In
[link] and
[link] , the proofs were fairly straightforward, since the functions with which we were working were linear. In
[link] , we see how to modify the proof to accommodate a nonlinear function.
Proving a statement about the limit of a specific function (geometric approach)
Prove that
Let
The first part of the definition begins “For every
so we must prove that whatever follows is true no matter what positive value of
ε is chosen. By stating “Let
we signal our intent to do so.
Without loss of generality, assume
Two questions present themselves: Why do we want
and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for
we will discover that
involves the quantity
Consequently, we need
In answer to the second question: If we can find
that “works” for
then it will “work” for any
as well. Keep in mind that, although it is always okay to put an upper bound on
ε , it is never okay to put a lower bound (other than zero) on
ε .
Choose
[link] shows how we made this choice of
This graph shows how we find δ geometrically for a given ε for the proof in
[link] .
We must show: If
then
so we must begin by assuming
We don’t really need
(in other words,
for this proof. Since
it is okay to drop
Hence,
Recall that
Thus,
and consequently
We also use
here. We might ask at this point: Why did we substitute
for
on the left-hand side of the inequality and
on the right-hand side of the inequality? If we look at
[link] , we see that
corresponds to the distance on the left of 2 on the
x -axis and
corresponds to the distance on the right. Thus,
We simplify the expression on the left:
Then, we add 2 to all parts of the inequality:
We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:
The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions. Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the primary tool used in proofs of statements about limits. For
[link] , we take on a purely algebraic approach.
![This graph shows how to find delta geometrically for a given epsilon for the above proof. First, the function f(x) = x^2 is drawn from [-1, 3]. On the y axis, the proposed limit 4 is marked, and the line y=4 is drawn to intersect with the function at (2,4). For a given epsilon, point 4 + epsilon and 4 – epsilon are marked on the y axis above and below 4. Blue lines are drawn from these points to intersect with the function, where pink lines are drawn from the point of intersection to the x axis. These lines land on either side of x=2. Next, we solve for these x values, which have to be positive here. The first is x^2 = 4 – epsilon, which simplifies to x = sqrt(4-epsilon). The next is x^2 = 4 + epsilon, which simplifies to x = sqrt(4 + epsilon). Delta is the smaller of the two distances, so it is the min of (2 – sqrt(4 – epsilon) and sqrt(4 + epsilon) – 2).](/ocw/mirror/col11964/m53493/CNX_Calc_Figure_02_05_003.jpg)
