```Date: Jun 13, 1996 4:07 PM
Author: Ted Alper
Subject: Re: How's this? Proof (hopefully) that pi is irrational.

In <tkidd.834615695@hubcap> tkidd@hubcap.clemson.edu (Travis Kidd) writes:>Yes.  But nowhere is either the numerator *or* the denominator 0 until the>limit is reached.  (I'll repeat what I just said in reply to another post...>I neglected to mention that k is positive.)  Therefore (I hope...I agree this >is the weakness in the proof), since substituting k for x yields equivalent >functions, the limit of the quotient would be 1.> But this is definitely not true. You seem to want to think thatif f and g are continuous functions,then          lim f(x)/g(x) = f(k)/g(k)         x->kbut it isn't true unless f(x)/g(x) is itself continuous at k,i.e. unless g(k) not equal to 0.it's not clear what you mean when you say"substituing k for x yields equivalent functions" -- once you substitute k for x, you no longer have functions, you have specific numbers.as someone else pointed out, choosing any k for which sin(k^2) = 0would work in your argument, whether or not k was an integer.Ted Alper
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