Using the epsilon-delta definition of a limit in calculus can be challenging. (That’s why, after using it for a few examples, we derive some easier techniques, and never use the definition directly unless we have to!) We’ll start with an overview of what the definition means, and then look at several examples of how it is applied to particular functions.
Humanizing the definition
We have had many questions in this area, but the following one provides the most thorough introduction. It comes from Rebecca in 2001:
Formal Definition of a Limit Could you please explain to me the formal definition of a limit. I need help specifically with finding a delta for a given epsilon and using the epsilon-delta definition of a limit. The problems tell me to find the limit and then find delta greater than 0 such that |f(x) - L| is less than 0.01 whenever 0 is less than |x - c| is less than delta. The other set of problems tells me to find the limit L and then use the epsilon-delta definition to prove that the limit is L. I have no idea how to do this at all - it's quite over my head. I am also taking this course by correspondence, so I have no one to help me. Thank you!
Doctor Fenton answered, taking this rather abstract definition and making it a little more human, before demonstrating how to apply it:
Thanks for writing to Dr. Math. This isn't an easy concept to explain, but see if the following helps. The definition of a limit is a very sophisticated idea, and it's important to be sure you understand what you're trying to do. The statement lim f(x) = L x->a means that by taking a value of x very close, but not equal, to a, the function value f(x) will be very close to L. We need an explicit test for this idea.
This is the basic concept behind a limit: that the function value \(f(x)\) approaches some value L as x approaches a value a. In symbols, \( \displaystyle \lim_{x\rightarrow a}f(x) = L \).
First of all, how do you TEST the statement "x is close to a"? Whether two numbers, say 5 and 6, are "close" is a matter of opinion, but given another number, say 5.5, we can definitely say that 5.5 is closer to 5 than 6 is. To decide if numbers are "close" we need a "standard of closeness," which can be any positive number. We will say that two numbers p and q are close according to the standard c if |p-q|<c. For brevity, I will just say that p and q are c-close. So, if c = .01, 5.2 and 5 are NOT .01-close, but 5.005 and 5 are .01-close, because |5.2-5| = .2 > .01, while |5.005-5| = .005 < .01 .Think of the function f(x) as a "machine": a scientific or graphing calculator is a good example. You push a key indicating the function f, such as "sin" or "log," and type in a value of x, push "enter," and the display changes to the value f(x) of the function f at x. Imagine some machine generating random values of x that can be put into the machine f. You are the "input quality control inspector;" you must decide whether to let the machine evaluate f(x), and you must base your decision solely on whether x passes an "input closeness" test, that is, whether x is close to a.At the output end, an "output quality control inspector" will be judging your work. She will have her own standard of closeness, and she will test the output for closeness to L. If it passes, you get paid; if it fails, you will be fired.So, you will have a positive number for an input closeness standard (the traditional name for the input closeness standard is the Greek letter delta, which I will write d), and the output inspector will have her own standard of closeness, traditionally called epsilon, which I will write e.
This convention of using δ (delta) and ε (epsilon) in the definition of limits goes back to Cauchy in 1823; it is believed that they stand for “difference” (in the function input) and “error” (in the output).
A simple example: linear functions
There is one final consideration. Consider two functions f1 and f2, where
f1(x) = x+1
and f2(x) = 2x+1.
Clearly, if x is very close to 2, then f1(x) is very close to 3, while f2(x) is very close to 5, but we can be more precise. We can write
|f1(x)-3| = |x+1-3|
= |x-2| .
This equation says that the output closeness which the output inspector will measure is exactly equal to the input closeness, for the function f1. For the function f2, we have
|f2(x)-7| = |2x+1-5|
= |2x-4|
= 2|x-2|
which says that the "output closeness" is twice the input closeness.
In both cases, it is clear that there must be a relation between the input and output closeness standards. For f1, if the output inspector has a standard of closeness smaller than your input standard, some inputs will pass your test but fail the output test, and you will be fired. In fact, if the output inspector is out to get you fired, and if she knows your input standard, she can come in after you have left, run numbers that pass your input test through the machine until an f(x) comes out that is not exactly equal to L, and then choose her output standard so small that the output test will fail for that x. This test will only be fair if the output inspector publicly announces her output closeness standard, epsilon, before you have to choose your input standard, delta.
So, if you are the inspector for function f1, and the output inspector announces that epsilon = .01 will determine which outputs pass, you know you may simply take your input closeness standard, delta, also equal to .01, since you know, by your analysis of the machine, that the output closeness is equal to the input closeness, so if x passes the input test with delta = .01, then f1(x) will pass the output test |f1(x)-3|<.01 . Now, if the output inspector retires, and a new output inspector arrives determined to improve the quality of output, so that from now on, epsilon will be .001, then you know you can keep your job if you pick delta = .001. If you then get promoted to becoming input inspector for f2, where the output inspector has a closeness standard of epsilon = .0001, then your analysis of the machine f2 shows that your input had better pass |x-2| < .00005, so you take delta = .00005.The key is that no matter what output closeness standard, epsilon, is chosen, it is always POSSIBLE to find an input closeness standard, delta, that will guarantee that if the input x is delta-close to a,|x-a|<d , then the output f(x) will be epsilon-close to L, |f(x)-L|<e.That is, given any positive e, there exists a d such that any x (different from a) which satisfies |x-a|<d will produce a value f(x) that satisfies |f(x)-L|<e .
In proper symbols, this says that \(\displaystyle\lim_{x\rightarrow a}f(x) = L\) when, for any positive number ε, there exists a positive number δ such that for any \(x\ne a\) such that \(\left|x-a\right|<\delta\), the function value will satisfy \(\left|f(x)-L\right|<\epsilon\).
How do we apply this definition?
In a limit proof from the definition, your objective is to find a strategy for choosing d after e is given. For most of the problems usually encountered, you can find a relation between the input closeness and the output closeness of the form |f(x)-L|<= C|x-a| . If you can find such an inequality, then it is easy to choose d when e is given. For a function satisfying such an inequality, the output closeness is no more than C times the input closeness, so you can guarantee that the output closeness |f(x)-L| < e if the quantity C|x-a|<e, since C|x-a| is known to be larger than |f(x)-L|. Since C|x-a|<e if |x-a|<e/C, this suggests that we choose d as e/C. Then, if |x-a|<d = e/C, then |f(x)-L| <= C|x-a| < C[e/C] = e , and f(x) passes the output closeness test.
A harder example: x2
Now Doctor Fenton gives an example a little harder than the linear functions \(f_1\) and \(f_2\).
How do we find such an inequality for a given function? The functions f1 and f2 above show that it is easy (actually an equality) for linear functions f(x) = mx+b. What about f(x) = x^2 near a = 2? We would expect that
lim x^2 = 4
x->2 .
Look at the output closeness:
|f(x)-L| = |x^2 - 4|
= |x+2||x-2| .
Here, the output closeness is a variable multiple of the input closeness. However, we are only interested in x's that are close to 2, so we can make a preliminary restriction to consider only those x's that are at least within 1 unit of 2. That is, we first require
|x-2| < 1 .
If |x-2| < 1, then -1 < x-2 < 1, or 1 < x < 3, and
3 < x+2 < 5,
so that the factor
|x+2| < 5.
Then, for x such that |x-2| < 1,
|x^2-4| < 5|x-2|.
Given e>0, choose d so that both
|x-2| < 1 (to make the inequality true)
and so that
|x-2| < e/5, to make |f(x)-L| < e.
The restriction to require x to be within 1 units of 2 is arbitrary; we just wanted some restriction that would allow us to be sure that \(|x+2|\) never gets larger than a fixed number. In effect, then, we are taking \(\delta\) to be the minimum of 1 and \(\epsilon/5\).
For other introductions to the definition of a limit, see
Formal Definition of a Limit (Doctor Benway, 1998) Limit of x sin(1/x) (Doctor Mitteldorf, 2002)
The latter uses the image of a Devil’s Advocate in the role of the Output Inspector.
For four more explanations of the limit of the square function, see a further discussion below, and also these pages (and a post about it, coming soon — this is a popular topic!):
A Limit Proof Using Estimation (Doctor Rob, 1998) Epsilon/Delta Definition of Limits (Doctors Jerry and Hans, 1999) Delta-Epsilon Proofs and Arbitrary Epsilon Choice (Doctor Peterson, 2002) Definition of the Limit (Doctor Peterson, 2002)
For another example of a limit of a linear function, see
Epsilon - Delta Proofs (Doctor Jerry, 1996)
Proving a limit: 1/x
A 2004 question examines a detail of the technique that was just used, in two examples. The second part is about the answer above, but first we’ll look at a different kind of function, namely 1/x:
Epsilon-Delta Proofs I have two questions on epsilon-delta proofs, one from your site and one found in the book "Calculus 8th edition: Varberg, Purcell, Rigdon". I hope you can help me with them. First, an example from the book: lim 1/x = 1/c x->c Proof: We need 0 < |x - c| < delta so that |1/x - 1/c| < epsilon. After doing some calculations, it comes to: (1/|x|)(1/|c|)(|x - c|)<epsilon Now they say in the book, "The factor 1/|x| is troublesome, especially if x is near 0. To that end note that: |c| = |c + x - x| <= |c - x| + |x| |x| <= |c| - |x - c| Thus if we choose delta <= |c|/2, we succeed in making |x| <= |c|/2. Finally, if we also require delta <= (epsilon)(c)^2/2 then (1/|x|)(1/|c|)(|x - c|)<(1/|c|/2)(1/|c|)(epsilon)(c)^2/2 = epsilon" My question is that I don't understand all those things in the quotes, especially how they got delta <= |c|/2. So I hope you guys can clear that up for me, please.
I answered:
The idea in your first question is that you have to find conditions so that
(1/|x|)(1/|c|)(|x - c|) < epsilon
when |x - c| is small enough, but 1/|x| can get very large if x is small. If c is near zero, then, the expression on the left can get large if you aren't careful to keep x away from 0.
So we look at a number line:
|=======| <-- keep x in here,
<------+---+---+------> and it will stay away from 0
0 c/2 c
In order to keep x away from 0, we just have to keep it close enough to c, namely closer than |c/2|. That's the basic idea. The details are a matter of proving that this will do the trick. The triangle inequality is used to prove that if |x-c| < |c|/2, then |x| > |c|/2, which is obvious from my number line picture! Then we find that under those conditions,
(1/|x|)(1/|c|)(|x-c|) < (1/|c|/2)(1/|c|)delta = 2/c^2 delta
so that we will be within epsilon as long as delta is chosen small enough so that the right side is less than epsilon.
Is that enough to guide you through what they're doing?
More on x2
The second part referred to Doctor Fenton’s answer above about the square function:
You have given a proof regarding lim x->c for x^2 = 4. I don't
understand why the following is done:
|f(x)-L| = |x^2 - 4|
= |x+2||x-2|
Here, the output closeness is a variable multiple of the input
closeness. However, we are only interested in x's that are close to
2, so we can make a preliminary restriction to consider only those
x's that are at least within 1 unit of 2. That is, we first require
|x-2| < 1.
What does it mean to let |x+2| < 5? Is this the "upper bound" that
I've been hearing about? What does upper bound mean? I hope you guys
can help me clear up all these doubtful questions. Thank you very
much for this service that is helping so many out there.
I replied:
Turning to your second question, it's similar to the other. If
|x-2| < delta, then
|f(x) - L| < |x+2| delta
so that as long as we can be sure that |x+2| is kept small enough
(within bounds, that is, so that it never exceeds some fixed number),
we can promise that the right side will be less than epsilon. Again,
we look at a number line:
|=======| <-- keep x within here, and
<------+---+---+---+----> x+2 will be no greater than 5
0 1 2 3
We've arbitrarily chosen the numbers here--we are talking about
keeping x close to 2 in the first place; we arbitrarily choose to make
sure that delta is no greater than 1, and see that under those
conditions, x+2 will be no greater than 3+2, which is 5. That puts a
bound on |x+2|; that is, we know the highest it could possibly get.
So as long as
|x-2| < delta < 1
we know that
|x+2| < 5
and therefore
|f(x)-L| < |x+2| |x-2| < 5 delta
So by requiring delta to be less than both 1 and epsilon/5, we know
that we can keep |f(x)-L| less than epsilon. The algebra just proves
that this all really works.
Proving a limit: rational functions and others
Finally, here is a question from 1998, about several different types of functions:
Delta-Epsilon Limit Proofs Thanks for helping me last time with the question you helped me figure out. Here are four more from my homework which I just couldn't figure: 22. For the limit as x approaches 2 of (4x+1)/(3x-4) = 4.5, illustrate the definition of a limit by finding the values of delta that correspond to epsilon = 0.5 and epsilon = 0.1. 30. Prove that the limit as x approaches 4 of (5-2x) = -3 using the epsilon, delta definition of a limit. Include a diagram. 34. Prove that the limit as x goes to 2 of (x^2+x-6)/(x-2) using the epsilon, delta definition of a limit. 42. Prove that the limit as x goes to -4 of (x^2-1) = 15 using the epsilon, delta definition of a limit. I hope this isn't too much, but it's these sorts of problems that I seem to be having the most trouble with. If you can help, thanks.
The first question is not an actual proof, but just asks for deltas that will work for a couple specific values of epsilon, as practice.
Doctor Jerry briefly answered each question:
Question 22:
-----------
Write (4x+1)/(3x-4) - 4.5 = (4x+1)/(3x-4) - 9/2 and combine the fractions:
|(4x+1)/(3x-4) - 4.5| = (19/2)|x-2|/|3x-4|
When x is within 1 of 2, that is |x-2| < 1, the denominator |3x-4| is not smaller than 1. So:
(19/2)|x-2|/|3x-4| < (19/2)|x-2| if delta is less than 1
So, to force (19/2)|x-2| to be less than 0.5:
|x-2| < 0.5*2/19.
So, we can take delta to be anything less than 1/19, say 1/20.
The initial work, filled out a bit, looks like this:
\(\displaystyle \frac{4x+1}{3x-4} – \frac{9}{2} = \frac{2(4x+1)}{3x-4} – \frac{9(3x-4)}{2(3x-4)} = \frac{8x+2-27x+36}{2(3x-4)} = \frac{-19x+38}{2(3x-4)} = -\frac{19}{2}\frac{x-2}{3x-4}\)
so that
\(\displaystyle \left|\frac{4x+1}{3x-4} – \frac{9}{2}\right| = \frac{19}{2}\frac{|x-2|}{|3x-4|}\).
Then we simplify things by assuming (as is true in the two cases requested) that we will be choosing a delta less than 1, so that x is between 1 and 3, 3x – 4 is between -1 and 5 the denominator |3x – 4| is between … oops! I suspect that Doctor Jerry read it as 4x – 3, because this doesn’t do what he said. We can, instead, assume delta is less than 1/2, so that 3x – 4 will be between 1/2 and 7/2. Since the denominator will then be more than 1/2, replacing it with 1/2 will decrease the denominator, and therefore increase the fraction. So we get
\(\displaystyle \frac{19}{2}\frac{|x-2|}{|3x-4|} < 19|x-2|\)
So now, if we want our error to be less than 0.5, we can just ensure that \(19|x-2|\) is less than 0.5 by making \( |x-2|\) less than 1/19. Any value of delta less than this will be sufficient. And, in fact, when \(x\) is 2 1/20, that is, 2.05, we find that
\(\displaystyle \left|\frac{4x+1}{3x-4} – \frac{9}{2}\right| = \left|\frac{4(2.05)+1}{3(2.05)-4} – \frac{9}{2}\right| = 0.22 < 0.5\).
Note that this is not the very largest value \(x\) could have to make the function value below 0.5; we only guarantee that it will be. One could answer this question by actually solving \(\displaystyle \left|\frac{4x+1}{3x-4} – \frac{9}{2}\right| = 0.5\); the solutions will be \(x=\frac{21}{11}=1.909…\) and \(x=\frac{17}{8}=2.125\), so that our delta could be as large as 0.125. But, again, we don’t need that to prove a limit.
The same sort of thing can be done when epsilon is 0.1.
The next example is linear, so it is quite easy:
Question 30: ----------- Since |-3-(5-2x)| = |-8+2x| = 4|x-2|, to force 4|x-2| < E, it is enough to take |x-2| < E/4.
The next is a rational function like #22, but is much easier, because it can be simplified to a linear function:
Question 34: ----------- Since we won't allow x to be 2, we can say that: (x^2+x-6)/(x-2) = x+3 and so the limit would be 5 as x->2. Since |5-(x+3)| = |x-2|, to make |x-2| < E, it is enough to take |x-2| < E.
Finally, we have a quadratic function, which works much like the square function we looked at before:
Question 42: ----------- Note that |15-(x^2-1)|=|16-x^2|=|4-x||4+x|. Let's first control the annoying factor |4+x|. We do this by making a preliminary condition on delta. If delta is less than one, then if |x-4| < delta, we see that |x+4| is smaller than 5. So: |15-(x^2-1)| = |16-x^2| = |4-x||4+x| < 5|x-4| To force 5|x-4| < E or |x-4| < E/5, it is enough to take delta = E/5 and delta less than 1. So, the "formula" for delta is: If E>0 is given, choose delta to be less than E/5 and less than 1.
These examples should give a good sense of how basic limit proofs work.