Epsilon-Delta (
) proof is a proof of a formula on limits based on the epsilon-delta definition. Most students often have difficulties to quickly mastering the epsilon-delta definition and how to use it for proofs! Styles in proof writing usually vary greatly, and no two people will ever be expected to write exact same proof but the logical structure of any two or more proofs should be more or less the same.
In an
-proof, it is customary to first do some calculations to find the number
, since the proof consists of specifying a value for
in terms of
and showing that the implication in the limit definition holds for this value of
. These calculations to find
, may not necessarily be part of the proof.
Let's consider two things, before we look for an actual example:
Absolute Value
If
is a real number, the absolute value of
is the distance from
to
, written as
. We can alternatively define an absolute value of a real number
,
Therefore, if
is any real number, we have
Here, we can naturally, think of
as the distance from
to
.
However, the following are two important equivalences involving absolute value:
.
Here, the symbol
meas "is equivalent to". In words:
is less than
units from
if and only if the difference
is between
and
if and only if
is in the interval (
).
Epsilon-Delta Definition of a Limit
Suppose that
and
are real numbers and
)
is a function defined in an open interval containing
, except perhaps at
. If for every positive number
there exists a positive number
(which depends on
) such that
 - L|< \epsilon)
,
then we say that
is the limit of
as
approaches
and we write
 = L)
.
Intuitively, this definition says that, If
is a function defined for all values of
near
, except perhaps at
, and if
is a real number such that the values of
get closer and closer to
as the values of
are chosen closer and closer to
, then we say
is the limit of
as
gets closer and closer to
and we write
.
Example
Show that
 = 1)
Proof
Let
and define
. Suppose
. Then, we have
 - 1| = |3x-6| = 3|x-2| < 3(\epsilon/3) = \epsilon)
(since we have made an assumption that
and
).
Therefore we have shown that
implies
 - 1|<\epsilon)
,
and by definition
