Limit of x sin(1/x)
Date: 04/23/2002 at 05:53:03 From: Rohan Hewson Subject: Limit of x sin(1/x) How can I prove that lim x sin(1/x) = 0? x->0 I graphed the function y = x sin(1/x) on a graphics calculator. As x went to +-infinity, y went to 1. As x went to 0, y oscillated around the x axis in the same fashion as sin(1/x) does, but with one difference: as x got closer to 0, the function oscillated less and less. I assumed from the graph that the function had a limit at x=0 of 0, but since it involves sin(1/0) I can not prove this using the basic trigonometric limits (sin x/x and (1 - cos x)/x), L'Hopital's rule or by rearranging the equation. Can you help? Rohan Hewson
Date: 04/23/2002 at 06:00:57 From: Doctor Mitteldorf Subject: Re: Limit of x sin(1/x) Dear Rohan, Go back to the definition of a limit. (Have you studied the formal definition of a limit?) The formal definition is that for every epsilon there exists a delta such that whenever x is within delta of zero, the absolute value of your function x sin(1/x) is less than epsilon. In other words, you have to supply a delta for x that guarantees the smallness of |x sin(1/x)|. In fact, since you know that however much sin(1/x) oscillates, it always has an absolute value <=1, you can just say delta=epsilon, and prove that |x sin(1/x)|<=epsilon. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
Date: 04/27/2002 at 04:12:46 From: Rohan Hewson Subject: Limit of x sin(1/x) I have not learnt this 'delta and epsilon' definition of a limit. I am in Year 12 (last year high school) and my calculus textbook describes lim f(x) as 'the number the function approaches as x approaches a'. x->a I have learnt how to rearrange equations that return 0/0, e.g. (x^2-25)/(x-5) at x=5 can be rearranged to x+5, etc. I have also learnt the two basic trigonometric limits and L'Hopital's rule. Could you explain the 'delta and epsilon' definition of a limit? Rohan Hewson
Date: 04/27/2002 at 05:57:43 From: Doctor Mitteldorf Subject: Re: epsilon / delta definition of a limit Rohan- The delta-epsilon definition is pretty abstract, but in fact it's the simplest definition you could come up with if you tried yourself to formalize your intuitions about a limit. What does it mean that f->0? Well, it can't mean that f=0. But it must mean that f gets closer and closer to zero - arbitrarily close. There is no small number epsilon, no matter how small epsilon is, where f doesn't become smaller than that epsilon. So the definition must be: whatever number epsilon you give me, no matter how small, I can guarantee you that f is always smaller than that. I can guarantee you that the absolute value of f is smaller than epsilon. Now, what does it mean to guarantee? Certainly not ALL values of f are smaller than this tiny number. But "beyond a certain point" they must be. What do we mean by "beyond a certain point"? It must mean "whenever x is less than a certain number," which we'll call delta. So now we have it. If I claim that f(x)->0 when x->0, and you say it doesn't, then here's how we decide: For any number epsilon that you specify, no matter how small, I claim that I can choose another number delta (I get to pick it - it can be as small as I like) such that whenever |x|<delta, I can demonstrate to you that |f(x)|<epsilon. That's it. That's the formal definition. You, playing Devil's Advocate, get to pick the epsilon, and can make it as tiny as you want. If it's my responsibility to show that this is the limit, then I get to go second. Using your epsilon, I come up with a delta, as small as I like. My burden of proof is to guarantee that every value of x that obeys |x| less than my delta corresponds to an f(x) such that the absolute value of f(x) is less than the epsilon you've chosen. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
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