Expert: Paul Klarreich - 10/4/2006

Question
Hello Paul
My name is Mary and I am studying Calculus 1.  The problem that I'm having is understanding how you can find out if a function is continuous or not.  Here is the problem:

Given f(x)=(x-4)/(x^2)-16, find all points where f is not defined (and therefore not continuous).  For each such point, tell whether or not the discontinuity is removable.

I am stumped on the entire "continuous", "Not continuous" problems.

Thanks Paul

Answer
Mary Asks in Category Calculus ...
 
Subject:  Continuous or not
 
Question:  Hello Paul
My name is Mary and I am studying Calculus 1.  The problem that I'm having is understanding how

you can find out if a function is continuous or not.  Here is the problem:

Given f(x)=(x-4)/(x^2)-16, find all points where f is not defined (and therefore not continuous).  

For each such point, tell whether or not the discontinuity is removable.

I am stumped on the entire "continuous", "Not continuous" problems.

Thanks Paul
----------------------------------------------------
Hi, Mary,

In general, 'continuous' means there are no holes or breaks in the graph.  You could draw the entire graph without having to lift your pencil.  Any place where you would have to lift your pencil is a discontinuity.  There is also a precise definition for 'continuity at a point':

f(x) is continuous at  x = a if all of these three conditions are satisfied:

A.  f(a) exists. [If there is no value at x=a, you can't plot it and you will have to lift.

B. lim f(x) exists.  
 x->a
 [If f(x) does not approach some single number, you can't get across x=a without some jump.]

C. Both of those are the same. [IF not, the function is approaching some number, but when it gets there, it isn't there.  Well, YOU find a better way to say it.]

Now, in general, we don't often have C, but we could.  More often, we don't have A or B.

Here they are again, without the commentary.

A.  f(a) exists.

B. lim f(x) exists.  
 x->a

C. Both of those are the same.

Sometimes the conditions are just written as a single statement:

lim  f(x) = f(a)   [ABC all together]
x->a

which embodies all three.  You can't say two things are equal unless they both exist.  This one statement says:  When  x is near a,  f(x) is near f(a).  

And that may be a convenient way to think about it.

If we don't have B, i.e. the limit does not exist, then we will have a big space in the graph.

But if we have B and not A, then the only problem in the graph is that tiny missing hole at x=a.  

Perhaps we can REDEFINE the function to plug the hole.  If we can, then this is a REMOVABLE DISCONTINUITY.

What about your example?
        x - 4
f(x) = --------
      x^2 - 16

There will be a problem for any x that makes the denominator zero.  What are these values?

x^2 - 16 = 0  -->  x = 4  or x = -4

Does lim f(x) exist?  Do this computation:
   x->4

           x - 4          1
f1(x) = -------------- = -----
       (x - 4)(x + 4)   x + 4

All we did was factor and cancel the factor of  x-4.  But it is illegal to cancel a factor that

could be zero, and  x-4 is certainly zero when x=4; that's why I'm writing f1(x) for this.

Is  f1 the same as f?  Almost.  It is exactly the same unless x = 4.  That means if we are interested in  lim(x->4) f(x), we will get the same answer if we do lim(x->4) f1(x), because

lim (.....)  
x->4

has nothing to do with  x EQUALS 4.

But lim(x->4) f1(x) is easy.
          1
f1(4) = ------ = 1/8
       4 + 4

So lim(x->4) f(x) = 1/8, and B is satisfied at x = 4.  But  f(4) is undefined.  Since only A is the difficulty, we REDEFINE f(x) to be:

F(x) = (same as f(x)) when x /= 4,  and = 1/8 when x = 4.

This new function is continuous at x = 4, so we have removed the discontinuity.  Understand, however, that we had to change the function.

So how about  x = -4?  If we try that in our f1(x), we have:
           1       1
f1(-4) = ------- = ---
        - 4 + 4    0

So not only is  f(-4) undefined, so is f1(-4) and the limit does not exist.  This discontinuity is not removable.  We cannot just redefine f(-4) and get our missing point.

Hope this helps.  Send along your examples that involve 'Regional' functions.  Those are the ones

that say:

      | Something,  when  x < something
f(x) = | Something else,  when  ... <= x < ...
      | yet another thing,  when  x > ...