Calculus I, Chapter 3: The Average Rate of Change and Limits
Previous Section: 3.11 The Intermediate Value Theorem
To close this chapter, we end with the formal definition of a limit. At this point we have been studying limits with the goal of identifying how a rate of change works. We stated that a limit is what a function approaches as its input approaches a specific value. But, what does it mean to approach a specific value? This was a real problem whose historical context will be discussed in the next section. In this section we aim to understand how to prove that a limit exists and that all of the math we have been doing is valid.
Visualizing a Limit
Let’s begin by identifying what it means to approach a limit as best we can. In the previous section we used the Intermediate Value Theorem to estimate some values of functions. By making the intervals smaller and smaller we effectively shrink the range of outputs. This process is leading to a more specific answer. The same idea applies to showing that a limit approaches as value. Using the graph of the function, we can draw a box to represent the bounds of the input and output intervals. We can define whether or not a limit exists using this box.
Take the following two sets of graphs as an example of this process:
Using our knowledge from previous sections, we know that the first set of images is of a graph that is continuous and therefore a limit exists at any point, including the one we enclosed. The second set of images does not have a limit at the point because it is not continuous there. The difference lies in the fact that the box is able to shrink, according to the input, so that the output is also shrinking. In the first set of images we are able to zoom in infinitely while in the second set, no matter how close the input values are, the outputs will never get closer. That is why the limit does not exist on the second set.
Defining a Limit
In order to construct our definition we need to name some parts of the process. Let’s define the following things:
c The value that the input approaches, the specific x-value we want
L The value that we want to show that the output approaches, the specific y-value
δ delta, The distance that the x-values that define our ‘box’ can be from c
ε epsilon, The distance that the y-values that define our ‘box’ can be from L
Combining some of these:
|x-c| is the distance from the current, random x-value and the specific x-value (c) that we want to get closer to. Similarly, |f(x)-L| is the distance from the current output and the specific y-value we are claiming is the limit.
If |x-c|<δ then we are claiming the current x-value is close enough to c to be in the box whose width is 2δ. Similarly, if |f(x)-L|<ε then the y-value is in the box whose height is 2ε. Recall that the goal is to have x approach c, or to make the box zoom in. From the pictures above, we know that the goal is to have it be that the y-values have to get closer and if they don’t then the limit does not exist. So we can say that, like the Intermediate Value Theorem, there is a degree of certainty that we may want to achieve. Since ε defines how far from the limit we can be, we want to say that for the limit to exist all ε values must have some corresponding δ values to pair with it. That is to say, if we want to be within 6 units of the limit, then ε=6 and there is some distance from c called δ that defines which x-values we can plug in so that we can achieve that. This is a fairly complicated idea so take a moment to wrestle with what that means. In a lot of words, what we are saying is the following: If, for all epsilon ranges there is a delta range so that if the chosen x-value is within delta units of the goal x-value implies that the chosen y-value is within epsilon units of the goal y-value, then the limit as x approaches c of the function is L. That is quite a mouthful. Here is the logical format: If [ ∀ε>0 ∃δ>0 s.t.( |x-c|<δ → |f(x)-L|<ε )] then lim┬(x→c)〖f(x)〗 = L And finally, the formal definition: Let f : D → R be a function and let c∈D and L∈R. Then f has a limit L at c if for any ε>0, there exists δ>0 such that for all x∈D with 0<|x-c|<δ implies |f(x) – L|<ε.
Here is a visual of a linear function with each of the parts labelled:
A Skeleton
As said previously, a skeleton or some sort of scaffolding is incredibly helpful when learning how to use a specific type of proof. In order to create a skeleton for this proof, called the Epsilon-Delta Proof of a limit, we can use the definition and break it into pieces.
Proof. Always initiate a formal proof with the statement ‘Proof.’
Let f(x) = _ on Define the function and its domain.
Let ε be an arbitrary positive number make sure ε is an undefined distance
Choose δ = , then δ>0 because Choose specific δ (done later), show it fits
Then for all x∈, with 0<|x- | < δ Choose x-values in the appropriate range |f(x) – L | = _______ Show that the y-values are in the necessary range
… will be shown algebraically here
… This will be where you choose δ
< ε to show, no matter what, there is an ε Thus, (lim)┬(x→)〖〗 exists and is equal to __. Conclusion with answer.
There are a couple of things to note. The blanks are intended to represent spots that will change based on the specific limit being evaluated. The steps with the ellipses (…) will vary in technique, strategy, and number of steps depending on the type of function. This method of proof is pretty advanced and is above the level of your typical high school or differential calculus college course. As such, it may a bit of overkill depending on the expectations of your studies. This depth is meant to introduce formal proofs in a context that is accessible and appropriate for the content.
Linear Epsilon-Delta Proofs
Example 1: Show that the limit as x approaches 3 for the function f(x) = x+1 exists and is equal to 4.
Proof
Let f(x) = x+1 on (-∞,∞)
Let ε be an arbitrary positive number
Choose δ = ε, then δ>0 because ε>0
Then for all x∈(-∞,∞), with 0<|x- 3| < δ
|f(x) – L | = |x+1 – 4|
= | x – 3|
< δ
= ε
Thus, lim┬(x→3)〖x+1〗 exists and is equal to 4.
For this first example, the step with |x-3| is the fundamental step in proving this limit exists. The strategy for all of these problems involves manipulating the equation so that the |x-c| appears. This is less than δ because we claimed that as a starting point. The next trick involves turning δ into ε. The easiest way to do this is to take the expression as is and set it equal to ε then solve to find what δ should equal. This is then filled in on third line and reasons are given as to why it is an acceptable choice. This will always involve ε.
Example 2: Use the Epsilon-Delta Proof to show that lim┬(x→-4)〖3x+7〗=-5
Proof
Let f(x) = 3x+7 on (-∞,∞)
Let ε be an arbitrary positive number
Choose δ = ε, then δ>0 because ε>0
Then for all x∈(-∞,∞), with 0<|x+4| < δ
|f(x) – L | = |3x+7+5|
= |3x+12|
= 3|x+4|
< 3δ
=3(ε/3)
=ε
Thus, lim┬(x→-4)〖3x+7〗 exists and is equal to -5.
Polynomial Epsilon-Delta Proofs
When raising the power of the polynomial we run into an issue with the step that introduces delta. The best way to approach this is to use algebra to manipulate the expression in order to create multiple deltas, equal in number to the power of the polynomial. In order to do this we will need to factor and use the triangle inequality to construct the necessary delta terms. This can be tricky initially, so be careful as you read the following examples. There is a second issue that arises with the delta step: delta will have an exponent. This can be resolved by choosing a delta sufficiently small. This is desirable anyways because we want the box to zoom in, so if we say that delta is small then we are just moving ahead the process. The number to choose is that delta is less than or equal to 1. This reason for this is the following algebra:
δ≤1 implies that δ^2≤δ because we can multiply both sides of the inequality by δ since it is positive.
Triangle Inequality: |a+b|≤|a|+|b|
Example 3: Show f(x) = x^2 has a limit at the point x=2.
Proof
Let f(x) = x^2 on (-∞,∞)
Let ε be an arbitrary positive number
Choose δ = min{1,ε/5}, then δ>0 because ε>0
Then for all x∈(-∞,∞), with 0<|x-2| < δ
|f(x) – L | = |x^2-4|
= |x-2||x+2|
=|x-2||x-2+4|
≤|x-2| ( |x-2|+|4|)
=|x-2| ( |x-2|+4)
< δ(δ+4)
=δ^2+4δ
≤δ+4δ
=5δ
=ε
Thus, lim┬(x→2)〖x^2 〗 exists and is equal to 4.
Rational Epsilon-Delta Proofs
To prove that a limit exists when using a function with rational components we utilize an algebraic trick with inequalities. Generally, when dividing we making the term on top smaller. This is true when the denominator is greater than 1. If you don’t divide by that term it is actually larger than if we had divided it. That is to say a/b1. In these proofs it is acceptable to change the values of expressions, as long as each consecutive term is larger or equal in size to the previous term! So, in many cases we can actually just remove the denominator and it makes the problem easier to solve. The only requirement is to show that the denominator is larger than 1.
Example 4: Prove that lim┬(x→-2)〖(x^2-2x-7)/(x-5)〗 exists
Proof
Let f(x) = lim┬(x→-2)〖(x^2-2x-7)/(x-5)〗 on (-∞,5)∪(5,∞)
Let ε be an arbitrary positive number
Choose δ = min{1,ε/48}, then δ>0 because ε>0
Then for all x∈(-∞,5)∪(5,∞), with 0<|x+2| < δ |f(x) – L | = |(x^2-2x-7)/(x-5)-(-1)/7| = |7(x^2-2x-7)/(7(x-5))+(x-5)/(7(x-5))| = |(7x^2-14x-49+x-5)/(7(x-5))| = |(7x^2-13x-54)/(7(x-5))| <|7x^2-13x-54| because lim┬(x→-2) 7(x-5)=-49 and |-49|>1
=|x+2||7x-27|
=|x+2||7x+14-41|
≤|x+2| ( 7|x-2|+|-41|)
=|x+2| ( 7|x-2|+41)
< δ(7δ+41)
=7δ^2+41δ
≤7δ+41δ
=48δ
=ε
Thus, lim┬(x→-2)〖(x^2-2x-7)/(x-5)〗 exists and is equal to (-1)/7.
Rational Epsilon-Delta Proofs
With Rational functions we use the conjugate method that was previously used to solve the limits. Combine that with the previous method to remove the bottom of the conjugate and we can say that √a-b= (√a-b)(√a+b)/(√a+b)<(√a-b)(√a+b) as long as √a+b>1.
Example 5: Show that lim┬(x→4)〖√(2x+7)〗 exists.
Proof
Let f(x) = √(2x+7) on (-3.5,∞)
Let ε be an arbitrary positive number
Choose δ=ε/2, then δ>0 because ε>0
Then for all x∈ (-3.5,∞), with 0<|x-4| < δ |f(x) – L | = |√(2x+7)-√15| < |√(2x+7)-√15||√(2x+7)+√15| because |√(2x+7)+√15|>1
= |2x+7-15|
= |2x-8|
< 2δ
=ε
Thus, lim┬(x→4)√(2x+7) exists and is equal to √15.
Trigonometric Epsilon-Delta Proofs
Using the information from the Squeeze Theorem, we know that each of the base trigonometric functions are less than or equal to 1. This is sufficient for handling trigonometric proofs.
Example 6:
Proving a Limit Does Not Exist, The Disproof
To show a limit we used the following definition:
lim┬(x→c)〖f(x)〗 exists and is equal to L if
∀ε>0∃δ>0 s.t.|x-c|<δ→|f(x)-L|<ε To say that this is NOT true is to say that NOT all epsilons have a delta. That is to say, there is an epsilon that does not have a delta. Further, There exists an epsilon such that for all delta the following statement is false. This would be a statement of the form: lim┬(x→c)〖f(x)〗 does not exist if ∃ε>0∀δ>0 ¬(|x-c|<δ→|f(x)-L|<ε) … to elaborate even further lim┬(x→c)〖f(x)〗 does not exist if ∃ε>0∀δ>0 (|x-c|<δ ∧ ¬|f(x)-L|<ε)
FINALLY
lim┬(x→c)〖f(x)〗 does not exist if
∃ε>0∀δ>0 (|x-c|<δ ∧ |f(x)-L|≥ε)
So… in English, again, we have
There exists some epsilon, such that, for all delta with |x-c| less than delta, the |f(x)-L| is not less than epsilon.
To do a proof of this we can look at the following example:
Let’s show that the limit lim┬(x→2)x does NOT equal 1.
Choose ε = 5, then ε>0 because 5>0
Let δ>0
Let x be in the domain with 0 < | x – 2 | < δ
And |f(x) – L | = |x – 1|
= |x – 2 + 1|
≤|x – 2| + |1|
< δ + 1