Evaluate the Limit

Date: 12/03/97 at 19:39:26
From: Justin Yackoski
Subject: Limit of (1-cos x)/(x^2) as X->0

In Calculus today we were trying to evalulate the limit of

(1-cos x)
---------
   x^2

as x approaches zero.

I hope that looks right on your screen. The book said the answer was 
1/2 but I graphed it on my TI-82 and after zoom-boxing quite a few 
times (until the x values were expressed as N * 10^A with A being -3 
or -4, I found that it seemed to oscillate infinately and there was no 
limit. I also tried to calculate the value of y when x was 1*10^-7 
and other very small positive and negative numbers. Anything that 
small came out as zero. I wasn't sure if the book was right and I'm 
not seeing something, or if it doesn't exist, or if it is zero.  I 
assume the zero result when very extremely small numbers are used is 
because the calculator rounds and makes it zero. 

Thanks.

Date: 12/03/97 at 22:28:09
From: Doctor Pete
Subject: Re: Limit of (1-cos x)/(x^2) as X->0

Hi,

First of all, assuming no error on your part of entering the data into 
your calculator, I think you are right to be suspicious of the answers 
it gives you. Calculators are very simple computational tools, and I 
will say that whether it's an ancient 8-digit calculator, or a TI-92, 
the bottom line is that they compute with fixed numerical accuracy.  
Whether it is 8, 10, 12, or 15 digits of accuracy, it is still fixed.  
It's when you try to evaluate limits on such machines that their 
shortcomings become most apparent.

Well, here's how you would evaluate the limit using mathematics.  I 
will write it as

     Limit[(1-Cos[x])/x^2, x->0].

Now, notice that (1-Cos[x]) -> 0 as x -> 0, and similarly, x^2 -> 0 
as x -> 0. So we have an indeterminate form of type 0/0. We apply 
L'Hopital's rule, which states that if f[x] -> 0, and g[x] -> 0 as 
x -> a for some differentiable functions f[x], g[x], then

     Limit[f[x]/g[x], x -> a] = Limit[f'[x]/g'[x], x->a];

hence with f[x] = 1-Cos[x], g[x] = x^2, we find

     Limit[(1-Cos[x])/x^2, x->0] = Limit[Sin[x]/(2x), x->0],

which again, is an indeterminate form of type 0/0.  So we apply 
L'Hopital's rule again, which gives

     Limit[Sin[x]/(2x), x->0] = Limit[Cos[x]/2, x->0] = 1/2.

Therefore the limit is 1/2.

As to why your calculator gave misleading information, I am not 
exactly sure why it gave the particular answers it did, though I am 
certain it has to do with its fixed precision. There are programs for 
the computer, such as Mathematica, that perform calculations with 
arbitrary precision (limited only by memory and disk space), or better 
yet, they do them symbolically - in a sense, they perform calculations 
and simplifications in much the same way as I have evaluated the limit 
using L'Hopital's rule. If you think this is interesting, check out 
the home page of Wolfram Research (which makes Mathematica):

     http://www.wri.com   

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/