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Chapter 2: Functions, Limits and Continuity

• Functions

• Limits

• Continuity

Chapter 2: Functions, Limits and Continuity 1

Functions

Functions are the major tools for describing the real world inmathematical terms.

Definition 1. A function f is a set of ordered pairs of numbers(x, y) satisfying the property that if (x, y1), (x, y2) ∈ f theny1 = y2 (that is, no two distinct ordered pairs in f have thesame first component).

If f is a function from a set X into a set Y , then we write

f : X → Y read as “f is a function from set X into set Y .

If (x, y) ∈ f , then we may write

y = f(x) (read as “y equals f of x”)


so that y is called the image of x under the function f ; x is a pre-imageof y under f . We also say that “y is the function value of x under f .”

In the function f : X → Y , the set X containing all of the firstcomponents of ordered pairs in f is called the domain of the function f ;the set Y is called the co-domain of f . The set of all second componentsof ordered pairs in f is called that range of f , that is,

range f = {y ∈ Y : y = f(x) for some x ∈ X}.

Clearly,range f ⊆ Y.


Example 1. Let f = {(x, y) : y =√x− 2}.

• The value of y that corresponds to x = 6 is y =√6− 2 =

√4 = 2.

Hence, f(6) = 2 and (6, 2) ∈ f .

• The domain of f is the set

dom f = [2,+∞)

and the range of f is the set

range f = R≥0 = [0,∞).


Definition 2. If f is a function, then the graph of f is the setof all points (x, y) in a given plane for which (x, y) ∈ f .

Example 2. The graph of f(x) = x− 2 is given by


Example 3. The graph of g(x) = x2 − 4 is given by


Example 4. The graph of h(x) =√x2 − 9 is given by


Example 5. The graph of h(x) =√x(x− 2) is given by


Example 6. The graph of F (x) =

3x− 1, if x < 24, if x = 27− x, if x > 2.

is given by


Example 7. The graph of G(x) =

{4− x2, if x ≤ 210, otherwise.

is given by

Vertical Line Test

A vertical line intersects the graph of a function f in at mostone point.


Limits

Example 8. How does the function f(x) =2x2 + x− 3

x− 1behave near x = 1?

x f(x) =2x2 + x− 3

x− 10 0.30.25 3.50.5 40.75 4.80.9 4.980.99 4.9980.999 4.99980.9999 4.999980.99999 4.999998

x less than 1

x f(x) =2x2 + x− 3

x− 12 71.75 6.51.5 6.01.25 5.51.1 5.21.01 5.021.001 5.0021.0001 5.00021.00001 5.00002

x greater than 1


Definition 3. (Informal Definition of Limit) Let f(x) bedefined on an open interval containing a, except possibly at aitself. If f(x) gets arbitrarily close to L for all x sufficientlyclose to a, we say that f approaches the limit L as x approachesa, and we write

limx→a

f(x) = L.

Example 9. Let f be defined by

f(x) =

x2 − 1

x− 1, if x 6= 1

1, if x = 1


Example 10. Find the following limits (if they exist).

1. limx→4

2

2. limx→−2

x

3. limx→0

(5x− 3)


Remark 1. A function may fail to have a limit at a point in its domain.

Example 11. Discuss the behavior of the following functions as x approachesa = 0.

1. U(x) =

{0, x < 01, x ≥ 0


2. g(x) =

{1/x, x 6= 00, x = 0


3. f(x) =

{0, x ≤ 0sin(1/x), x > 0


Limits from Graphs

For the given function f(x) with its graph shown, find the followinglimits.

Example 12. 1. limx→1

f(x)

2. limx→2

f(x)

3. limx→3

f(x)


Example 13. 1. limx→−2

f(x)

2. limx→−1

f(x)

3. limx→0

f(x)


Example 14. 1. limx→2

f(x)

2. limx→1

f(x)

3. limx→0

f(x)


Definition 4. (Formal Definition of Limit) Let f be a functiondefined at every number in some open interval containing a,except possibly at the number a itself. The limit of f(x) as xapproaches a is L, written as

limx→a

f(x) = L

if the following statement is true:Given any ε > 0, however small, there exists a δ > 0 such that

if 0 < |x− a| < δ then |f(x)− L| < ε.


Example 15. Let f(x) = 2x− 5.

1. Find a δ > 0 such that whenever 0 < |x−3| < δ whenever |f(x)−1| < εwhere ε = 0.1.

2. Show that limx→3

f(x) = 1.


Example 16. Let f(x) = x2.

1. Find a δ > 0 such that whenever 0 < |x−2| < δ whenever |f(x)−4| < εwhere ε = 0.3.

2. Show that limx→2

f(x) = 4.


The Limit TheoremsTheorem 1. (Limit of a Linear Function)If m and b are constants, then

limx→a

mx+ b = ma+ b.

Theorem 2. (Limit of a Constant)If c is a constant, then for any number a

limx→a

c = c.

Theorem 3. (Limit of the Identity Function)

limx→a

x = a.


Theorem 4. (Limit of the Sum and Difference of Two Functions)If limx→a

f(x) = L and limx→a

g(x) =M , then

limx→a

[f(x)± g(x)] = L±M.

Theorem 5. (Limit of the Sum and Difference of n Functions)If limx→a

f1(x) = L1, limx→a

f2(x) = L2, . . ., limx→a

fn(x) = Ln, then

limx→a

[f1(x)± f2(x)± · · · ± fn(x)] = L1 ± L2 ± · · · ± Ln.

Theorem 6. (Limit of the Product of Two Functions)If limx→a


g(x) =M , then

limx→a

[f(x)g(x)] = LM.


Theorem 7. (Limit of the Product of n Functions)If limx→a

f1(x) = L1, limx→a

f2(x) = L2, . . ., limx→a

fn(x) = Ln, then

limx→a

[f1(x) · f2(x) · · · fn(x)] = L1 · L2 · · ·Ln.

Theorem 8. (Limit of the nth Power of a Function)If limx→a

f(x) = L and n is any positive integer, then

limx→a

[f(x)]n = Ln.

Theorem 9. (Limit of the Quotient of Two Functions)If limx→a


g(x) =M , then

limx→a

[f(x)/g(x)] = L/M if M 6= 0.


Theorem 10. (Limit of the nth Root of a Function)If limx→a

f(x) = L and n is any positive integer, then

limx→a

n√f(x) =

n√L

with the restriction that if n is even, L > 0.Theorem 11.

For any real number a except 0

limx→a

1

x=

1

a

with the restriction that if n is even, L > 0.


Theorem 12.

For a > 0 and n a positive integer, or if a ≤ 0 and n is an odd positiveinteger, then

limx→a

n√x = n

√a.

Theorem 13.

limx→a

f(x) = L if and only if limx→a

[f(x)− L] = 0.

Theorem 14.

limx→a

f(x) = L if and only if limt→0

f(t+ a)] = L.

Theorem 15.

limx→a

f(x) = L1 and limx→a

f(x) = L2 implies L1 = L2.


Exercises:Find the indicated limit .

1. limx→−4

(5x+ 2)

2. limx→3

(2x2 − 4x+ 5)

3. limy→−1

(y3 − 2y2 + 3y − 4)

4. limx→2

3x+ 4

8x− 1

5. limx→−1

2x+ 1

x2 − 3x+ 4

6. limx→2

√x2 + 3x+ 4

x3 + 1

7. limx→−3

3

√5 + 2x

5− x

8. limz→−5

z2 − 25

z + 5

9. limx→1/3

3x− 1

9x2 − 1

10. limx→−1

√x+ 5− 2

x+ 1


One-Sided Limits

Definition 5. (Definition of Right-Hand Limit) Let f be afunction defined at every number in some open interval (a, c).The limit of f(x) as x approaches a from the right is L, writtenas

limx→a+

f(x) = L

if for any ε > 0, however small, there exists a δ > 0 such thatif 0 < x− a < δ then |f(x)− L| < ε.


-1 0 5 0 0 -1 0 8 1 -.25 1 .25 4 -.25 4 .25x0 x0 + δ

x•1.8 4 1.8

δ

for all x 6= x0 in here

.5 -3 2.5 -1.5 1 -1.2 1 -1.3 4 -1.3 4 -1.2 /

L− ε

L

L+ ε

f(x) lies in here

.6 3 .8 3 .8 7 .6 7 / .6 5 1 5 -.1 3 .1 3 -.1 5 .1 5 -.1 7 .1 7

f(x)•


Definition 6. (Definition of Left-Hand Limit) Let f be afunction defined at every number in some open interval (d, a).The limit of f(x) as x approaches a from the left is L, writtenas

limx→a+

f(x) = L

if for any ε > 0, however small, there exists a δ > 0 such thatif 0 < a− x < δ then |f(x)− L| < ε.


-1 0 5 0 0 -1 0 8 1 -.25 1 .25 4 -.25 4 .25

x0 − δ x0

x•1.8 4 1.8

δ

for all x 6= x0 in here

.5 -3 2.5 -1.5 1 -1.2 1 -1.3 4 -1.3 4 -1.2 /

L− ε

L

L+ ε

f(x) lies in here

.6 3 .8 3 .8 7 .6 7 / .6 5 1 5 -.1 3 .1 3 -.1 5 .1 5 -.1 7 .1 7

f(x)•


Example 17. Let sgn(x) =

−1, if x < 00, if x = 01, if x > 0.

.

Find limx→0+

sgn(x), limx→0−

sgn(x) and limx→0

sgn(x).


Theorem 16. limx→a

f(x) exists and is equal to L if and only if limx→a+

f(x)

and limx→a−

f(x) both exist and both are equal to L.

Example 18. Let f(x) be the function defined by f(x) =|x|x

. Then

limx→0+|x|x

= 1 and limx→0−|x|x

= −1. Using Theorem (16), we see that

limx→0|x|x

does not exist.


Example 19. Let f be defined by

f(x) =

x+ 5, if x < −3√9− x2, if − 3 ≤ x ≤ 3

3− x, if 3 < x.


Exercises:Sketch the graph of the function and find the indicated limit if it exists; ifit does not exist, state the reason.

1. g(t) =

3 + t2, if t < −20, if t = −211− t2, if t > −2

(a) limt→−2+

g(t)

(b) limt→−2−

g(t)

(c) limt→−2

g(t)


2. f(x) =

2x+ 3, if x < 14, if x = 1x2 + 2, if x > 1

(a) limx→1+

f(x)

(b) limx→1−

f(x)

(c) limx→1

f(x)

3. f(x) =

x+ 1, if x < −1x2, if − 1 ≤ x ≤ 12− x, if x > 1

(a) limx→−1+

f(x)

(b) limx→−1−

f(x)

(c) limx→−1

f(x)

(d) limx→1+

f(x)

(e) limx→1−

f(x)

(f) limx→1

f(x)


More Exercise

Find each of the following for the given function and specified a value.

limx→a−

f(x), limx→a+

f(x), and limx→a

f(x)

1. f(x) =

2, if x < 1

−1, if x = 1

−3, if 1 < x.

;

a = 1

2. f(x) =

{x+ 4, if x ≤ −44− x, otherwise.

;

a = −4

3. f(x) =

{x2, if x ≤ 2

8− 2x, otherwise.;

a = 2

4. f(x) =

2x+ 3, if x < 1

2, if x = 1

7− 2x, if 1 < x.

;

a = 1

5. f(x) =

x2 − 4, if x < 2

4, if x = 2

4− x2, if 2 < x.

;

a = 2

6. f(x) = |x− 5|;a = 0

7. f(x) =

2, if x < −2√4− x2, if − 2 ≤ x ≤ 2

−2, if 2 < x.

;

a = −2 and a = 2

8. f(x) =

√x2 − 9, if x ≤ −3√9− x2, if − 3 < x < 3√x2 − 9, if 3 ≤ x.

;

a = −3 and a = 3

9. f(x) =

3√x+ 1, if x ≤ −1√1− x2, if − 1 < x < 1

3√x− 1, if 1 ≤ x.

;

a = −1 and a = 1


Infinite Limits

One-sided infinite limits

Example 20. Find limx→1+

1

x− 1and lim

x→1−

1

x− 1.


Two-sided infinite limits

Example 21. Find limx→0

1

x2.


Definition 7. Infinite Limits

1. Let f be a function defined at every number in some open interval

containing the number a except possibly at the number a itself.

As x approaches a, f(x) increases without bound, which is written

limx→a f(x) = +∞, if for any number N > 0 there exists a δ > 0

such that

if 0 < |x− a| < δ then f(x) > N.

2. Let f be a function defined at every number in some open interval

containing the number a except possibly at the number a itself. As

x approaches a, f(x) decreases without bound, which is written

limx→a f(x) = −∞, if for any number N < 0 there exists a δ > 0

such that

if 0 < |x− a| < δ then f(x) < N.


Example 22. f(x) =2x

x− 1


Theorem 17. If r is any positive integer, then

(i) limx→0+

1

xr= +∞

(ii) limx→0−

1

xr=

{−∞, if r is odd+∞, if r is even.

Example 23. Find

(a) limx→0+

1

x3

(b) limx→0−

1

x5

(c) limx→0−

1

x6


Theorem 18. If a is any real number and if limx→a f(x) = 0 and limx→a g(x) = c,

where c is a constant not equal to 0, then

(i) if c > 0 and if f(x)→ 0 through positive values of f(x), then

limx→a

g(x)

f(x)= +∞

(ii) if c > 0 and if f(x)→ 0 through negative values of f(x), then

limx→a

g(x)

f(x)= −∞

(iii) if c < 0 and if f(x)→ 0 through positive values of f(x), then

limx→a

g(x)

f(x)= −∞

(iv) if c < 0 and if f(x)→ 0 through negative values of f(x), then

limx→a

g(x)

f(x)= +∞


The theorem is also valid if “x→ a” is replaced by “x→ a+” or “x→ a−.”

Example 24. Find limx→1−

2x

x− 1and lim

x→1+

2x

x− 1

Example 25. Find limx→3+

x2 + x+ 2

x2 − 2x− 3and lim

x→3−

x2 + x+ 2

x2 − 2x− 3

Example 26. Let f(x) =

√x2 − 4

x− 2and g(x) =

√4− x2

x− 2.

Find (a) limx→2+

f(x), (b) limx→2−

g(x)


Example 27. Let F (x) =x2 + x+ 2

x2 − 2x− 3. Find

1. limx→3+ F (x)

2. limx→3− F (x)

3. limx→−1+ F (x)

4. limx→−1− F (x)


Theorem 19.

(i) If limx→a

f(x) = +∞ and limx→a

g(x) = c where c is any constant, then

limx→a

[f(x) + g(x)] = +∞.

(ii) If limx→a

f(x) = −∞ and limx→a

g(x) = c where c is any constant, then

limx→a

[f(x) + g(x)] = −∞.

The theorem holds if “x→ a” is replaced by “x→ a+” or “x→ a−.”


Theorem 20. If limx→a

f(x) = +∞ and limx→a

g(x) = c where c is any

constant except 0, then

(i) if c > 0, limx→a f(x) · g(x) = +∞;

(ii) if c < 0, limx→a f(x) · g(x) = −∞.



Theorem 21. If limx→a

f(x) = −∞ and limx→a

g(x) = c where c is any

constant except 0, then

(i) if c > 0, limx→a f(x) · g(x) = −∞;

(ii) if c < 0, limx→a f(x) · g(x) = +∞.



Definition 8. The line x = a is a vertical asymptote ofthe graph of the function f if at least one of the followingstatements is true.

(i) limx→a+

f(x) = +∞

(ii) limx→a+

f(x) = −∞

(iii) limx→a−

f(x) = +∞

(iv) limx→a−

f(x) = −∞


Example 28. Find the vertical asymptotes of the graph of the function andsketch the graph.

(a) f(x) =2

x− 4

(b) f(x) =5

x2 + 8x+ 15

(c) f(x) =2

(x− 4)2


Limits at Infinity

Consider the function f(x) =2x2

x2 + 1. Observe that as x increases

(decreases) without bound, the value of the function approaches 2.

Thus, we write

limx→+∞

f(x) = 2 and limx→−∞

f(x) = 2.


Definition 9. (Definition of the Limit of f(x) as x Increaseswithout Bound) Let f be a function defined at every numberin some interval (a,+∞). The limit of f(x) as x increaseswithout bound is L, written as

limx→+∞

f(x) = L

if for any ε > 0, however small, there exists a number N > 0such that

if x > N then |f(x)− L| < ε.


Definition 10. (Definition of the Limit of f(x) as xDecreases without Bound) Let f be a function definedat every number in some interval (−∞, a). The limit of f(x)as x decreases without bound is L, written as

limx→−∞

f(x) = L

if for any ε > 0, however small, there exists a number N < 0such that

if x < N then |f(x)− L| < ε.


Theorem 22. If r is any positive integer, then

(i) limx→+∞1

xr= 0 (ii) limx→−∞

1

xr= 0

Example 29. Find limx→+∞4x− 3

2x+ 5.

Example 30. Find limx→+∞2x2 − x+ 5

4x3 − 1.

Example 31. Find limx→+∞3x+ 4√2x2 − 5

and limx→−∞3x+ 4√2x2 − 5

.

Example 32. Find limx→+∞x2

x+ 1.

Example 33. Find limx→+∞2x− x2

3x+ 5.


Horizontal Asymptotes

Definition 11. (Definition of a Horizontal Asymptote) Theline y = b is a horizontal asymptote of the graph of the functionf if at least one of the following statements is true:

(i) limx→+∞ f(x) = b and for some number N , if x > N ,then f(x) 6= b;

(ii) limx→−∞ f(x) = b and for some number N , if x < N ,then f(x) 6= b.

Example 34. Find the horizontal asymptotes of the graph of the function

f(x)) =x

√x2 + 1

.



Oblique Asymptotes

Definition 12. (Definition of a function Continuous at aNumber) The graph of the function f has the line y = mx+ bas an asymptote if either of the following statements is true:

(i) limx→∞[f(x) − (mx + b)] = 0 and for some numberM > 0, f(x) 6= mx+ b whenever x > M ;

(ii) limx→−∞[f(x) − (mx + b)] = 0 and for some numberM < 0, f(x) 6= mx+ b whenever x < M .

Part (i) of the definition indicates that for any ε > 0, there exists anumber N > 0 such that

if x > N then 0 < |f(x)− (mx+ b)| < ε,

that is, we can make the function value f(x) as close to the value f(x) as


close to the value of mx+ b as we please by taking x sufficiently large. Asimilar statement may be made for part (ii) of the definition.

The graph of a rational functionP (x)

Q(x), where the degree of the

polynomial P (x) is one more than the degree of Q(x) and Q(x) is not a

factor of P (x), has an oblique asymptote. To show this, we let f(x) =P (x)

Q(x)and divide P (x) by Q(x) to express f(x) as the sum of a linear functionand a rational function; that is,

f(x) = mx+ b+R(x)

Q(x)

where the degree of the polynomial is less than the degree of Q(x). Then

f(x)− (mx+ b) =R(x)

Q(x).


When the numerator and denominator of R(x)/Q(x) is divided by thehighest power of x appearing in Q(x), there will be a constant term in thedenominator and all other terms in the denominator and every term in thenumerator will be of the form k/xr where k is a constant and r is a positiveinteger. Therefore, as x→ +∞, the limit of the numerator will be zero and

the limit of the denominator will be a constant. Thus, limx→∞R(x)

Q(x)= 0.

Hence,limx→∞

[f(x)− (mx+ b) = 0.

From this we conclude that the line y = mx+ b is an oblique asymptote ofthe graph of f .


Example 35. Find the asymptotes of the graph of h(x) =x2 + 3

x− 1.


Continuity

Definition 13. (Definition of a function Continuous at aNumber) The function f is said to be continuous at thenumber a if and only if the following three conditions aresatisfied:

(i) f(a) exists;

(ii) limx→a f(x) exists;

(iii) limx→a f(x) = f(a).

If one or more of these three conditions fails to hold at a, thefunction f is said to be discontinuous at a.


Continuity of the function f(x) = x+ 1 at a = 0


Example 36. Consider the function f(x) = x+ 1 where x 6= 0. Then f isdiscontinuous at a = 0 since f(0) does not exist.


Example 37. Consider the function f(x) =

{x+ 1, if x 6= 02, otherwise.

where

x 6= 0. Then f is discontinuous at a = 0 since f(0) 6= limx→a f(x).

Removable Discontinuity


Remark: The discontinuity described in each of the previous examples iscalled a removable discontinuity for the reason that the function can beredefined so that f(0) = 1.

In general, if f is a function discontinuous at the number a butfor which limx→a f(x) exists. Then either f(a) does not exist or elselimx→a f(x) 6= f(a). Such discontinuity is a removable discontinuitybecause f is redefined at a so that f(a) is equal to limx→a f(x), the newfunction becomes continuous at a. If the discontinuity is not removable, itis called an essential discontinuity.


Example 38. The function f(x) =1

x2is discontinuous at a = 0.

Infinite Discontinuity

The discontinuity of this function at the number a = 0 is essential sincelimx→a f(x) does not exist. This kind of discontinuity is called an infinitediscontinuity.


Example 39. The function f(x) =

{1, if x ≥ 0−1, if x < 0

is discontinuous at

a = 0.

Jump Discontinuity

The discontinuity illustrated in this example is essential sincelimx→a+ f(x) 6= limx→a− f(x) and hence, limx→a f(x) does not exist.This discontinuity is called a jump discontinuity.


Theorem 23. If f and g are two functions continuous at the number a,then

(i) f + g is continuous at a;

(ii) f − g is continuous at a;

(iii) f · g is continuous at a;

(iv) f/g is continuous at a provided that g(a) 6= 0.

Theorem 24. A polynomial function is continuous at every number.

Theorem 25. A rational function is continuous at every number in itsdomain.


Theorem 26. If n is a positive integer and

f(x) = n√x

then

1. if n is odd, f is continuous at every number.

2. if n is even, f is continuous at every positive number.

Theorem 27. (Alternative Definition of Continuity)If the function f is continuous at the number a if f is defined on someopen interval containing a and if for any ε > 0 there exists a δ > 0 suchthat

if |x− a| < δ then |f(x)− f(a)| < ε.


Theorem 28. (Limit of a Composite Function)If limx→a

g(x) = b and if the function f is continuous at b,

limx→a

(f ◦ g)(x) = f(b)

or equivalently,limx→a

f(g(x)) = f( limx→a

g(x))

.

Theorem 29. (Continuity of a Composite Function)If the function g is continuous at a and the function f is continuous atg(a), then the composite function f ◦ g is continuous at a.


Definition 14. (Definition of Continuity on an OpenInterval) A function is said to be continuous on an openinterval if and only if it is continuous at every number in theopen interval.

Definition 15. (Definition of Right-Hand Continuity) Afunction is said to be continuous from the right at the numbera if and only if the following three conditions are satisfied:

1. f(a) exists;

2. limx→a+

f(x) exists;

3. limx→a+

f(x) = f(a).


Definition 16. (Definition of Left-Hand Continuity) Afunction is said to be continuous from the left at the number aif and only if the following three conditions are satisfied:

1. f(a) exists;

2. limx→a−

f(x) exists;

3. limx→a−

f(x) = f(a).



Definition 17. (Definition of Continuity on a ClosedInterval) A function whose domain includes the closed interval[a, b] is said to be continuous on [a, b] if and only if it iscontinuous on the open interval (a, b), as well as continuousfrom the right at a and continuous from the left at b.

In general, a function f is right-continuous (continuous from theright) at a point x = c in its domain if limx→c+ f(x) = f(c). It is left-continuous (continuous from the left) at a point x = c in its domain iflimx→c− f(x) = f(c). Thus, a function is continuous at a left endpoint a ofits domain if it is right-continuous at a and continuous at a right endpointb of its domain if it is left-continuous at b. A function is continuous at aninterior point c of its domain if and only if it is both right-continuous andleft-continuous at c.


Example 40. The function f(x) =√4− x2 is continuous at every point of

its domain [−2, 2]. This includes x = −2, where f is right-continuous andx = 2, where f is left-continuous.

Example 41. The unit function U(x) =

{1, if x ≥ 00, if x < 0

, is right-

continuous at x = 0, but is neither left-continuous nor continuous there.


Continuity TestA function f(x) is continuous at x = a if and only if it meetsthe following three conditions:

1. f(a) exists (c lies in the domain of f)

2. limx→a f(x) exists (f has a limit as x→ a)

3. limx→a f(x) = f(a) (the limit equals the function value)

For one-sided continuity and continuity at an endpoint, the limits in (2)and (3) of the test should be replaced by the appropriate one-sided limits.


Example 42. Consider the function y = f(x) in the given figure, whosedomain is the closed interval [0, 4]. Discuss the continuity of f at x =0, 1, 2, 3, 4.


y =√4− x2

Continuous on [−2, 2]


y =1

xContinuous on (−∞, 0) and (0,+∞)


y = U(x)

Continuous on (−∞, 0) and [0,+∞)


y = cos x

Continuous on (−∞,+∞)


Definition 18. (Definition of Continuity on a Half-OpenInterval)

(i) A function whose domain includes the interval half-openinterval to the right [a, b) is continuous on [a, b) if itis continuous on the open interval (a, b) and continuousfrom the right at a.

(ii) A function whose domain includes the interval half-openinterval to the left (a, b] is continuous on (a, b] if it iscontinuous on the open interval (a, b) and continuousfrom the left at b.


Example 43. Determine the largest interval (or union of intervals) on whichthe following function is continuous:

f(x) =

√25− x2

x− 3


The Intermediate Value Theorem

Theorem 30. (The Intermediate Value Theorem (IVT).)If the function f is continuous on the closed interval [a, b], and if f(a) 6=f(b), then for any number k between f(a) and f(b) there exists a numberc between a and b such that f(c) = k.


In terms of geometry, the IVT states that the graph of a continuousfunction on a closed interval must intersect every horizontal line y = kbetween the lines y = f(a) and y = f(b) at least once.

The IVT also assures us that if the function is continuous on the closedinterval [a, b], then f(x) assumes every value between f(a) and f(b) as xassumes values between a and b.


The following is a direct consequence of the IVT.

Theorem 31. (The Intermediate-Zero Theorem.)If the function f is continuous on the closed interval [a, b], and if f(a)and f(b) have opposite signs, then there exists a number c between a andb such that f(c) = 0; that is, c is a zero of f .


Continuity of the Trigonometric Functions

and the Squeeze Theorem

Theorem 32. (The Squeeze Theorem.)Suppose that the functions f , g, and h are defined on some open intervalI containing a except possibly at a itself, and that f(x) ≤ g(x) ≤ h(x)for all x ∈ I for which x 6= a. Also suppose that lim

x→af(x) and lim

x→ah(x)

exist and are equal to L. Then limx→a

g(x) an is equal to L.


Example 44. Let the functions f , g and h be defined by

f(x) = −4(x−2)2+3, g(x) =(x− 2)(x2 − 4x+ 7)

(x− 2), h(x) = 4(x−2)2+3.


Example 45. Given |g(x) − 2| ≤ 3(x − 1)2 for all x. Use the SqueezeTheorem to find lim

x→2g(x).

Example 46. Use the Squeeze Theorem to prove that limx→0

∣∣∣∣∣x sin 1

x

∣∣∣∣∣ = 0.


Consider the function f(x) =sinx

x. This function is not defined at

x = 0 but limx→0

f(x) exists and is equal to 1.


Theorem 33.

limt→0

sin t

t= 1

Proof. First assume that 0 < t <π/2. The figure shows the unit circlex2 + y2 = 1 and the shaded sectorBOP , where B is the point (1, 0)and P is the point (cos t, sin t). Thearea of a circular sector of radius rand central angle of radian measure tis determined by 1

2r2t; so if S square

units is the area of sector BOP , thenS = 1

2t. Consider now the triangleBOP with area K1 square units.Hence, K1 = 1

2|AP | · |OB| =12 sin t.


If K2 is the area of righttriangle BOT , where T is the point(1, tan t), then K2 =

12|BT | · |OB| =

12 tan t. Observe that K1 < S < K2

from which we obtain the inequality12 sin t <

12t <

12 tan t. Multiplying

each member of this inequality by2/ sin t, which is positive because0 < t < π/2, we obtain

1 <t

sin t<

1

cos t.

By taking the reciprocal of eachmember of this inequality, obtain

1 >sin t

t> cos t.


From cos t <sin t

t< 1 we obtain the inequality sin t < t. By replacing

t by 12t, we obtain

sin

(1

2t

)<

1

2t.

Squaring both sides of this inequality will yield

sin2(1

2t

)<

1

4t2 or

1− cos t

2<

1

4t2.

Thus, we obtain

1− 1

2t2 < cos t.

Using this inequality and

cos t <sin t

t


implies

1− 1

2t2 <

sin t

t< 1 if 0 < t < π/2.

Now if −π/2 < t < 0, then 0 < −t < −π/2 so that from the aboveinequality, we obtain

1− 1

2(−t)2 <

sin(−t)(−t)

< 1 if − π/2 < t < 0.

Hence, since sin(−t) = − sin t, we have

1− 1

2t2 <

sin t

t< 1 if − π/2 < t < 0.

Therefore,

1− 1

2t2 <

sin t

t< 1 if − π/2 < t < π/2 and t 6= 0.


Since

limt→

(1− 1

2t2) = 1 and lim

t→1 = 1

and by applying the Squeeze Theorem we obtain the desired result

limt→0

sin t

t= 1.


sin 3x

sin 5x.


Theorem 34. The sine function is continuous at 0.

Theorem 35. The cosine function is continuous at 0.

Theorem 36.

limt→0

1− cos t

t= 0


1− cosx

sinx.


2 tanx

x2.

Theorem 37. The sine and cosine functions are continuous at every realnumber.

Theorem 38. The tangent, cotangent, secant, and cosecant functions arecontinuous on their domains.


Chapter 2: Functions, Limits and Continuityfrancisjosephcampena.weebly.com/uploads/1/7/8/6/17869691/funcli… · xgreater than 1 Chapter 2: Functions, Limits and Continuity 11. De

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