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/