[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

Search:


View post   

>> No.11377660 [View]
File: 102 KB, 1275x600, dirichlet.png [View same] [iqdb] [saucenao] [google]
11377660

I tried using the epsilon/delta method, but had no luck. Since it's a modified version of the dirichlet function: f(x)={ 1, if x is rational; 0, if x is irrational}, multiply this function by x and you get spivak's modified function. Now h(x)=x has a limit everywhere, and f has no limit. Just assume that x goes to nonzero a. Using rules of limits, sup lim f(x) does not exist (true), and suppose that lim h(x) does exist (true) and is equal to nonzero b (just for arguments sake). Now for the sake of contradiction, assume that lim f(x)h(x) exists and is equal to c. By the usual rules of limits, lim (1/h(x))= 1/b (for nonzero b), and so lim f(x)h(x)(1/h(x))=c/b, that is, lim f(x)=c/b, but this is clearly a contradiction, since f(x) has no limit. This is just a rough proof, but have I got the general idea right? Is there a way to do this with epsilon/delta?

Navigation
View posts[+24][+48][+96]