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Mathematics
Milefoot.com
Mathematics
How To Construct a Delta-Epsilon Proof
The proof, using delta and epsilon, that a function has a limit will
mirror the definition of the limit. Therefore, we first recall the
definition:
We use MathJax
lim f (x) = L means that
x→c
for every ϵ > 0 , there exists a δ > 0 , such that for every x ,
the expression 0 < |x − c| < δ implies |f (x) − L| < ϵ .
Each phrase of the definition contributes to some aspect of the proof. Specifically:
The phrase "for every ϵ > 0 " implies that we have no control over epsilon, and
that our proof must work for every epsilon.
The phrase "there exists a δ > 0 " implies that our proof will have to give the
value of delta, so that the existence of that number is confirmed. Typically, the
value of delta will depend on the value of epsilon.
The phrase "such that for every x " implies that we cannot restrict the values of x
any further than the next restriction provides.
The phrase "the expression 0 < |x − c| < δ " is the starting point for a series
of implications (algebra steps) which will conclude with the final statement. The
expression |x − c| < δ means that the values of x will be close to c ,
specifically not more than (nor even equal to) delta units away. The expression
0 < |x − c| implies that x is not equal to c itself.
The phrase "implies |f (x) − L| < ϵ " is the conclusion of the series of
implications. Once this statement is reached, the proof will be complete.
Upon examination of these steps, we see that the key to the proof is the identification of
the value of delta. To find that delta, we typically begin with the final statement
|f (x) − L| < ϵ , and work backwards until we reach the form |x − c| < δ . So let's
consider some examples. Linear examples are the easiest. Non-linear examples exhibit
a few other quirks, and we will demonstrate them below also.
Example using a Linear Function
Prove, using delta and epsilon, that lim(5x − 7) = 13 .
x→4
We will place our work in a table, so we can provide a running commentary of our
thoughts as we work.
|f (x) − L| < ϵ
Before we can begin the proof, we must first
determine a value for delta. To find that
delta, we begin with the final statement and
work backwards.
|(5x − 7) − 13| < ϵ
We substitute our known values of f (x) and
L.
|5x − 20| < ϵ
Our short-term goal is to obtain the form
|x − c| < δ . So we begin by simplifying
inside the absolute value.
|5(x − 4)| < ϵ
|5||x − 4| < ϵ
|x − 4| <
ϵ
5
We now recall that we were evaluating a
limit as x approaches 4, so we now have the
form |x − c| < δ . Therefore, since c must
be equal to 4, then delta must be equal to
epsilon divided by 5 (or any smaller positive
value).
|x − c| < δ
δ=
In these three steps, we divided both sides of
the inequality by 5. Most often, these steps
will be combined into a single step.
However, when the slope of the linear
function is negative, you may want to do the
steps separately so as to avoid incorrectly
handling the negative sign.
ϵ
5
Now we are ready to write the proof. Once again, we will provide our running
commentary.
Proof.
Suppose ϵ > 0
has been provided.
Define δ =
ϵ
.
5
Since ϵ > 0 , then
we also have
This is always the first line of a delta-epsilon
proof, and indicates that our argument will
work for every epsilon.
Since the definition of the limit claims that a
delta exists, we must exhibit the value of
delta. We use the value for delta that we
found in our preliminary work above.
The definition does place a restriction on
what values are appropriate for delta (delta
must be positive), and here we note that we
δ > 0.
have chosen a value of delta that conforms
to the restriction.
Now, for every x ,
the expression
0 < |x − c| < δ
This is the next part of the wording from the
definition of the limit.
implies
|x − 4| <
ϵ
5
|5x − 20| < ϵ
We replace the values of c and delta by the
specific values for this problem. From here
on, we will be basically following the steps
from our preliminary work, but in reverse
order.
We multiplied both sides by 5. If the slope of
the original function was negative, we may
want to do this using more steps, so as to
introduce the negative sign correctly.
Now we break the expression into the two
|(5x − 7) − 13| < ϵ parts we need to exhibit, the original
function and the limit value.
lim(5x − 7) = 13 .
Since we began with c = 4 , and we
obtained the above limit statement, we have
met all of the requirements of the definition
of the limit, and obtained our final result.
Q.E.D.
This is an abbreviation for the Latin
expression "quod erat demonstrandum",
which means "which was to be
demonstrated". Some authors will include it
to denote the end of the proof.
Therefore,
x→4
Example using a Non-Linear Function
Prove, using delta and epsilon, that lim(3x 2 − 1) = 74.
x→5
|f (x) − L| < ϵ
|(3x2 − 1) − 74| < ϵ
Before we can
begin the proof,
we must first
determine a value
for delta. To find
that delta, we
begin with the
final statement
and work
backwards.
We substitute our
known values of
f (x) and L .
|3x2 − 75| < ϵ
Our short-term
goal is to obtain
the form
|x − c| < δ .
However, with
non-linear
functions, it is
easier to work
toward solving
for x by itself,
then introduce the
value of c . So we
begin by
simplifying inside
the absolute
value.
−ϵ < 3x2 − 75 < ϵ
With non-linear
functions, the
absolute values
will have to be
removed, since
the allowable
delta-distances
will be different
on the two sides
of the value
x = c.
75 − ϵ < 3x2 < 75 + ϵ
In these three
steps, we solve
for the variable x ,
by first adding 75
to each
expression, then
dividing each
expression by 3,
and finally taking
the square root of
each expression.
The square root
function is
increasing on all
real numbers, so
the inequality
does not change
direction. If you
are using a
decreasing
25 −
ϵ
ϵ
< x2 < 25 +
3
3
‾‾‾‾‾‾ϵ‾
‾‾‾‾‾‾ϵ‾
25 − < x < 25 +
√
√
3
3
−5 +
‾‾‾‾‾‾ϵ‾
‾‾‾‾‾‾ϵ‾
25 − < x − 5 < −5 + 25 +
√
√
3
3
|x − c| < δ
−δ < x − c < δ
function, the
inequality signs
will switch
direction. Notice
that the two ends
of the inequality
are no longer
opposites of one
another, which
means that
absolute values
could not be used
to write these as a
single inequality.
Also, the left
hand expression
can be undefined
for some values
of epsilon, so we
must be careful in
defining epsilon.
Since our shortterm goal was to
obtain the form
|x − c| < δ , and
the value of c is
5, we subtract 5
from each
expression.
Since the two
ends of the
expression above
are not opposites
of one another,
we cannot put the
expression back
into the form
|x − c| < δ .
Therefore, we
shall expand this
absolute value
expression
instead.
There are two
candidates for
delta, and we
need delta to be
less than or equal
to both of them.
Therefore, we
will require that
delta be equal to
the minimum of
the two
quantities. Notice
that since the leftend expression
‾‾‾‾‾‾ϵ‾
‾‾‾‾‾‾ϵ‾
was equivalent to
δ = min 5 − 25 − , −5 + 25 +
{
√
√
3
3 } negative delta, we
used its opposite
in our definition
of delta.
However, since
the first candidate
is undefined for
ϵ > 75 , we will
need to handle
the "large
epsilon" situation
by introducing a
second, smaller
epsilon in the
proof.
Now, we are ready to write the proof.
Proof.
Suppose ϵ > 0 has been provided.
Let ϵ2 = min{ϵ, 72} .
This is always
the first line of a
delta-epsilon
proof, and
indicates that our
argument will
work for every
epsilon.
To avoid an
undefined delta,
we introduce a
slightly smaller
epsilon when
needed. In this
example, the
value of 72 is
somewhat
arbitrary, but
does need to be
smaller than 75.
Since the
definition of the
limit claims that
a delta exists, we
must exhibit the
value of delta.
We use the value
for delta that we
Define
found in our
‾‾‾‾‾‾ϵ‾
‾‾‾‾‾‾ϵ‾
δ = min 5 − 25 − , −5 + 25 +
. preliminary work
{
√
√
3
3 } above, but based
on the new
second epsilon.
Therefore, this
delta is always
defined, as ϵ2 is
never larger than
72.
Since ϵ2 > 0 , then we also have δ > 0 .
The definition
does place a
restriction on
what values are
appropriate for
delta (delta must
be positive), and
here we note that
we have chosen
a value of delta
that conforms to
the restriction.
The result is not
real obvious, but
can be seen as
follows. Inside
the square root
expressions
above, when
subtracting from
25, the square
root will be
slightly smaller
than 5, so the
first delta
candidate is
positive. When
adding to 25, the
square root in the
second candidate
will be slightly
larger than 5, so
the second delta
candidate is also
positive.
Therefore, their
minimum is also
positive.
Now, for every x , the expression
0 < |x − c| < δ implies
This is the next
part of the
wording from
the definition of
the limit.
When we have
two candidates
for delta, we
need to expand
the absolute
value inequality
so we can use
both of them.
Then we replace
−δ < x − c < δ
the values of c
‾‾‾‾‾‾‾
ϵ2‾
‾‾‾‾‾‾‾
ϵ2‾ and delta by the
−5 + 25 −
< x − 5 < −5 + 25 +
√
√
3
3 specific values
for this problem.
From here on,
we will be
basically
following the
steps from our
preliminary
work, but in
reverse order.
‾‾‾‾‾‾‾
ϵ2‾
‾‾‾‾‾‾‾
ϵ2‾
25 −
< x < 25 +
√
√
3
3
25 −
ϵ2
ϵ2
< x2 < 25 +
3
3
75 − ϵ2 < 3x2 < 75 + ϵ2
−ϵ2 < 3x2 − 75 < ϵ2
We added 5 to
each expression,
then squared
each expression,
then multiplied
each by 3, then
subtracted 75.
Now we
recognize that
|3x2 − 75| < ϵ2 ≤ ϵ
the two ends of
our inequality
are opposites of
each other, so we
can write the
result as a single
absolute value
inequality.
Furthermore, ϵ2
is always less
than or equal to
the original
epsilon, by the
definition of ϵ2 .
|(3x2 − 1) − 74| < ϵ
Then we rewrite
our expression
so that the
original function
and its limit are
clearly visible.
Therefore, lim(3x 2 − 1) = 74.
Having reached
the final
statement that
|f (x) − L| < ϵ ,
we have finished
demonstrating
the items
required by the
definition of the
limit, and
therefore we
have our result.
Q.E.D.
Our proof is
complete.
x→5