathlab/handouts.html"
Delta Epsilon Proofs.pdf 참조했음
"This handout is available at: www.uvu.edu/mathlab/handouts.html"