Calculus:  The Language of Change

Exercise set 5.5.1-CD
Continuity and the Derivative

by Brandon W. Buster

1.

Part 1.

Suppose that f[x] is smooth on an interval around a so that  the "microscope" increment equation is valid.  Suppose that x≈a so that,  x=a + δx for δx≈0.  Show that f[x]=f[a + δx]≈f[a]; in other words, show that smooth real functions are continuous at real points.

Solution to Part 1.

We know that δx = a - x, so we use the approximation
	f[a]=f[x + δx]= f[x]+ f'[x]δx + ϵδx.
where [f'[x] + ϵ]δx is a medium value times a small value which gives us a small value.  Therefore, f[a]≈f[x].

Part 2.

Consider the real function f[x] = (1/x), which is undefined at x = 0.  We could extend the definition by simply assigning f[0] = 0.  Show that this function is not continuous at x = 0, but is continuous at every other real x.

Solution to Part 2.

Let a = 0 and x = a + δx for δx ≈ 0.  If f[x] is continuous at 0, then f[0] ≈ f[x].  We are given that f[0] = 0.  If δx ≈ 0 then f[x] is very large and cannot be "close" to zero.  So f[0] is not close to f[x] and f[x] is not continuous at zero.  Evaluate the cell below to see this discontinuity in the graph.

Clear[j, x]; j[x_] := (1/x); j[x]Plot[j[x], {x, -10, 10}, PlotStyle -> {Thickness[0.01], Blue}];

Part 3.

Give an intuitive graphical description of the definition of continuity in terms of powerful microscopes and explain why it follows that smooth functions must be continuous.

Solution to Part 3.

Graphically, a function is continuous if f[x] is close to f[a] when x is close to a, for every x ≈ a, f[x] is defined and f[x] ≈ f[a].  So for any magnification we should be able to see both f[x] and f[a].  From Chapter 3 we know that a highly magnified view of a smooth graph appears as a straight line.  Since we can see f[x] for any x in our window, it follows that smooth functions must be continuous.

Part 4.

The function f[x] = (Sqrt[x]) is defined for x ≥ 0; there is nothing wrong with f[0].  However, our increment computation for (Sqrt[x]) above was not valid at x = 0 because a microscopic view of the graph focused at x = 0 looks like a vertical ray (or half-line).  Explain why this is so, but show that f[x] is still continuous "from the right;" that is, if 0≺ x ≈ 0, then (Sqrt[x]) ≈ 0 ,but (((Sqrt[x]))/x)is very large.

Solution to Part 4.

Evaluate the cell below to see what happens to our magnified view of f[x] near x = 0 as we increase the magnification.

\(\(f[x_] := Sqrt[x]; 
x = 0.001; 
SetOptions[Plot, Ticks\  -> \ None]; 
    
Print["Magnifying near x = ", x]; 

    Do[delta\  = \ 1/\((1.5^n)\); 
\ \ 	
      \(Plot[f[x + dx] - f[x], {dx, \(- .001\), delta}, 
\ \ 		
        PlotRange\  -> \ {\(-0.001\), delta}, 
        PlotStyle -> {Thickness[0.01], Blue}, AxesOrigin -> {0, 0}]; \)

      	, {n, 0, 16}]
\)\)

We see that the view becomes more and more vertical as we zoom in.  From Example 5.6 we know that
	f'[x] = (1/(2(Sqrt[x])))
So for positive x ≈ 0, f'[x] becomes large.  Additionally, (((Sqrt[x]))/x) = (1/((Sqrt[x]))) becomes very large as x becomes very small.

Clear[k, x]; \(k[x_] := \@x; 
k[x]\)Plot[k[x], {x, 0, 10}, PlotStyle -> {Thickness[0.01], Red}];

Clear[k, x]; \(k[x_] := \@x\/x; 
k[x]\)Plot[k[x], {x,  .001, 10}, PlotStyle -> {Thickness[0.01], Red}];