Continuity of f(x) = sin(x)/x at x = 0

Date: 01/20/2004 at 11:52:00
From: Zaheer
Subject: continuity


Given f(x) = ((sin x)/x if x is not equal to 0) 
             (      1   if x is  equal    to 0)


Please tell me how f(x) is continuous at 0?  I think that we have to
draw a graph of sinx/x and then see whether it is continuous at zero
or not.

Date: 01/21/2004 at 19:12:25
From: Doctor Fenton
Subject: Re: continuity

Hi Zaheer,

Thanks for writing to Dr. Math.  Drawing a graph can suggest whether
the function is continuous, but it does not constitute a proof.  A
function f(x) is continuous at x = a if

   lim f(x) = f(a)   ,
  x->a

so to show that f(x) = sin(x)/x is continuous at x = 0, when f(0) is
defined as 1, you must show that

        sin(x)
   lim  ------ = 1 .
  x->0     x

This is usually "proved" by a geometric argument (one can argue about
the rigor of the following argument):

                                   B   D
                                   .  |
                        .          : .|
                .                  :  .
       .    @                      :  .
     O ----------------------------+----
                                   A  C

If @ (for the Greek letter theta) is angle AOB, and arc BC is an arc
of the unit circle, so that OB and OA are unit radii, then (using
radian measure) the area of sector COB is @/2.  AB is sin(@), and OA
is cos(@), so

   Area(triangle AOB) < area(sector COB)

    (1/2)*(OA)*(AB)   < area(sector COB)

    (1/2)cos(@)sin(@) < (1/2)@

         cos(@)sin(@) <  @ ,

or
       sin(@)      1
       ------ < ------  .
          @     cos(@)

If OBD is a line, then CD = tan(@), and

   Area(sector COB) < Area(triangle COB) ,

                                     sin(@)
             (1/2)@ < (1/2)tan(@) = --------
                                    2*cos(@) ,

so
                   sin(@)
          cos(@) < ------   .
                      @ 

Combining the inequalities,

                   sin(@)      1
          cos(@) < ------ < ------ 
                      @     cos(@)    .

Since 

     lim cos(@) = 1   ,
    @->0

then
          sin(@)
     lim  ------ = 1 .
    @->0     @


If you have any questions, please write back and I will try to explain
further.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/