δ- Proofs
Sudesh Kalyanswamy
Example 1. We know limx→0 x2 = 0. Find δ so that if −δ < x < δ, then f (x) = x2 is
between −.01 and .01.
Solution 1. Suppose δ = .1. The condition −δ < x < δ is precisely that |x| < δ (this
is important to understand!). To say that f (x) is between −.01 and .01 is to say that
|x2 − 0| < .01. We write it like this only to compare to the statement in the limit definition
which we care most about in this problem: If |x − a| < δ, then |f (x) − L| < . So in this
case, = .01, a = 0, and L = 0. So suppose |x| < .1. Then
|x2 − 0| = |x2 | = |x|2 < (.1)2 = .01 = .
Thus the given value of δ works.
The problem with this, of course, is that this didn’t tell us how to find δ. So let’s work
through that:
Example 2. limx→0 2x = 0. Find δ so that −δ < x < δ ensures f (x) is between −.002 and
.002.
Solution 2. Let’s break this down into steps:
(1) Identify : If no error for f (x) is given, then use a general . In this case, we want
f (x) between ±.002. The represents the error from the limit, which in this case is
just = .002.
(2) Write down the statement you want: We want to find δ so that If |x − a| < δ
then |f (x) − L| < . In this case, we want that if |x − 0| < δ, then |2x − 0| < .002.
(We are using that a = 0, L = 0, and = .002.)
(3) Work with the |f (x) − L| term to somehow find δ: We have this |2x − 0| = |2x|
term on the left. We want this to be smaller than .002 = . Well x can be anything
between −δ and δ, so let’s first ask “How large can |2x| be?” Well |2x| = 2|x|, and we
are assuming |x| < δ (this is why we do step (2) and write down what we want; the
first part is IF |x| < δ!!!). So the largest 2|x| can be is 2δ since |x| can be no bigger
than δ. So
2|x| < 2δ.
Now we want |2x| < , so if we make 2δ < , then we are done, since this is bigger
than |2x|. So
2δ < .002 =⇒ δ < .001.
So the largest δ which will work is δ = .001 , but any positive δ less than this will
work as well.
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The moral of this example is that you work with the |f (x) − L| term and, just as we did
with the infinite limits and the M-term there, we ask what’s the largest |f (x) − L| term can
be, and get this entirely in terms of constants and δ’s, no x’s.
Example 3. We know limx→3 2x + 1 = 7. Find the largest δ so that 3 − δ < x < 3 + δ
ensures 2x + 1 is between 6.9 and 7.1.
Solution 3. Again, let’s work through the steps in the previous example:
(1) Here, = .1, this is the error from the limit (in this case, 7) which we are allowing.
(2) We want: If |x − 3| < δ, then |2x + 1 − 7| < .1 (the 3 − δ < x < 3 + δ is exactly that
|x − 3| < δ).
(3) Start from the |f (x) − L| term:
|2x + 1 − 7| = |2x − 6| = 2|x − 3|.
Since the largest |x − 3| can be is δ (the “if” above), 2|x − 3| < 2δ. Thus
2|x − 3| < 2δ < = .1 =⇒ δ < .1/2 = .05.
Thus the largest δ which works is δ = .05 .
Example 4. Find the largest δ corresponding to general > 0 for limx→1 (x − 1) sin(x) = 0.
Solution 4. Here, no was given. In your head, you can imagine it is .1 or .01, and work
it out as if it were that. But ultimately, we just say “pick > 0.” This means give me any
error you want. We want to show that IF|x − 1| < δ, then |(x − 1) sin(x) − 0| < . (Again,
you can write down = .1 as an example.) Well, working with the |f (x) − L| term, we have
|(x − 1) sin(x)| = |x − 1|| sin(x)|. So we want the largest this can be in terms of δ. Well the
|x − 1| < δ (we are assuming this). However, we cannot do that with the | sin(x)|. In fact,
it will be hard to do this because it is hard to relate sin(x) to x − 1. However, thankfully,
| sin(x)| ≤ 1. So
|x − 1|| sin(x)| < δ · 1 = δ.
Now we want this less than , so we can say δ < . Therefore the largest δ which will work
is δ = .
Example 5. We have limx→1
is between .9 and 1.1.
1
x
= 1. Find the largest δ such that 1 − δ < x < δ ensures
1
x
Solution 5. In this problem, = .1. We want IF |x − 1| < δ, then x1 − 1 < = .1. In this
problem, it will be important that |x − 1| < δ is the same as 1 − δ < x < 1 + δ. Also note
that because x1 has that asymptote at x = 0, we can assume δ < 1, so that our interval
1 on
the x-axis is fully contained on the positive side of the x-axis. Ok, so we have the x − 1
term. Combine the fractions to get
1
− 1 = 1 − x = 1 − x = |1 − x| = |x − 1| ,
x
x x x |x|
|x|
2
where the last step comes from the fact that |1 − x| = | − (x − 1)| = |x − 1|. We want to
know the largest this can be. Well a fraction is biggest when the numerator is biggest and
the denominator is smallest. The numerator is |x − 1|, and the largest this can be is δ. For
the denominator, the smallest x can be is 1 − δ (this is where we use 1 − δ < x < 1 + δ). So
|x − 1|
δ
<
< .1 = .
|x|
1−δ
Solve for δ, and you get δ < .1/1.1. So the largest δ is δ = .1/1.1 .
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