Calculus Question

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PDF – Section 1.0: Slopes and VelocitiesPDF – Section 1.1: Limit of a FunctionPDF – Section 1.2: Limit PropertiesPDF – Section 1.3: Continuous Functions PDF – Section 1.4: Formal Definition of a Limit I have a quiz about these subjects due to some issues at work right now i could not take it and need some help

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Chapter 1, Section 1.4: Formal Definition of a Limit, remixed by Jeff Eldridge from work by Dale
Hoffman, is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License.
© Mathispower 4u. UMGC has modified this work and it is available under the original license.
98
contemporary calculus
1.4
This intuitive and graphical understanding of limit was sufficient for the first 100plus years of the development of calculus (from Newton and Leibniz in the late
1600s to Cauchy in the early 1800s) and it
is sufficient for using and understanding
the results in beginning calculus.
Definition of Limit
It may seem strange that we have been using and calculating the values
of limits for quite a while without having a precise definition of “limit,”
but the history of mathematics shows that many concepts—including
limits—were successfully used before they were precisely defined or
even fully understood. We have chosen to follow the historical sequence,
emphasizing the intuitive and graphical meaning of limit because most
students find these ideas and calculations easier than the definition.
Mathematics, however, is more than a collection of useful tools, and
part of its power and beauty comes from the fact that in mathematics
terms are precisely defined and results are rigorously proved. Mathematical tastes (what is mathematically beautiful, interesting, useful)
change over time, but because of careful definitions and proofs, the
results remain true—everywhere and forever. Textbooks seldom give
all of the definitions and proofs, but it is important to mathematics that
such definitions and proofs exist.
The goal of this section is to provide a precise definition of the limit
of a function. The definition will not help you calculate the values
of limits, but it provides a precise statement of what a limit is. The
definition of limit is then used to verify the limits of some functions
and prove some general results.
The Intuitive Approach
The precise (“formal”) definition of limit carefully states the ideas that
we have already been using graphically and intuitively. The following
side-by-side columns show some of the phrases we have been using to
describe limits, and those phrases—particularly the last ones—provide
the basis on which to build the definition of limit.
A Particular Limit
lim 2x − 1 = 5
General Limit
lim f ( x ) = L
“as the values of x approach 3, the values of 2x − 1
approach (are arbitrarily close to) 5”
“as the values of x approach a, the values of f ( x )
approach (are arbitrarily close to) L”
“when x is close to 3 (but not equal to 3), the value
of 2x − 1 is close to 5”
“when x is close to a (but not equal to a), the value
of f ( x ) is close to L”
“we can guarantee that the values of 2x − 1 are as
close to 5 as we want by restricting the values of x
to be sufficiently close to 3 (but not equal to 3)”
“we can guarantee that the values of f ( x ) are as close
to L as we want by restricting the values of x to be
sufficiently close to a (but not equal to a)”
x→a
x →3
Let’s examine what the last phrase (“we can. . . ”) means for the Particular Limit in the previous discussion.
limits and continuity
Example 1. We know lim 2x − 1 = 5 and need to show that we can
x →3
guarantee that the values of f ( x ) = 2x − 1 are as close to 5 as we want
by restricting the values of x to be sufficiently close to 3.
What values of x guarantee that f ( x ) = 2x − 1 is within:
(a) 1 unit of 5?
(c) E units of 5?
(b) 0.2 units of 5?
Solution. (a) “Within 1 unit of 5” means between 5 − 1 = 4 and 5 + 1 =
6, so the question can be rephrased as “for what values of x is
y = 2x − 1 between 4 and 6: 4 < 2x − 1 < 6?” We want to know which values of x ensure the values of y = 2x − 1 are in the the shaded band in the uppermost margin figure. The algebraic process is straightforward: 4 < 2x − 1 < 6 ⇒ 5 < 2x < 7 ⇒ 2.5 < x < 3.5 We can restate this result as follows: “If x is within 0.5 units of 3, then y = 2x − 1 is within 1 unit of 5.” (See second margin figure) Any smaller distance also satisfies the guarantee: for example, “If x is within 0.4 units of 3, then y = 2x − 1 is within 1 unit of 5.” (See third margin figure) (b) “Within 0.2 units of 5” means between 5 − 0.2 = 4.8 and 5 + 0.2 = 5.2, so the question can be rephrased as “for which values of x is y = 2x − 1 between 4.8 and 5.2: 4.8 < 2x − 1 < 5.2?” Solving for x, we get 5.8 < 2x < 6.2 and 2.9 < x < 3.1. “If x is within 0.1 units of 3, then y = 2x − 1 is within 0.2 units of 5.” (See fourth margin figure.) Any smaller distance also satisfies the guarantee. Rather than redoing these calculations for every possible distance from 5, we can do the work once, generally: (c) “Within E unit of 5” means between 5 − E and 5 + E, so the question becomes, “For what values of x is y = 2x − 1 between 5 − E and 5 + E: 5 − E < 2x − 1 < 5 + E?” Solving 5 − E < 2x − 1 < 5 + E for x, we get: 6 − E < 2x < 6 + E ⇒ 3− E E < x < 3+ 2 2 “If x is within E2 units of 3, then y = 2x − 1 is within E units of 5.” (See last figure.) Any smaller distance also works. J Part (c) of Example 1 illustrates the power of general solutions in mathematics. Rather than redoing similar calculations every time someone demands that f ( x ) = 2x − 1 be within some given distance of 5, we did the calculations once. And then we can quickly respond for any given distance. For the question “What values of x guarantee that f ( x ) = 2x − 1 is within 0.4, 0.1 or 0.006 units of 5?” we can answer, “If 0.1 0.006 x is within 0.2 (= 0.4 2 ), 0.05 (= 2 ) or 0.003 (= 2 ) units of 3.” 99 100 contemporary calculus Practice 1. Knowing that lim 4x − 5 = 3, determine which values of x x →2 guarantee that f ( x ) = 4x − 5 is within (a) 1 unit of 3. (b) 0.08 units of 3. (c) E units of 3. The same ideas work even if the graphs of the functions are not straight lines, but the calculations become more complicated. Example 2. Knowing that lim x2 = 4, determine which values of x x →2 guarantee that f ( x ) = x2 is within: (a) 1 unit of 4. (b) 0.2 units of 4. State each answer in the form: “If x is within f ( x ) is within units of 4.” units of 2, then Solution. (a) If x2 is within 1 unit of 4 (and x is near 2, hence positive) √ √ then 3 < x2 < 5 so 3 < x < 5 or 1.732 < x < 2.236. The interval √ containing these x values extends from 2 − 3 ≈ 0.268 units to √ the left of 2 to 5 − 2 ≈ 0.236 units to the right of 2. Because we want to specify a single distance on each side of 2, we can pick the smaller of the two distances, 0.236, and say: “If x is within 0.236 units of 2, then f ( x ) is within 1 unit of 4.” (b) Similarly, if x2 is within 0.2 units of 4 (and x is near 2, so x > 0)
√
√
then 3.8 < x2 < 4.2 so 3.8 < x < 4.2 or 1.949 < x < 2.049. The √ interval containing these x values extends from 2 − 3.8 ≈ 0.051 √ units to the left of 2 to 4.2 − 2 ≈ 0.049 units to the right of 2. Again picking the smaller of the two distances, we can say: “If x is within 0.049 units of 2, then f ( x ) is within 1 unit of 4.” J The situation in Example 2—with different distances on the left and right sides—is very common, and we always pick our single distance to be the smaller of the distances to the left and right. By using the smaller distance, we can be certain that if x is within that smaller distance on either side, then the value of f ( x ) is within the specified distance of the value of the limit. √ Practice 2. Knowing that lim x = 3, determine which values of x x → 9 √ guarantee that f ( x ) = x is within: (a) 1 unit of 3. (b) 0.2 units of 3. State each answer in the form: “If x is within f ( x ) is within units of 3.” units of 9, then The same ideas can also be used when the function and the specified distance are given graphically, and in that case we can give the answer graphically. limits and continuity Example 3. In the margin figure, lim f ( x ) = 3. Which values of x x →2 guarantee that y = f ( x ) is within E units (given graphically) of 3? State your answer in the form: “If x is within (show a distance D graphically) of 2, then f ( x ) is within E units of 3.” Solution. The solution process requires several steps: (i) Use the given distance E to find the values 3 − E and 3 + E on the y-axis. (See margin.) (ii) Sketch the horizontal band with lower edge at y = 3 − E and upper edge at y = 3 + E. (iii) Find the first locations to the right and left of x = 2 where the graph of y = f ( x ) crosses the lines y = 3 − E and y = 3 + E, and at these locations draw vertical line segments extending to the x-axis. (iv) On the x-axis, graphically determine the distance from 2 to the vertical line on the left (labeled DL ) and from 2 to the vertical line on the right (labeled DR ). (v) Let the length D be the smaller of the lengths DL and DR . If x is within D units of 2, then f ( x ) is within E units of 3. J Practice 3. In the last margin figure, lim f ( x ) = 1.8. Which values of x x →3 guarantee that y = f ( x ) is within E units (given graphically) of 1.8? The Formal Definition of Limit The ideas from the previous examples and practice problems, restated for general functions and limits, provide the basis for the definition of limit given below. The use of the lowercase Greek letters e (epsilon) and δ (delta) in the definition is standard, and this definition is sometimes called the “epsilon-delta” definition of a limit. Definition of lim f ( x ) = L: x→a For every given number e > 0 there is a number δ > 0 so that
if
then
x
f (x)
is within δ units of a (and x 6= a)
is within e units of L
Equivalently: | f ( x ) − L| < e whenever 0 < | x − a| < δ In this definition, e represents the given distance on either side of the limiting value y = L, and δ is the distance on each side of the point x = a on the x-axis that we have been finding in the previous 101 102 contemporary calculus examples. This definition has the form of a “challenge and response”: for any positive challenge e (make f (c) within e of L), there is a positive response δ (start with x within δ of a and x 6= a). Example 4. As seen in the second margin figure, lim f ( x ) = L, with x→a a value for e given graphically as a length. Find a length for δ that satisfies the definition of limit (so “if x is within δ of a, and x 6= a, then f ( x ) is within e of L”). Solution. Follow the steps outlined in Example 3. The length for δ is shown in the third margin figure, and any shorter length for δ also satisfies the definition. J Practice 4. In the bottom margin figure, lim f ( x ) = L, with a value for x→a e given graphically. Find a length for δ that satisfies the definition of limit. Example 5. Prove that lim 4x − 5 = 7. x →3 Solution. We need to show that “for every given e > 0 there is a δ > 0
so that if x is within δ units of 3 (and x 6= 3) then 4x − 5 is within e
units of 7.”
Actually, there are two things we need to do. First, we need to find a
value for δ (typically depending on e) and, second, we need to show
that our δ really does satisfy the “if. . . then. . . ” part of the definition.
Finding δ is similar to part (c) in Example 1 and Practice 1: Assume
4x − 5 is within e units of 7 and solve for x. If 7 − e < 4x − 5 < 7 + e then 12 − e < 4x < 12 + e ⇒ 3 − 4e < x < 3 + 4e so x is within 4e units of 3. Put δ = 4e . To show that δ = 4e satisfies the definition, we merely reverse the order of the steps in the previous paragraph. Assume that x is within δ units of 3. Then 3 − δ < x < 3 + δ, so: 3− e e < x < 3+ 4 4 12 − e < 4x < 12 + e 7 − e < 4x − 5 < 7 + e so we can conclude that f ( x ) = 4x − 5 is within e units of 7. This formally verifies that lim 4x − 5 = 7. J x →3 Practice 5. Prove that lim 5x + 3 = 23. x →4 The method used to prove the values of the limits for these particular linear functions can also be used to prove the following general result about the limits of linear functions. Theorem: lim mx + b = ma + b x→a limits and continuity 103 Proof. Let f ( x ) = mx + b. Case 1: m = 0. Then f ( x ) = 0x + b = b is simply a constant function, and any value for δ > 0 satisfies the definition. Given any value of
e > 0, let δ = 1 (any positive value for δ works). If x is is within 1 unit
of a, then f ( x ) − f ( a) = b − b = 0 < e, so we have shown that for any e > 0 there is a δ > 0 that satisfies the limit definition.
Case 2: m 6= 0. For any e > 0, put δ = |me | > 0. If x is within δ = |me |
of a then
a−
e
e
e
e
e
< x < a+ ⇒ < x−a < ⇒ | x − a| < |m| |m| |m| |m| |m| Then the distance between f ( x ) and L = ma + b is: | f ( x ) − L| = |(mx + b) − (ma + b)| = |mx − ma| e = |m| · | x − a| < |m| =e |m| so f ( x ) is within e of L = ma + b. In each case, we have shown that “given any e > 0, there is a δ > 0”
that satisfies the rest of the limit definition.
If there is even a single value of e for which there is no δ, then we
say that the limit “does not exist.”
Example 6. With f ( x ) defined as:
(
f (x) =
2 if x < 1 4 if x > 1
use the limit definition to prove that lim f ( x ) does not exist.
x →1
Solution. One common proof technique in mathematics is called “proof
by contradiction” and that is the method we use here:
• We assume that the limit does exist and equals some number L.
• We show that this assumption leads to a contradiction
• We conclude that the assumption must have been false.
We therefore conclude that the limit does not exist.
First, assume that the limit exists: lim f ( x ) = L for some value for L.
x →1
Let e = 12 . Then, because we are assuming that the limit exists, there is
a δ > 0 so that if x is within δ of 1 then f ( x ) is within e of L.
Next, let x1 be between 1 and 1 + δ. Then x1 > 1 so f ( x1 ) = 4. Also,
x1 is within δ of 1 so f ( x1 ) = 4 is within 12 of L, which means that L is
between 3.5 and 4.5: 3.5 < L < 4.5. Let x2 be between 1 and 1 − δ. Then x2 < 1, so f ( x2 ) = 2. Also, x2 is within δ of 1 so f ( x2 ) = 2 is within 12 of L, which means that L is between 1.5 and 2.5: 1.5 < L < 2.5. The definition says “for every e” so we can certainly pick 12 as our e value; why we chose this particular value for e shows up later in the proof. 104 contemporary calculus These inequalities provide the contradiction we hoped to find. There is no value L that satisfies both 3.5 < L < 4.5 and 1.5 < L < 2.5, so our assumption must be false: f ( x ) does not have a limit as x → 1. J 1 does not exist. x →0 x Practice 6. Use the limit definition to prove that lim Proofs of Two Limit Theorems We conclude with proofs of two parts of the Main Limit Theorem so you can see how such proofs proceed—you have already used these theorems to evaluate limits. There are rigorous proofs of all of the other limit properties in the Main Limit Theorem, but they are somewhat more complicated than the proofs given here. Theorem: If then lim f ( x ) = L x→a lim k · f ( x ) = kL x→a Proof. Case k = 0: The theorem is true but not very interesting: lim k · f ( x ) = lim 0 · f ( x ) = lim 0 = 0 = 0 · L = kL x→a x→a x→a Case k 6= 0: Because lim f ( x ) = L, then, by the definition, for every x→a e > 0 there is a δ > 0 so that | f ( x ) − L| < e whenever | x − a| < δ. For any e > 0, we know |ek| > 0, so pick a value of δ that satisfies
| f ( x ) − L| < |ek| whenever | x − a| < δ. When | x − a| < δ (“x is within δ of a”) then | f ( x ) − L| < |ek| (“ f ( x ) is within |ek| of L”) so |k| · | f ( x ) − L| < e ⇒ |k · f ( x ) − k · L| < e (that is, k · f ( x ) is within e of k · L). Theorem: If then lim f ( x ) = L x→a and lim g( x ) = M x→a lim [ f ( x ) + g( x )] = L + M. x→a Proof. Given any e > 0, we know 2e > 0, so there is a number δ f > 0
such that when | x − a| < δ f then | f ( x ) − L| < 2e (“if x is within δ f of a, then f ( x ) is within 2e of L”). Likewise, there is a number δg > 0 such that when | x − a| < δg then g ( | x ) − L| < 2e (“if x is within δg of a, then g( x ) is within 2e of L”). Let δ be the smaller of δ f and δg . If | x − a| < δ then | f ( x ) − L| < 2e and | g( x ) − M | < 2e so: |( f ( x ) + g( x )) − ( L + M))| = |( f ( x ) − L) + ( g( x ) − M)| e e + =e 2 2 so f ( x ) + g( x ) is within e of L + M whenever x is within δ of a. ≤ | f ( x ) − L| + | g( x ) − M| < limits and continuity 105 1.4 Problems In problems 1–4, state each answer in the form “If x is within units of. . . ” 9. Knowing that lim x3 = 8, what values of x guarx →2 antee that f ( x ) = x3 is within: 1. Knowing that lim 2x + 1 = 7, what values of x x →3 (a) 0.5 units of 8? guarantee that f ( x ) = 2x + 1 is within: (a) 1 unit of 7? (c) 0.04 units of 7? (b) 0.6 units of 7? (d) e units of 7? x →16 (a) 1 unit of 4? x →1 (b) 0.1 units of 4? guarantee that f ( x ) = 3x + 2 is within: (b) 0.6 units of 5? (d) e units of 5? √ 1 + x = 2, what values of x √ guarantee that f ( x ) = 1 + x is within: 11. Knowing that lim (c) 0.09 units of 5? √ x = 4, what values of x √ guarantee that f ( x ) = x is within: 10. Knowing that lim 2. Knowing that lim 3x + 2 = 5, what values of x (a) 1 unit of 5? (b) 0.05 units of 8? x →3 (a) 1 unit of 2? 3. Knowing that lim 4x − 3 = 5, what values of x (b) 0.0002 units of 2? x →2 guarantee that f ( x ) = 4x − 3 is within: (a) 1 unit of 5? (c) 0.08 units of 5? (b) 0.4 units of 5? (d) e units of 5? 4. Knowing that lim 5x − 3 = 2, what values of x x →1 guarantee that f ( x ) = 5x − 3 is within: (a) 1 unit of 2? (c) 0.01 units of 2? (b) 0.5 units of 2? (d) e units of 2? 5. For problems 1–4, list the slope of each function f and the δ (as a function of e). For these linear functions f , how is δ related to the slope? 12. You need to cut four pieces of wire (exactly the same length after the cut) and form them into a square. If the area of the square must be within 0.06 inches of 100 inches, then each piece of wire must be within how many inches of 10? 13. You need to cut four pieces of wire (exactly the same length after the cut) and form them into a square. If the area of the square must be within 0.06 inches of 25 inches, then each piece of wire must be within how many inches of 5? 6. You have been asked to cut two boards (exactly the same length after the cut) and place them end to end. If the combined length must be within 0.06 inches of 30 inches, then each board must be within how many inches of 15? 7. You have been asked to cut three boards (exactly the same length after the cut) and place them end to end. If the combined length must be within 0.06 inches of 30 inches, then each board must be within how many inches of 10? 2 8. Knowing that lim x = 9, what values of x guarx →3 antee that f ( x ) = x2 is within: (a) 1 unit of 9? (b) 0.2 units of 9? In problems 14–17, lim f ( x ) = L and the function x→a f and a value for e are given graphically. Find a length for δ that satisfies the limit definition for the given function and value of e. 14. 106 contemporary calculus 18. Redo each of problems 14–17 taking a new value of e to be half the value of e given in the problem. In problems 19–22, use the limit definition to prove that the given limit does not exist. (Find a value for e > 0 for which there is no δ that
satisfies the definition.)
15.
19. With f ( x ) defined as:
(
f (x) =
4
3
if x < 2 if x > 2
show that lim f ( x ) does not exist.
x →2
16.
20. Show that lim b x c does not exist.
x →3
21. With f ( x ) defined as:
(
f (x) =
x
6−x
if x < 2 if x > 2
show that lim f ( x ) does not exist.
x →2
22. With f ( x ) defined as:
(
f (x) =
17.
x+1
x2
if x < 2 if x > 2
show that lim f ( x ) does not exist.
x →2
23. Prove: If lim f ( x ) = L and lim g( x ) = M then
x→a
x→a
lim [ f ( x ) − g( x )] = L − M.
x→a
1.4 Practice Answers
1. (a) 3 − 1 < 4x − 5 < 3 + 1 ⇒ 7 < 4x < 9 ⇒ 1.75 < x < 2.25: “x within 14 unit of 2.” (b) 3 − 0.08 < 4x − 5 < 3 + 0.08 ⇒ 7.92 < 4x < 8.08 ⇒ 1.98 < x < 2.02: “x within 0.02 units of 2.” (c) 3 − E < 4x − 5 < 3 + E ⇒ 8 − E < 4x < 8 + E ⇒ 2 − E4 < x < 2 + E4 : “x within E4 units of 2.” limits and continuity 107 √ 2. “Within 1 unit of 3”: If 2 < x < 4, then 4 < x < 16, which extends from 5 units to the left of 9 to 7 units to right of 9. Using the smaller √ of these two distances from 9: “If x is within 5 units of 9, then x is within 1 unit of 3.” √ “Within 0.2 units of 3”: If 2.8 < x < 3.2, then 7.84 < x < 10.24, which extends from 1.16 units to the left of 9 to 1.24 units to the right of 9. “If x is within 1.16 units of 9, then x is within 0.2 units of 3. 3. 4. 5. Given any e > 0, take δ = 5e . If x is within δ = 5e of 4, then
4 − 5e < x < 4 + 5e so: − e e < x − 4 < ⇒ −e < 5x − 20 < e ⇒ −e < (5x + 3) − 23 < e 5 5 so, finally, f ( x ) = 5x + 3 is within e of L = 23. We have shown that “for any e > 0, there is a δ > 0 (namely δ = 5e )”
so that the rest of the definition is satisfied.
6. Using “proof by contradiction” as in the solution to Example 6:
1
(i) Assume that the limit exists: lim = L for some value of L. Let
x →0 x
e = 1. Since we’re assuming that the limit exists, there is a δ > 0
so that if x is within δ of 0 then f ( x ) = 1x is within e = 1 of L.
(ii) Let x1 be between 0 and 0 + δ and also require that x1 < 12 . Then 0 < x1 < 12 so f ( x1 ) = x1 > 2. Because x1 is within δ of 0,
1
f ( x1 ) > 2 is within e = 1 of L, so L > 2 − e = 1: that is, 1 < L. Let x2 be between 0 and 0 − δ and also require x2 > − 12 . Then
0 > x2 > 21 so f ( x2 ) = x12 < −2. Since x2 is within δ of 0, f ( x2 ) < −2 is within e = 1 of L, so L < −2 + e = −1 ⇒ −1 > L.
(iii) The two inequalities derived above provide the contradiction we
were hoping to find. There is no value L that satisfies both 1 < L and L < −1, so we can conclude that our assumption was false and that f ( x ) = 1x does not have a limit as x → 0. This is a much more sophisticated (= harder) problem. The definition says “for every e” so we can pick e = 1. For this particular limit, the definition fails for every e > 0.
Chapter 1, Section 1.3: Continuous Functions, remixed by Jeff Eldridge from work by Dale Hoffman, is
licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. © Mathispower 4u.
UMGC has modified this work and it is available under the original license.
84
contemporary calculus
1.3
Continuous Functions
In section 1.2 we saw a few “nice” functions whose limits as x → a
simply involved substituting a into the function: lim f ( x ) = f ( a).
x→a
Functions whose limits have this substitution property are called continuous functions and such functions possess a number of other useful
properties.
In this section we will examine what it means graphically for a function to be continuous (or not continuous), state some properties of continuous functions, and look at a few applications of these properties—
2x + 1
including a way to solve horrible equations such as sin( x ) =
.
x−2
Definition of a Continuous Function
We begin by formally stating the definition of this new concept.
Definition of Continuity at a Point:
A function f is continuous at x = a if and only if
lim f ( x ) = f ( a).
x→a
The graph in the margin illustrates some of the different ways a
function can behave at and near a point, and the accompanying table
contains some numerical information about the example function f and
its behavior. We can conclude from the information in the table that f
is continuous at 1 because lim f ( x ) = 2 = f (1).
x →1
We can also conclude that f is not continuous at 2 or 3 or 4, because
lim f ( x ) 6= f (2), lim f ( x ) 6= f (3) and lim f ( x ) 6= f (4).
x →2
a
f ( a)
1
2
3
4
2
1
2
undefined
lim f ( x )
x→a
2
2
DNE
2
x →3
x →4
Graphical Meaning of Continuity
When x is close to 1, the values of f ( x ) are close to the value f (1), and
the graph of f does not have a hole or break at x = 1. The graph of f is
“connected” at x = 1 and can be drawn without lifting your pencil. At
x = 2 and x = 4 the graph of f has “holes,” and at x = 3 the graph has
a “break.” The function f is also continuous at 1.7 (why?) and at every
point shown except at 2, 3 and 4.
Informally, we can say:
• A function is continuous at a point if the graph of the function
is connected there.
• A function is not continuous at a point if its graph has a hole
or break at that point.
limits and continuity
Sometimes the definition of “continuous” (the substitution condition
for limits) is easier to use if we chop it into several smaller pieces and
then check whether or not our function satisfies each piece.
f is continuous at a if and only if:
(i) f is defined at a
(ii) the limit of f ( x ), as x → a, exists
(so the left limit and right limits exist and are equal)
(iii) the value of f at a equals the value of the limit as x → a:
lim f ( x ) = f ( a)
x→a
If f satisfies conditions (i), (ii) and (iii), then f is continuous at a. If
f does not satisfy one or more of the three conditions at a, then f is not
continuous at a.
For f ( x ) in the figure on the previous page, all three conditions are
satisfied for a = 1, so f is continuous at 1. For a = 2, conditions (i)
and (ii) are satisfied but not (iii), so f is not continuous at 2. For a = 3,
condition (i) is satisfied but (ii) is violated, so f is not continuous at 3.
For a = 4, condition (i) is violated, so f is not continuous at 4.
A function is continuous on an interval if it is continuous at every
point in the interval.
A function f is continuous from the left at a if lim f ( x ) = f ( a)
x → a−
and is continuous from the right at a if lim f ( x ) = f ( a).
x → a+
Example 1. Is the function


 x+1
f (x) =
2


1
x −3
if x ≤ 1
if 1 < x ≤ 2 if x > 2
continuous at x = 1? At x = 2? At x = 3?
Solution. We could answer these questions by examining the graph
of f ( x ), but let’s try them without the benefit of a graph. At x = 1,
f (1) = 2 and the left and right limits are equal:
lim f ( x ) = lim ( x + 1) = 2 = lim 2 = lim f ( x )
x →1−
x →1−
x →1+
x →1+
and their common limit matches the value of the function at x = 1:
lim f ( x ) = 2 = f (1)
x →1
so f is continuous at 1.
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contemporary calculus
At x = 2, f (2) = 2, but the left and right limits are not equal:
1
= lim f ( x )
− 3 x →2+
so f fails condition (ii) and is therefore not continuous at 2. We can,
however, say that f is continuous from the left at 2 (but not from the
right at 2).
1
At x = 3, f (3) = , which is undefined, so f is not continuous at 3
0
because it fails condition (i).
J
lim f ( x ) = lim 2 = 2 6= −1 = lim
x →2−
x →2+ x
x →1−
Example 2. Where is f ( x ) = 3×2 − 2x continuous?
Solution. By the Substitution Theorem for Polynomial and Rational
Functions, lim P( x ) = P( a) for any polynomial P( x ) at any point a,
x→a
so every polynomial is continuous everywhere. In particular, f ( x ) =
3×2 − 2x everywhere.
J
Example 3. Where are the functions g( x ) =
continuous?
x+5
x2 + 4x − 5
and h( x ) = 2
x−3
x − 4x + 3
Solution. g( x ) is a rational function, so by the Substitution Theorem
for Polynomial and Rational Functions it is continuous everywhere
except where its denominator is 0: g is continuous everywhere except
3. The graph of g (see margin) is “connected” everywhere except at 3,
where it has a vertical asymptote.
We can write the rational function h( x ) as:
h( x ) =
( x − 1)( x + 5)
( x − 1)( x − 3)
and note that its denominator is 0 at x = 1 and x = 3, so h is continuous
everywhere except 3 and 1. The graph of h (see margin) is “connected”
everywhere except at 3, where it has a vertical asymptote, and 1, where
J
it has a hole: f (1) = 00 is undefined.
Example 4. Where is f ( x ) = b x c continuous?
Solution. The graph of y = b x c seems to be “connected” except at
each integer, where there is a “jump” (see margin).
If a is an integer, then lim b x c = a − 1 and lim b x c = a so lim b x c
x → a−
x → a+
x→a
is undefined, and b x c is not continuous at x = a.
If a is not an integer, then the left and right limits of b x c, as x → a,
both equal b ac so: lim b x c = a = b ac, so b x c is continuous at x = a.
x→a
Summarizing: b x c is continuous everywhere except at the integers.
In fact, f ( x ) = b x c is continuous from the right everywhere and is
continuous from the left everywhere except at the integers.
J
Practice 1. Where is f ( x ) =
|x|
continuous?
x
limits and continuity
Why Do We Care Whether a Function Is Continuous?
There are several reasons for us to examine continuous functions and
their properties:
• Many applications in engineering, the sciences and business are
continuous or are modeled by continuous functions or by pieces of
continuous functions.
• Continuous functions share a number of useful properties that do
not necessarily hold true if the function is not continuous. If a
result is true of all continuous functions and we have a continuous
function, then the result is true for our function. This can save us
from having to show, one by one, that each result is true for each
particular function we use. Some of these properties are given in the
remainder of this section.
• Differential calculus has been called the study of continuous change,
and many of the results of calculus are guaranteed to be true only for
continuous functions. If you look ahead into Chapters 2 and 3, you
will see that many of the theorems have the form “If f is continuous
and (some additional hypothesis), then (some conclusion).”
Combinations of Continuous Functions
Not only are most of the basic functions we will encounter continuous
at most points, so are basic combinations of those functions.
Theorem:
If
f ( x ) and g( x ) are continuous at a
and k is any constant
then
the elementary combinations of f and g
• k · f (x)
• f ( x ) + g( x )
• f ( x ) − g( x )
• f ( x ) · g( x )
•
f (x)
g( x )
(as long as g( a) 6= 0)
are continuous at a.
The continuity of a function is defined using limits, and all of these
results about simple combinations of continuous functions follow from
the results about combinations of limits in the Main Limit Theorem.
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contemporary calculus
Our hypothesis is that f and g are both continuous at a, so we can
assume that
lim f ( x ) = f ( a)
x→a
and
lim g( x ) = g( a)
x→a
and then use the appropriate part of the Main Limit Theorem.
For example,
lim [ f ( x ) + g( x )] = lim f ( x ) + lim g( x ) = f ( a) + g( a)
x→a
x→a
x→a
so f + g is continuous at a.
Practice 2. Prove: If f and g are continuous at a, then k · f and f − g
are continuous at a (where k a constant).
Composition of Continuous Functions:
g( x ) is continuous at a and
f ( x ) is continuous at g( a)
If
then
lim f ( g( x )) = f ( lim g( x )) = f ( g( a))
x→a
x→a
so f ◦ g( x ) = f ( g( x )) is continuous at a.
The proof of this result involves some technical details, but just
formalizes the following line of reasoning:
The hypothesis that “g is continuous at a” means that if x is close
to a then g( x ) will be close to g( a). Similarly, “ f is continuous at g( a)”
means that if g( x ) is close to g( a) then f ( g( x )) = f ◦ g( x ) will be close
to f ( g( a)) = f ◦ g( a). Finally, we can conclude that if x is close to a,
then g( x ) is close to g( a) so f ◦ g( x ) is close to f ◦ g( a) and therefore
f ◦ g is continuous at x = a.
The next theorem presents an alternate version of the limit condition
for continuity, which we will use occasionally in the future.
Theorem:
lim f ( x ) = f ( a) if and only if lim f ( a + h) = f ( a)
x→a
h →0
Proof. Let’s define a new variable h by h = x − a so that x = a + h
(see margin figure). Then x → a if and only if h = x − a → 0, so
lim f ( x ) = lim f ( a + h) and therefore lim f ( x ) = f ( a) if and only if
x→a
h →0
x→a
lim f ( a + h) = f ( a).
h →0
We can restate the result of this theorem as:
A function f is continuous at a if and only if lim f ( a + h) = f ( a).
h →0
limits and continuity
89
Which Functions Are Continuous?
Fortunately, the functions we encounter most often are either continuous everywhere or continuous everywhere except at a few places.
Theorem: The following functions are continuous everywhere
(b) sin( x ) and cos( x )
(a) polynomials
(c) | x |
Proof. (a) This follows from the Substitution Theorem for Polynomial
and Rational Functions and the definition of continuity.
(b) The graph of y = sin( x ) (see margin) clearly has no holes or breaks,
so it is reasonable to think that sin( x ) is continuous everywhere.
Justifying this algebraically, for every real number a:
Recall the angle addition formula for
sin(θ ) and the results from section 1.2
that lim cos(h) = 1 and lim sin(h) = 0.
lim sin( a + h) = lim sin( a) cos(h) + cos( a) sin(h)
h →0
h →0
= lim sin( a) · lim cos(h) + lim cos( a) · lim sin(h)
h →0
h →0
h →0
h →0
= sin( a) · 1 + cos( a) · 0 = sin( a)
so f ( x ) = sin( x ) is continuous at every point. The justification for
f ( x ) = cos( x ) is similar.
(c) For f ( x ) = | x |, when x > 0, then | x | = x and its graph (see margin)
is a straight line and is continuous because x is a polynomial.
When x < 0, then | x | = − x and it is also continuous. The only questionable point is the “c