"극한의 엄밀한 정의 - 엡실론과 델타"의 두 판 사이의 차이

수학노트
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">이 항목의 스프링노트 원문주소</h5>
  
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">개요</h5>
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<h5>예</h5>
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<math>\lim_{(x,y)\to(3,2)}\frac{y}{x-1}=1</math> 의 증명
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First we prove a set of inequalities.
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Assume <math>\sqrt{(x-3)^2+(y-2)^2}<\delta</math>.
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(1) <math>|x-3|\leq \sqrt{(x-3)^2+(y-2)^2}<\delta</math> implies <math>|x-3|<\delta</math>
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(2) <math>|y-2|\leq \sqrt{(x-3)^2+(y-2)^2}<\delta</math> implies <math>|y-2|<\delta</math>
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(3) <math>|y-x+1|=|(y-2)-(x-3)|\leq |y-2|+|x-3|<2\delta</math>
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(4) By rewriting <math>|x-3|<\delta</math> as <math>|(x-1)-2|<\delta</math>, we get  <math>2-\delta<|x-1|<2+\delta</math>.
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Let <math>\epsilon>0</math> be given. Let <math>\delta</math> be the minimum of  <math>\{1,{\epsilon}/2\}</math>.
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Then from above inequalities, we can say
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<math>|y-x+1|<2\delta\leq \epsilon</math> by (3)
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<math>|x-1|>2-\delta\geq 1</math> (by (4)) so <math>\frac{1}{|x-1|}<1</math>.
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<math>|\frac{y}{x-1}-1|=|\frac{y-x+1}{x-1}|<|y-x+1|<\epsilon</math>
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Therefore,
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<math>\lim_{(x,y)\to(3,2)}\frac{y}{x-1}=1</math>.
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<h5>역사</h5>
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* http://www.google.com/search?hl=en&tbs=tl:1&q=
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* [[수학사연표 (역사)|수학사연표]]
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*  
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<h5>메모</h5>
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<h5>관련된 항목들</h5>
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* [[05 수열의 극한]]
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">수학용어번역</h5>
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* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
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* 발음사전 http://www.forvo.com/search/
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
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* [http://www.nktech.net/science/term/term_l.jsp?l_mode=cate&s_code_cd=MA 남·북한수학용어비교]
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* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
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<h5>사전 형태의 자료</h5>
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* http://ko.wikipedia.org/wiki/
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* http://en.wikipedia.org/wiki/
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* http://www.proofwiki.org/wiki/
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* http://www.wolframalpha.com/input/?i=
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* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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** http://www.research.att.com/~njas/sequences/?q=
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<h5>관련논문</h5>
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* [http://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit http://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit]
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* http://www.jstor.org/action/doBasicSearch?Query=
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* http://www.ams.org/mathscinet
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* http://dx.doi.org/
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<h5>관련도서</h5>
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*  도서내검색<br>
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** http://books.google.com/books?q=
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** http://book.daum.net/search/contentSearch.do?query=
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*  도서검색<br>
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** http://books.google.com/books?q=
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** http://book.daum.net/search/mainSearch.do?query=
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** http://book.daum.net/search/mainSearch.do?query=
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<h5>관련기사</h5>
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*  네이버 뉴스 검색 (키워드 수정)<br>
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** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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** http://news.search.naver.com/search.naver?where=news&x=0&y=0&sm=tab_hty&query=
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<h5>링크</h5>
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*  구글 블로그 검색<br>
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** http://blogsearch.google.com/blogsearch?q=
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* [http://navercast.naver.com/science/list 네이버 오늘의과학]
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* [http://www.ams.org/mathmoments/ Mathematical Moments from the AMS]
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* [http://betterexplained.com/ BetterExplained]
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* [http://www.exampleproblems.com/ http://www.exampleproblems.com]

2012년 2월 5일 (일) 05:21 판

이 항목의 스프링노트 원문주소

 

 

개요

 

 

 

예

\(\lim_{(x,y)\to(3,2)}\frac{y}{x-1}=1\) 의 증명

 

 

First we prove a set of inequalities.

Assume \(\sqrt{(x-3)^2+(y-2)^2}<\delta\).

(1) \(|x-3|\leq \sqrt{(x-3)^2+(y-2)^2}<\delta\) implies \(|x-3|<\delta\)

(2) \(|y-2|\leq \sqrt{(x-3)^2+(y-2)^2}<\delta\) implies \(|y-2|<\delta\)

(3) \(|y-x+1|=|(y-2)-(x-3)|\leq |y-2|+|x-3|<2\delta\)

(4) By rewriting \(|x-3|<\delta\) as \(|(x-1)-2|<\delta\), we get  \(2-\delta<|x-1|<2+\delta\).

 

Let \(\epsilon>0\) be given. Let \(\delta\) be the minimum of  \(\{1,{\epsilon}/2\}\).

Then from above inequalities, we can say

\(|y-x+1|<2\delta\leq \epsilon\) by (3)

\(|x-1|>2-\delta\geq 1\) (by (4)) so \(\frac{1}{|x-1|}<1\).

\(|\frac{y}{x-1}-1|=|\frac{y-x+1}{x-1}|<|y-x+1|<\epsilon\)

Therefore,

\(\lim_{(x,y)\to(3,2)}\frac{y}{x-1}=1\).

 

 

역사

 

 

 

메모

 

 

관련된 항목들

 

 

수학용어번역

 

 

사전 형태의 자료

 

 

관련논문

 

 

관련도서

 

 

관련기사

 

 

링크
원본 주소 "https://wiki.mathnt.net/index.php?title=극한의_엄밀한_정의_-_엡실론과_델타&oldid=5275"