/export/home/lstout/MathematicaNotebooks/LimitsTutoriallnk7.html

T.2) Formal definition of lim x->a f[x]

In the examples in the Basics section, we saw that the value of
[Graphics:LimitsTutorialgr35.gif] f[x] should depend on what values f[x] has for points near, but not equal to, a. Once again we want to quantify this.

Also, recall that one way we found limits in the Basics notebook was by using algebraic simplification to eliminate division by 0. When an algebraic simplification can be used to eliminate division by 0 all that is changed in the definition of the function is the definition at the point where division by 0 occurred, not the values for nearby points where both the original function and the simplified function are both defined.
If we define [Graphics:LimitsTutorialgr36.gif] f[x] in such a way that x is never equal to a, we will be able to use algebraic simplification to reduce difficult limits to simple ones.

We can say that x is within a tolerance delta of a by saying
| x - a | < delta
and we can say that x is not equal to a by saying
0 < | x - a | .
So saying that x is within delta of a but not equal to a can be said with 0 < | x - a | < delta .

Our object is to make f[x] close to L. We can quantify that with a tolerance by making | f[x] - L | < epsilon .

Putting these together we get the following definition:

Definition: The limit of f[x] as x goes to a is L if and only if for any epsilon > 0 there is a delta such that
if 0 < | x - a | < delta , then | f[x] - L| < epsilon .

T.2.a)

Take f[x] = 2 x - 3. Use the definition of limit to prove that
[Graphics:LimitsTutorialgr37.gif] f[x] = -1.
Use plots to illustrate what's going on in this proof for epsilon = 0.1.

Answer:

Let's first make a plot that shows the statement, [Graphics:LimitsTutorialgr38.gif] f[x] = -1, is reasonable:

Clear[f];
f[x_] = 2 x -3;
fplot = Plot[f[x], {x, 0.5, 
1.5},AspectRatio->Automatic]

The graph certainly appears to go through {1, -1}. So saying that [Graphics:LimitsTutorialgr41.gif] f[x] = -1 seems reasonable. To prove the limit is actually -1, we need to show that given an epsilon > 0, we can find always find a delta that works. Note that the value of delta can depend on the value of epsilon , but it cannot depend on the value of x.

Step 1) Find an appropriate delta .

Here f[x] changes 2 times as fast as x does so any error in an x value gets multiplied by 2. Given an epsilon > 0, if we think of epsilon as giving the allowed error in approaching L with f[x], it becomes clear that we want delta = epsilon /2.

Step 2) Show that the chosen delta actually works.

Now if
0 < | x-1 | < delta = epsilon /2,
then
2 | x - 1| < epsilon ==> | 2 x - 2 | < epsilon ,
so | 2 x - 3 - (- 1) | < epsilon as needed.

Here's what's going on graphically when epsilon = 0.1. We first focus our attention on the x's that are within delta units of 1. These are the x's that lie inside the green delta -tube shown below:

epsilon = 0.1;
delta = epsilon/2;

deltatube = 
   
Graphics[
		{{CobaltGreen,Line[{{1 - delta, 0}, 
       	{1 - 
delta, f[1 - delta]}}]},
		{CobaltGreen,Line[{{1 + delta, 0}, 
		
{1 + delta, f[1 + delta]}}]}},
		ColorOutput->RGBColor];
		

Show[fplot,deltatube]

The definition of limit, and the proof we've just written say the following: If x lies within the green delta -tube and x is not equal to 1, then we know (even before we plot it) that f[x] will lie within the epsilon -tube created by the lines y = -1 + epsilon and y = -1 - epsilon . Here's a plot to confirm this fact:

epsilontube = Plot[{-1 + epsilon,-1 - epsilon},
				
   {x, 0.5, 1.5}, 			PlotStyle->{Red}, 
DisplayFunction->Identity];

Show[fplot, deltatube, 
epsilontube]

The graph clearly shows that if x is not equal to 1 and if x lies in the green delta -tube, then f[x] lies in the red epsilon -tube.

You should play around with changing the value of epsilon in the above graphs to see what happens. Don't make epsilon to small, or the picuture will be hard to see.

T.2.b)

Take f[x] = x^2 + x + 1. Use the definition of limit to prove that
f[x] = 3.
Use plots to illustrate what's going on in this proof for = 0.1.

T.2.c)

Take f[x] = e^(-x^2). Use the definition of limit to prove that
f[x] = 1.
Use plots to illustrate what's going on in this proof for = 0.1.

T.2.d)

Take f[x] = (x^2 -1) / (x - 1). Use the definition of limit to prove that f[x] = 2.
Use plots to illustrate what's going on in this proof for = 0.1.

T.2.e)

Take f[x] = Sin[1/x]. Use the definition of limit to prove that [Graphics:LimitsTutorialgr84.gif] f[x] cannot exist.

Answer:

We looked at this function back in B.4). Here are two plots to look at:

Clear[f];
f[x_]=  Sin[1/x];
Plot[f[x], {x, -.1, 0.1}]

Plot[f[x], {x, -0.01, 0.01}]

As we zoom in on x = 0, we get more and more "waves" piling up. Furthermore, the waves are remaining constant in height---all the peaks are at y = 1 and the dips are at y = -1. This kind of wild wiggling implies there should be no limit.

To prove this fact from the definition of limit, note that any interval containing 0 will contain points of the form
1/( /2 + 2k ) and 1/(3 /2 + 2k )
if we take k large enough. But if we evaluate f[x] at these points we get
f[ 1/( /2 + 2k )] = Sin[ /2 + 2k ] = Sin[ /2] = 1
and
f[1/(3 /2 + 2k )] = Sin[3 /2 + 2k ] = Sin[3 /2] = -1

Since these two values are 2 units apart, there is no possible L within epsilon = .5 of both of them. Thus no limit exists.