Saladsamurai said:
BUT the bigger question is WHY did we decide to put it into that form in the first place?
Well you kind of answered this yourself here:
Saladsamurai said:
I believe that in general, the proof is that given some \varepsilon>0 , there exists some corresponding \delta such that for all 'x' 0<|x-x_o|<\delta \Rightarrow |f(x)-L|<\varepsilon
There is no possible way you're going to find a delta that works by guessing (well sometimes you do, but generally this does not work). You always start by determining how close you want f(x) to be to the limit L, and then determine how sufficiently close you need x to be to x_0 so that the degree of closeness between f(x) and L is actually achieved. Thus, we pick an arbitrary \varepsilon > 0, because if the limit really does exist (which it should), then we can make f(x) as close to L = 1 as we like, which is the same thing as saying that we can make|f(x) - 1| < \varepsilon for any \varepsilon > 0 (if you don't understand this part
geometrically and
intuitively, then revisit that before working algebraically).
Now that we have picked an arbitrary positive epsilon, we need to determine how close x must be to x_0 = 1 so that |f(x) - 1| < \varepsilon holds. Similar to what was just discussed in the above paragraph, making x close to x_0 = 1 means making |x - 1| small. How small? Small enough so that |f(x) - 1| < \varepsilon is satisfied. If we have any hope of doing this, we have to somehow relate the expression f(x) - 1 to x - 1, and the easiest way to do this is to work backwards . Again, we have already prescribed how close we want f(x) and L to be, i.e. we have picked an arbitrary epsilon so that |f(x) - L| < epsilon, so trying to manipulate f(x) -1 to get some info on how close x must be to 1 (how small |x-1| needs to be) makes more sense than doing it the other way around. Now I retrace everything you did in the first post without the algebraic mistake at one step (hopefully you have a better idea of what our goal is):
|f(x) - L| < \varepsilon \Leftrightarrow |\frac{1}{x} - 1| <\varepsilon \Leftrightarrow -\varepsilon < \frac{1}{x} - 1 < \varepsilon \Leftrightarrow 1 - \varepsilon < \frac{1}{x} < 1 + \varepsilon.
At this point, there is a slight hitch, because even though we normally care about very small epsilon, epsilon can be much larger than 1, and in this case, 1 - \varepsilon < 0 which means that flipping the inequality 1 - \varepsilon < \frac{1}{x} can lead to a false statement. But for now, we may temporarily require that 0 < \varepsilon < 1 and further manipulate the last expression. It might be that in the end, we can require x to be sufficiently close to 1 so that any epsilon will work. Note that so far, all of our inequalities are reversible, which means that we can start with the last chain of inequalities and conclude the very first. We can reverse our steps because all we've done is basic algebra. Continuing from above under the temporary assumption 0 < \varepsilon < 1, we have
1 - \varepsilon < \frac{1}{x} < 1 + \varepsilon \Leftrightarrow \frac{1}{1+\varepsilon} < x < \frac{1}{1 - \varepsilon} \Leftrightarrow \frac{1}{1+\varepsilon} - 1 < x - 1 < \frac{1}{1 - \varepsilon} - 1 \Leftrightarrow -\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1 - \varepsilon}.
Now we know that if x - 1 satisfies the last chain of inequalities and 0 < epsilon < 1, then f(x) will be within epsilon of L (i.e. |f(x) - L| < epsilon). However, we generally would like to have symmetry when talking about how close we want a point x to be to another point y on the real line. For instance, |f(x) - L| < epsilon means L - epsilon < f(x) < L + epsilon, which means that when we let epsilon become smaller and smaller, we are "closing in" on the limit L and requiring that f(x) be in the open interval (L - epsilon, L + epsilon). We have satisfied our goal of writing f(x) - 1 in terms of x - 1, but remember the reason we wanted to do this in the first place is so that we can require |x-1| to be small enough so that our function f will map these x values to f(x) such that |f(x) - 1| < epsilon. But remember requiring |x - 1| to be small enough means that we want |x-1| < [some positive number], similarly to what was done for the degree of closeness between f(x) and L = 1. So let d > 0 be a positive number. How do we choose d so that |x-1| < d implies that f(x) is within epsilon of L? Well we have
-\frac{\varepsilon}{1 + \varepsilon} < x - 1 < \frac{\varepsilon}{1 - \varepsilon},
which tells us something about how the distance relationship between x and 1. It tells us that x is between two quantities:
1 -\frac{\varepsilon}{1 + \varepsilon} < x < 1 + \frac{\varepsilon}{1 - \varepsilon},
On the other hand, writing out |x-1| < d as 1 - d < x < 1 + d tells that x is between 1 - d and 1 + d, and we note the symmetry discussed above. But
1 -\frac{\varepsilon}{1 + \varepsilon} < x < 1 + \frac{\varepsilon}{1 - \varepsilon},
is not very symmetrical. To remedy this, if we choose the smaller of the two expressions:
\frac{\varepsilon}{1 + \varepsilon} and \frac{\varepsilon}{1 - \varepsilon},
as our number d, will this work? In general yes, since if we have 1 - A < x < 1 + B for two distinct positive numbers A and B, then if A < B, then 1 - A < 1 + A < 1 + B, so if something is true for 1 - A < x < 1 + B, then it is also true for 1 - A < x < 1 + A, since 1 + A < 1 + B (and we can choose d = A). Similarly you can work out the case when A > B > 0 to find that if something is true for x such that 1 - A < x < 1 + B then it is also true for x with 1 - A < 1 - B < x < 1 + B (and here d = B). By some basic algebra, we find that in this case,
\frac{\varepsilon}{1 + \varepsilon} < \frac{\varepsilon}{1 - \varepsilon}
for 0 < epsilon < 1. Now choose d = \frac{\varepsilon}{1 + \varepsilon}.
By what we just said, it should be the case that
1 - \frac{\varepsilon}{1 + \varepsilon} < x < 1 + \frac{\varepsilon}{1 + \varepsilon} < 1 + \frac{\varepsilon}{1 - \varepsilon},[/itex] <br />
<br />
and noting that last chain of inequalities from when we manipulated f(x) - 1 to something involving x - 1, our choice of d should suffice to show that f(x) is within epsilon of 1. Let us check that it does. We have<br />
<br />
|x-1| &lt; d = \frac{\varepsilon}{1 + \varepsilon} \Rightarrow - d &lt; x - 1 &lt; d \Rightarrow -\frac{\varepsilon}{1 + \varepsilon} &lt; x - 1 &lt; \frac{\varepsilon}{1 + \varepsilon} \Rightarrow 1 - \frac{\varepsilon}{1 + \varepsilon} &lt; x &lt; 1 + \frac{\varepsilon}{1 + \varepsilon} \Rightarrow \frac{1}{\varepsilon + 1} &lt; x &lt; \frac{2\varepsilon + 1}{\varepsilon + 1}.<br />
<br />
Note that our choice of d has gotten rid of earlier problems involving taking reciprocals of both sides of an inequality since we restricted our inequality for the sake of symmetry. Note that x is clearly positive, so we may take reciprocals, taking care to flip the inequality sign as well:<br />
<br />
\frac{1}{\varepsilon + 1} &lt; x &lt; \frac{2\varepsilon + 1}{\varepsilon + 1} \Rightarrow \frac{\varepsilon + 1}{2\varepsilon + 1} &lt; \frac{1}{x} &lt; \varepsilon + 1 \Rightarrow \frac{\varepsilon + 1}{2\varepsilon + 1} - 1 &lt; \frac{1}{x} - 1 &lt; \varepsilon + 1 - 1 \Rightarrow \frac{-\varepsilon}{2\varepsilon + 1} &lt; \frac{1}{x} - 1 &lt; \varepsilon.<br />
<br />
We just have to check that <br />
<br />
\frac{-\varepsilon}{1 + 2\varepsilon} &gt; -\varepsilon \Leftrightarrow \frac{\varepsilon}{1 + 2\varepsilon} &lt; \varepsilon \Leftrightarrow \varepsilon &lt; \varepsilon + 2\varepsilon ^2,<br />
<br />
and this last inequality is clearly true. Hence <br />
<br />
-\varepsilon &lt; \frac{-\varepsilon}{2\varepsilon + 1} &lt; \frac{1}{x} - 1 &lt; \varepsilon <br />
<br />
so -\varepsilon &lt; \frac{1}{x} - 1 &lt; \varepsilon \Rightarrow |\frac{1}{x}-1| &lt; \varepsilon \Rightarrow |f(x) - 1| &lt; \varepsilon, <br />
<br />
which is what we wanted.
I have read your post many times. And I think I get it with the exception of that one part.
