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Prove X4 is Continuous Using Epsilon Delta

Learning Objectives

  • 2.5.1 Describe the epsilon-delta definition of a limit.
  • 2.5.2 Apply the epsilon-delta definition to find the limit of a function.
  • 2.5.3 Describe the epsilon-delta definitions of one-sided limits and infinite limits.
  • 2.5.4 Use the epsilon-delta definition to prove the limit laws.

By now you have progressed from the very informal definition of a limit in the introduction of this chapter to the intuitive understanding of a limit. At this point, you should have a very strong intuitive sense of what the limit of a function means and how you can find it. In this section, we convert this intuitive idea of a limit into a formal definition using precise mathematical language. The formal definition of a limit is quite possibly one of the most challenging definitions you will encounter early in your study of calculus; however, it is well worth any effort you make to reconcile it with your intuitive notion of a limit. Understanding this definition is the key that opens the door to a better understanding of calculus.

Quantifying Closeness

Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance between two points a and b on a number line is given by | a b | . | a b | .

  • The statement | f ( x ) L | < ε | f ( x ) L | < ε may be interpreted as: The distance between f ( x ) f ( x ) and L is less than ε.
  • The statement 0 < | x a | < δ 0 < | x a | < δ may be interpreted as: x a x a and the distance between x and a is less than δ.

It is also important to look at the following equivalences for absolute value:

  • The statement | f ( x ) L | < ε | f ( x ) L | < ε is equivalent to the statement L ε < f ( x ) < L + ε . L ε < f ( x ) < L + ε .
  • The statement 0 < | x a | < δ 0 < | x a | < δ is equivalent to the statement a δ < x < a + δ a δ < x < a + δ and x a . x a .

With these clarifications, we can state the formal epsilon-delta definition of the limit.

Definition

Let f ( x ) f ( x ) be defined for all x a x a over an open interval containing a. Let L be a real number. Then

lim x a f ( x ) = L lim x a f ( x ) = L

if, for every ε > 0 , ε > 0 , there exists a δ > 0 , δ > 0 , such that if 0 < | x a | < δ , 0 < | x a | < δ , then | f ( x ) L | < ε . | f ( x ) L | < ε .

This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if we break it down phrase by phrase. The statement itself involves something called a universal quantifier (for every ε > 0 ), ε > 0 ), an existential quantifier (there exists a δ > 0 ), δ > 0 ), and, last, a conditional statement (if 0 < | x a | < δ , 0 < | x a | < δ , then | f ( x ) L | < ε ). | f ( x ) L | < ε ). Let's take a look at Table 2.9, which breaks down the definition and translates each part.

Definition Translation
1. For every ε > 0 , ε > 0 , 1. For every positive distance ε from L,
2. there exists a δ > 0 , δ > 0 , 2. There is a positive distance δ δ from a,
3. such that 3. such that
4. if 0 < | x a | < δ , 0 < | x a | < δ , then | f ( x ) L | < ε . | f ( x ) L | < ε . 4. if x is closer than δ δ to a and x a , x a , then f ( x ) f ( x ) is closer than ε to L.

Table 2.9 Translation of the Epsilon-Delta Definition of the Limit

We can get a better handle on this definition by looking at the definition geometrically. Figure 2.39 shows possible values of δ δ for various choices of ε > 0 ε > 0 for a given function f ( x ) , f ( x ) , a number a, and a limit L at a. Notice that as we choose smaller values of ε (the distance between the function and the limit), we can always find a δ δ small enough so that if we have chosen an x value within δ δ of a, then the value of f ( x ) f ( x ) is within ε of the limit L.

There are three graphs side by side showing possible values of delta, given successively smaller choices of epsilon. Each graph has a decreasing, concave down curve in quadrant one. Each graph has the point (a, L) marked on the curve, where L is the limit of the function at the point where x=a. On either side of L on the y axis, a distance epsilon is marked off  - namely, a line is drawn through the function at y = L + epsilon and L – epsilon. As smaller values of epsilon are chosen going from graph one to graph three, smaller values of delta to the left and right of point a can be found so that if we have chosen an x value within delta of a, then the value of f(x) is within epsilon of the limit L.

Figure 2.39 These graphs show possible values of δ δ , given successively smaller choices of ε.

Example 2.39 shows how you can use this definition to prove a statement about the limit of a specific function at a specified value.

Example 2.39

Proving a Statement about the Limit of a Specific Function

Prove that lim x 1 ( 2 x + 1 ) = 3 . lim x 1 ( 2 x + 1 ) = 3 .

Analysis

In this part of the proof, we started with | ( 2 x + 1 ) 3 | | ( 2 x + 1 ) 3 | and used our assumption 0 < | x 1 | < δ 0 < | x 1 | < δ in a key part of the chain of inequalities to get | ( 2 x + 1 ) 3 | | ( 2 x + 1 ) 3 | to be less than ε. We could just as easily have manipulated the assumed inequality 0 < | x 1 | < δ 0 < | x 1 | < δ to arrive at | ( 2 x + 1 ) 3 | < ε | ( 2 x + 1 ) 3 | < ε as follows:

0 < | x 1 | < δ | x 1 | < δ δ < x 1 < δ ε 2 < x 1 < ε 2 ε < 2 x 2 < ε ε < 2 x 2 < ε | 2 x 2 | < ε | ( 2 x + 1 ) 3 | < ε . 0 < | x 1 | < δ | x 1 | < δ δ < x 1 < δ ε 2 < x 1 < ε 2 ε < 2 x 2 < ε ε < 2 x 2 < ε | 2 x 2 | < ε | ( 2 x + 1 ) 3 | < ε .

Therefore, lim x 1 ( 2 x + 1 ) = 3 . lim x 1 ( 2 x + 1 ) = 3 . (Having completed the proof, we state what we have accomplished.)

After removing all the remarks, here is a final version of the proof:

Let ε > 0 . ε > 0 .

Choose δ = ε / 2 . δ = ε / 2 .

Assume 0 < | x 1 | < δ . 0 < | x 1 | < δ .

Thus,

| ( 2 x + 1 ) 3 | = | 2 x 2 | = | 2 ( x 1 ) | = | 2 | | x 1 | = 2 | x 1 | < 2 · δ = 2 · ε 2 = ε . | ( 2 x + 1 ) 3 | = | 2 x 2 | = | 2 ( x 1 ) | = | 2 | | x 1 | = 2 | x 1 | < 2 · δ = 2 · ε 2 = ε .

Therefore, lim x 1 ( 2 x + 1 ) = 3 . lim x 1 ( 2 x + 1 ) = 3 .

The following Problem-Solving Strategy summarizes the type of proof we worked out in Example 2.39.

Problem-Solving Strategy

Problem-Solving Strategy: Proving That lim x a f ( x ) = L lim x a f ( x ) = L for a Specific Function f ( x ) f ( x )

  1. Let's begin the proof with the following statement: Let ε > 0 . ε > 0 .
  2. Next, we need to obtain a value for δ . δ . After we have obtained this value, we make the following statement, filling in the blank with our choice of δ δ : Choose δ = _______. δ = _______.
  3. The next statement in the proof should be (at this point, we fill in our given value for a):
    Assume 0 < | x a | < δ . 0 < | x a | < δ .
  4. Next, based on this assumption, we need to show that | f ( x ) L | < ε , | f ( x ) L | < ε , where f ( x ) f ( x ) and L are our function f ( x ) f ( x ) and our limit L. At some point, we need to use 0 < | x a | < δ . 0 < | x a | < δ .
  5. We conclude our proof with the statement: Therefore, lim x a f ( x ) = L . lim x a f ( x ) = L .

Example 2.40

Proving a Statement about a Limit

Complete the proof that lim x −1 ( 4 x + 1 ) = −3 lim x −1 ( 4 x + 1 ) = −3 by filling in the blanks.

Let _____.

Choose δ = _______. δ = _______.

Assume 0 < | x _______ | < δ . 0 < | x _______ | < δ .

Thus, | ________ ________ | = _____________________________________ ε . | ________ ________ | = _____________________________________ ε .

Checkpoint 2.27

Complete the proof that lim x 2 ( 3 x 2 ) = 4 lim x 2 ( 3 x 2 ) = 4 by filling in the blanks.

Let _______.

Choose δ = _______ . δ = _______ .

Assume 0 < | x ____ | < ____ . 0 < | x ____ | < ____ .

Thus,

| _______ ____ | = ______________________________ ε . | _______ ____ | = ______________________________ ε .

Therefore, lim x 2 ( 3 x 2 ) = 4 . lim x 2 ( 3 x 2 ) = 4 .

In Example 2.39 and Example 2.40, the proofs were fairly straightforward, since the functions with which we were working were linear. In Example 2.41, we see how to modify the proof to accommodate a nonlinear function.

Example 2.41

Proving a Statement about the Limit of a Specific Function (Geometric Approach)

Prove that lim x 2 x 2 = 4. lim x 2 x 2 = 4.

Checkpoint 2.28

Find δ corresponding to ε > 0 ε > 0 for a proof that lim x 9 x = 3 . lim x 9 x = 3 .

The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions. Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the primary tool used in proofs of statements about limits. For Example 2.42, we take on a purely algebraic approach.

Example 2.42

Proving a Statement about the Limit of a Specific Function (Algebraic Approach)

Prove that lim x −1 ( x 2 2 x + 3 ) = 6 . lim x −1 ( x 2 2 x + 3 ) = 6 .

Checkpoint 2.29

Complete the proof that lim x 1 x 2 = 1 . lim x 1 x 2 = 1 .

Let ε > 0 ; ε > 0 ; choose δ = min { 1 , ε / 3 } ; δ = min { 1 , ε / 3 } ; assume 0 < | x 1 | < δ . 0 < | x 1 | < δ .

Since | x 1 | < 1 , | x 1 | < 1 , we may conclude that −1 < x 1 < 1 . −1 < x 1 < 1 . Thus, 1 < x + 1 < 3 . 1 < x + 1 < 3 . Hence, | x + 1 | < 3 . | x + 1 | < 3 .

You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits.

Proving Limit Laws

We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.

Definition

The triangle inequality states that if a and b are any real numbers, then | a + b | | a | + | b | . | a + b | | a | + | b | .

Proof

We prove the following limit law: If lim x a f ( x ) = L lim x a f ( x ) = L and lim x a g ( x ) = M , lim x a g ( x ) = M , then lim x a ( f ( x ) + g ( x ) ) = L + M . lim x a ( f ( x ) + g ( x ) ) = L + M .

Let ε > 0 . ε > 0 .

Choose δ 1 > 0 δ 1 > 0 so that if 0 < | x a | < δ 1 , 0 < | x a | < δ 1 , then | f ( x ) L | < ε / 2 . | f ( x ) L | < ε / 2 .

Choose δ 2 > 0 δ 2 > 0 so that if 0 < | x a | < δ 2 , 0 < | x a | < δ 2 , then | g ( x ) M | < ε / 2 . | g ( x ) M | < ε / 2 .

Choose δ = min { δ 1 , δ 2 } . δ = min { δ 1 , δ 2 } .

Assume 0 < | x a | < δ . 0 < | x a | < δ .

Thus,

0 < | x a | < δ 1 and 0 < | x a | < δ 2 . 0 < | x a | < δ 1 and 0 < | x a | < δ 2 .

Hence,

| ( f ( x ) + g ( x ) ) ( L + M ) | = | ( f ( x ) L ) + ( g ( x ) M ) | | f ( x ) L | + | g ( x ) M | < ε 2 + ε 2 = ε . | ( f ( x ) + g ( x ) ) ( L + M ) | = | ( f ( x ) L ) + ( g ( x ) M ) | | f ( x ) L | + | g ( x ) M | < ε 2 + ε 2 = ε .

We now explore what it means for a limit not to exist. The limit lim x a f ( x ) lim x a f ( x ) does not exist if there is no real number L for which lim x a f ( x ) = L . lim x a f ( x ) = L . Thus, for all real numbers L, lim x a f ( x ) L . lim x a f ( x ) L . To understand what this means, we look at each part of the definition of lim x a f ( x ) = L lim x a f ( x ) = L together with its opposite. A translation of the definition is given in Table 2.10.

Definition Opposite
1. For every ε > 0 , ε > 0 , 1. There exists ε > 0 ε > 0 so that
2. there exists a δ > 0 , δ > 0 , so that 2. for every δ > 0 , δ > 0 ,
3. if 0 < | x a | < δ , 0 < | x a | < δ , then | f ( x ) L | < ε . | f ( x ) L | < ε . 3. There is an x satisfying 0 < | x a | < δ 0 < | x a | < δ so that | f ( x ) L | ε . | f ( x ) L | ε .

Table 2.10 Translation of the Definition of lim x a f ( x ) = L lim x a f ( x ) = L and its Opposite

Finally, we may state what it means for a limit not to exist. The limit lim x a f ( x ) lim x a f ( x ) does not exist if for every real number L, there exists a real number ε > 0 ε > 0 so that for all δ > 0 , δ > 0 , there is an x satisfying 0 < | x a | < δ , 0 < | x a | < δ , so that | f ( x ) L | ε . | f ( x ) L | ε . Let's apply this in Example 2.43 to show that a limit does not exist.

Example 2.43

Showing That a Limit Does Not Exist

Show that lim x 0 | x | x lim x 0 | x | x does not exist. The graph of f ( x ) = | x | / x f ( x ) = | x | / x is shown here:

A graph of a function with two segments. The first exists for x<0, and it is a line with no slope that ends at the y axis in an open circle at (0,-1). The second exists for x>0, and it is a line with no slope that begins at the y axis in an open circle (1,0).

One-Sided and Infinite Limits

Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality 0 < x a < δ 0 < x a < δ replaces 0 < | x a | < δ , 0 < | x a | < δ , which ensures that we only consider values of x that are greater than (to the right of) a. Similarly, in the definition of the limit from the left, the inequality δ < x a < 0 δ < x a < 0 replaces 0 < | x a | < δ , 0 < | x a | < δ , which ensures that we only consider values of x that are less than (to the left of) a.

Definition

Limit from the Right: Let f ( x ) f ( x ) be defined over an open interval of the form ( a , b ) ( a , b ) where a < b . a < b . Then,

lim x a + f ( x ) = L lim x a + f ( x ) = L

if for every ε > 0 , ε > 0 , there exists a δ > 0 δ > 0 such that if 0 < x a < δ , 0 < x a < δ , then | f ( x ) L | < ε . | f ( x ) L | < ε .

Limit from the Left: Let f ( x ) f ( x ) be defined over an open interval of the form ( b , c ) ( b , c ) where b < c . b < c . Then,

lim x a f ( x ) = L lim x a f ( x ) = L

if for every ε > 0 , ε > 0 , there exists a δ > 0 δ > 0 such that if δ < x a < 0 , δ < x a < 0 , then | f ( x ) L | < ε . | f ( x ) L | < ε .

Example 2.44

Proving a Statement about a Limit From the Right

Prove that lim x 4 + x 4 = 0 . lim x 4 + x 4 = 0 .

Checkpoint 2.30

Find δ δ corresponding to ε for a proof that lim x 1 1 x = 0 . lim x 1 1 x = 0 .

We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have lim x a f ( x ) = + , lim x a f ( x ) = + , we want the values of the function f ( x ) f ( x ) to get larger and larger as x approaches a. Instead of the requirement that | f ( x ) L | < ε | f ( x ) L | < ε for arbitrarily small ε when 0 < | x a | < δ 0 < | x a | < δ for small enough δ , δ , we want f ( x ) > M f ( x ) > M for arbitrarily large positive M when 0 < | x a | < δ 0 < | x a | < δ for small enough δ . δ . Figure 2.43 illustrates this idea by showing the value of δ δ for successively larger values of M.

Two graphs side by side. Each graph contains two curves above the x axis separated by an asymptote at x=a. The curves on the left go to infinity as x goes to a and to 0 as x goes to negative infinity. The curves on the right go to infinity as x goes to a and to 0 as x goes to infinity. The first graph has a value M greater than zero marked on the y axis and a horizontal line drawn from there (y=M) to intersect with both curves. Lines are drawn down from the points of intersection to the x axis. Delta is the smaller of the distances between point a and these new spots on the x axis. The same lines are drawn on the second graph, but this M is larger, and the distances from the x axis intersections to point a are smaller.

Figure 2.43 These graphs plot values of δ δ for M to show that lim x a f ( x ) = + . lim x a f ( x ) = + .

Definition

Let f ( x ) f ( x ) be defined for all x a x a in an open interval containing a. Then, we have an infinite limit

lim x a f ( x ) = + lim x a f ( x ) = +

if for every M > 0 , M > 0 , there exists δ > 0 δ > 0 such that if 0 < | x a | < δ , 0 < | x a | < δ , then f ( x ) > M . f ( x ) > M .

Let f ( x ) f ( x ) be defined for all x a x a in an open interval containing a. Then, we have a negative infinite limit

lim x a f ( x ) = lim x a f ( x ) =

if for every M > 0 , M > 0 , there exists δ > 0 δ > 0 such that if 0 < | x a | < δ , 0 < | x a | < δ , then f ( x ) < M . f ( x ) < M .

Section 2.5 Exercises

In the following exercises, write the appropriate ε δ ε δ definition for each of the given statements.

176 .

lim x a f ( x ) = N lim x a f ( x ) = N

177.

lim t b g ( t ) = M lim t b g ( t ) = M

178 .

lim x c h ( x ) = L lim x c h ( x ) = L

179.

lim x a φ ( x ) = A lim x a φ ( x ) = A

The following graph of the function f satisfies lim x 2 f ( x ) = 2 . lim x 2 f ( x ) = 2 . In the following exercises, determine a value of δ > 0 δ > 0 that satisfies each statement.

A function drawn in quadrant one for x > 0. It is an increasing concave up function, with points approximately (0,0), (1, .5), (2,2), and (3,4).

180 .

If 0 < | x 2 | < δ , 0 < | x 2 | < δ , then | f ( x ) 2 | < 1 . | f ( x ) 2 | < 1 .

181.

If 0 < | x 2 | < δ , 0 < | x 2 | < δ , then | f ( x ) 2 | < 0.5 . | f ( x ) 2 | < 0.5 .

The following graph of the function f satisfies lim x 3 f ( x ) = −1 . lim x 3 f ( x ) = −1 . In the following exercises, determine a value of δ > 0 δ > 0 that satisfies each statement.

A graph of a decreasing linear function, with points (0,2), (1,1), (2,0), (3,-1), (4,-2), and so on for x >= 0.

182 .

If 0 < | x 3 | < δ , 0 < | x 3 | < δ , then | f ( x ) + 1 | < 1 . | f ( x ) + 1 | < 1 .

183.

If 0 < | x 3 | < δ , 0 < | x 3 | < δ , then | f ( x ) + 1 | < 2 . | f ( x ) + 1 | < 2 .

The following graph of the function f satisfies lim x 3 f ( x ) = 2 . lim x 3 f ( x ) = 2 . In the following exercises, for each value of ε, find a value of δ > 0 δ > 0 such that the precise definition of limit holds true.

A graph of an increasing linear function intersecting the x axis at about (2.25, 0) and going through the points (3,2) and, approximately, (1,-5) and (4,5).

[T] In the following exercises, use a graphing calculator to find a number δ δ such that the statements hold true.

186 .

| sin ( 2 x ) 1 2 | < 0.1 , | sin ( 2 x ) 1 2 | < 0.1 , whenever | x π 12 | < δ | x π 12 | < δ

187.

| x 4 2 | < 0.1 , whenever | x 8 | < δ | x 4 2 | < 0.1 , whenever | x 8 | < δ

In the following exercises, use the precise definition of limit to prove the given limits.

188 .

lim x 2 ( 5 x + 8 ) = 18 lim x 2 ( 5 x + 8 ) = 18

189.

lim x 3 x 2 9 x 3 = 6 lim x 3 x 2 9 x 3 = 6

190 .

lim x 2 2 x 2 3 x 2 x 2 = 5 lim x 2 2 x 2 3 x 2 x 2 = 5

191.

lim x 0 x 4 = 0 lim x 0 x 4 = 0

192 .

lim x 2 ( x 2 + 2 x ) = 8 lim x 2 ( x 2 + 2 x ) = 8

In the following exercises, use the precise definition of limit to prove the given one-sided limits.

193.

lim x 5 5 x = 0 lim x 5 5 x = 0

194 .

lim x 0 + f ( x ) = −2 , where f ( x ) = { 8 x 3 , if x < 0 4 x 2 , if x 0 . lim x 0 + f ( x ) = −2 , where f ( x ) = { 8 x 3 , if x < 0 4 x 2 , if x 0 .

195.

lim x 1 f ( x ) = 3 , where f ( x ) = { 5 x 2 , if x < 1 7 x 1 , if x 1 . lim x 1 f ( x ) = 3 , where f ( x ) = { 5 x 2 , if x < 1 7 x 1 , if x 1 .

In the following exercises, use the precise definition of limit to prove the given infinite limits.

196 .

lim x 0 1 x 2 = lim x 0 1 x 2 =

197.

lim x −1 3 ( x + 1 ) 2 = lim x −1 3 ( x + 1 ) 2 =

198 .

lim x 2 1 ( x 2 ) 2 = lim x 2 1 ( x 2 ) 2 =

199.

An engineer is using a machine to cut a flat square of Aerogel of area 144 cm2. If there is a maximum error tolerance in the area of 8 cm2, how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to δ , δ , ε, a, and L?

200 .

Use the precise definition of limit to prove that the following limit does not exist: lim x 1 | x 1 | x 1 . lim x 1 | x 1 | x 1 .

201.

Using precise definitions of limits, prove that lim x 0 f ( x ) lim x 0 f ( x ) does not exist, given that f ( x ) f ( x ) is the ceiling function. (Hint: Try any δ < 1 .) δ < 1 .)

202 .

Using precise definitions of limits, prove that lim x 0 f ( x ) lim x 0 f ( x ) does not exist: f ( x ) = { 1 if x is rational 0 if x is irrational . f ( x ) = { 1 if x is rational 0 if x is irrational . (Hint: Think about how you can always choose a rational number 0 < r < d , 0 < r < d , but | f ( r ) 0 | = 1 .) | f ( r ) 0 | = 1 .)

203.

Using precise definitions of limits, determine lim x 0 f ( x ) lim x 0 f ( x ) for f ( x ) = { x if x is rational 0 if x is irrational . f ( x ) = { x if x is rational 0 if x is irrational . (Hint: Break into two cases, x rational and x irrational.)

204 .

Using the function from the previous exercise, use the precise definition of limits to show that lim x a f ( x ) lim x a f ( x ) does not exist for a 0 . a 0 .

For the following exercises, suppose that lim x a f ( x ) = L lim x a f ( x ) = L and lim x a g ( x ) = M lim x a g ( x ) = M both exist. Use the precise definition of limits to prove the following limit laws:

205.

lim x a ( f ( x ) + g ( x ) ) = L + M lim x a ( f ( x ) + g ( x ) ) = L + M

206 .

lim x a [ c f ( x ) ] = c L lim x a [ c f ( x ) ] = c L for any real constant c (Hint: Consider two cases: c = 0 c = 0 and c 0 .) c 0 .)

207.

lim x a [ f ( x ) g ( x ) ] = L M . lim x a [ f ( x ) g ( x ) ] = L M . (Hint: | f ( x ) g ( x ) L M | = | f ( x ) g ( x ) L M | = | f ( x ) g ( x ) f ( x ) M + f ( x ) M L M | | f ( x ) | | g ( x ) M | + | M | | f ( x ) L | .) | f ( x ) g ( x ) f ( x ) M + f ( x ) M L M | | f ( x ) | | g ( x ) M | + | M | | f ( x ) L | .)

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Source: https://openstax.org/books/calculus-volume-1/pages/2-5-the-precise-definition-of-a-limit

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