Benginning Calculus Lecture notes 2 - limits and continuity

Beginning Calculus- Limits and Continuity -

Shahrizal Shamsuddin Norashiqin Mohd Idrus

Department of Mathematics,FSMT - UPSI

(LECTURE SLIDES SERIES)

VillaRINO DoMath, FSMT-UPSI

(D1) Limits and Continuity 1 / 54

The Limit of a Function Limits of Trigonometric Functions Infinite Limits and Limits at Infinity Continuity

Learning Outcomes

Determine the existence of limits of functions

Compute the limits of functions

Determine the continuity of functions.

Connect the idea of limits and continuity of functions.




Limits

Definition 1

The limit of f (x), as x approaches a, equals L, denoted by

limx→a

f (x) = L or f (x)→ L as x → a (1)

if the values of f (x) moves arbitrarily close to L as x moves suffi cientlyclose to a (on either side of a ) but not equal to a.




Example

limx→2

(x2 − x + 2

)= 4

0 2 40

5

10

x

y x < 2 f (x) x > 2 f (x)1.0 2.000000 3.0 8.0000001.5 2.750000 2.5 5.7500001.9 3.710000 2.1 4.3100001.99 3.970100 2.01 4.0301001.995 3.985025 2.005 4.0150251.999 3.997001 2.001 4.003001




Example

Estimate the value of limt→0

√t2 + 9− 3t2

.

f

t0.10.0010.00010.00001−0.00001−0.0001−0.001−0.1

=

1t2

(√t2 + 9− 3

)0.166 620.166 670.166 670.166 670.166 670.166 670.166 670.166 62




Example - continue

4 2 0 2 4

0.12

0.13

0.14

0.15

0.16

limt→0

√t2 + 9− 3t2

=16




Example

f (x) = x + 1.

1 1 2 3 4 51

1

2

3

4

5

x

y

limx→2

f (x) = 3




Example

g (x) ={x + 1 if x ≤ 2(x − 2)2 + 3 if x > 2

1 1 2 3 4 51

1

2

3

4

5

x

y

limx→2

g (x) = 3




Example

h (x) ={x + 1 if x < 2(x − 2)2 + 3 if x > 2

1 1 2 3 4 51

1

2

3

4

5

x

y

limx→2

h (x) = 3, eventhough h is not defined at x = 2.




One-Sided Limits

Left-hand limit of flimx→a−

f (x) = L (2)

Right-hand limit of flimx→a+

f (x) = L (3)

limx→a

f (x) = L⇔ f limx→a−

f (x) = limx→a+

f (x) = L. (4)




Example

f (x) ={x + 1 if x ≤ 2(x − 2)2 + 1 if x > 2

1 1 2 3 4 51

1

2

3

4

5

x

y

limx→2−

f (x) = 3 and limx→2+

f (x) = 1

limx→2

f (x) does not exist (DNE), eventhough f is defined at x = 2.




Example

1 1 2 3 4 51

1

2

3

4

5

x

y

Find:

f (2) and f (4)limx→2−

f (x) , limx→2+

f (x) , limx→2

f (x)

limx→4−

f (x) , limx→4+

f (x) limx→4

f (x)




Properties of Limits

Suppose that limx→a

f (x) and limx→a

g (x) exists. Then,

1. limx→a

(cf (x)) = c limx→a

f (x) , for any constant c

2. limx→a

[f (x)± g (x)] = limx→a

f (x)± limx→a

g (x)

3. limx→a

[f (x) g (x)] =[limx→a

f (x)] [limx→a

g (x)]

4. limx→a

[f (x)g (x)

]=limx→a

f (x)

limx→a

g (x)provided that lim

x→ag (x) 6= 0




Properties of Limits - continue

5. limx→a

x = a

6. limx→a

c = c , for any constant c .

7. limx→a

[f (x)]n =[limx→a

f (x)]nwhere n ∈ Z+.

8. limx→a

n√x = n√a where n ∈ Z+ (If n is even, we assume that a > 0 ).

9. limx→a

n√f (x) = n

√limx→a

f (x) where n ∈ Z+. (If n is even, we assume

that limx→a

f (x) > 0 ).




Direct Substitution Property

If f is a polynomial or a rational function and a is in the domain of f ,then

limx→a

f (x) = f (a) (5)




Example

limx→5

(2x2 − 3x + 4

)= lim

x→5

(2x2)− limx→5

3x + limx→5

4

= 2 limx→5

x2 − 3 limx→5

x + limx→5

4

= 2(52)− 3 (5) + 4

= 39




Example

limx→−2

x3 + 2x2 − 15− 3x =

limx→−2

(x3 + 2x2 − 1

)limx→−2

(5− 3x)

=limx→−2

x3 + 2 limx→−2

x2 − limx→−2

1

limx→−2

5− 3 limx→−2

x

=(−2)3 + 2 (−2)2 − 1

5− 3 (−2) = − 111




Definition 2

If f (x) = g (x) when x 6= a, then limx→a

f (x) = limx→a

g (x) , provided that

the limits exist.




Example

limx→1

x2 − 1x − 1 . For x 6= 1,

x2 − 1x − 1 =

(x − 1) (x + 1)x − 1 = x + 1

limx→1

x2 − 1x − 1 = lim

x→1x + 1 = 2

1 1 2 3 4 51

1

2

3

4

5

x

y




Example

limh→0

(3+ h)2 − 9h

. For h 6= 0,

(3+ h)2 − 9h

=9+ 6h+ h2 − 9

h= 6+ h

limh→0

(3+ h)2 − 9h

= limh→0

6+ h = 6




Example

limx→2|x − 2|x − 2 .

For x − 2 > 0, |x − 2| = x − 2.

limx→2|x − 2|x − 2 = lim

x→2x − 2x − 2 = lim

x→21 = 1

For x − 2 < 0, |x − 2| = − (x − 2) = 2− x .

limx→2|x − 2|x − 2 = lim

x→2− (x − 2)x − 2 = lim

x→2−1 = −1

limx→2|x − 2|x − 2 DNE




Remark 1

limθ→0

cos θ − 1θ

= 0

Rewrite:1− cos θ

θto make the numerator stays positive.

θ1

O

A

BC

BC = 1− cos θ, arclength AB = θ.

1− cos θ

θ→ 0 as θ → 0




Remark 2

limθ→0

sin θ

θ= 1

θ1

O

A

BC

AC = sin θ, arclength AB = θ

sin θ

θ→ 1 as θ → 0.

Principle: Short pieces of curves are nearly straight.




Example

limθ→0

tan θ

θ

tan θ

θ=

sin θ

cos θθ

=sin θ

θ cos θ=sin θ

θ· 1cos θ

limθ→0

tan θ

θ= lim

θ→0sin θ

θ· lim

θ→01cos θ

= 1 · 1 = 1




Example

limθ→0

sin 2θ

tan θ

sin 2θ

tan θ=

sin 2θ

θtan θ

θ

=

2 sin 2θ

2θtan θ

θ

limθ→0

sin 2θ

tan θ= lim

θ→0

2 sin 2θ

2θtan θ

θ

=limθ→0

2 sin 2θ

2θ

limθ→0

tan θ

θ

=21= 2




Infinite Limits

Definition 3

Let f defined on both sides of a, except possibly at a itself. Then

limx→a

f (x) = ∞ or limx→a

f (x) = −∞ (6)

means that the values of f (x) can be made arbitrarily large (as large aspossible) by taking x suffi ciently close to a, but not equal to a. x = a isthe vertical asymptote.

y

x

y = f(x)

x = aa

y

x

y = f(x)

x = a

a




Example

limx→3+

2xx − 3 = +∞ and lim

x→3−2xx − 3 = −∞

5 5 10

5

5

10

x

y

x = 3

The vertical asymptote is at x = 3.




Example

f (x) = tan x =sin xcos x

The vertical asymptote can be obtained by setting cos x = 0, that is,

x =π

2x = (2n+ 1)

π

2, n ∈ Z

x

y




Limits at Infinity

Definition 4 (Limits at Infinity)

(a) Let f be a function defined on some interval (a,∞) . Then

limx→∞

f (x) = L (7)

means that the values of f (x) can be made arbitrarily close to L bytaking x suffi ciently large.

(b) Let f be a function defined on some interval (−∞, a) . Then

limx→−∞

f (x) = L (8)

means that the values of f (x) can be made arbitrarily close to L bytaking x suffi ciently large negative.




Horizontal Asymptotes

The line y = L is called a horizontal asymptote of the curve y = f (x) ifeither

limx→∞

f (x) = L or limx→−∞

f (x) = L (9)




Example

f (x) =x2 − 1x2 + 1

limx→∞

f (x) = 1 = limx→−∞

f (x)

10 5 5 10

1

1

2

x

y

No vertical asymtote.The horizontal asymptote is y = 1.




Example

f (x) =1x.

limx→0−

1x= −∞, lim

x→0+1x= +∞

limx→∞

1x= 0 = lim

x→−∞

1x

Vertical asymtote at x = 0The horizontal asymptote at y = 0.

4 2 2 4

4

2

2

4

x

y




Example - Finding the Asymptotes

f (x) =3x2 − x − 25x2 + 4x + 1

limx→∞

3x2 − x − 25x2 + 4x + 1

= limx→∞

3x2

x2− xx2− 2x2

5x2

x2+4xx2+1x2

= limx→∞

3− 1x− 2x2

5+4x+1x2

=

limx→∞

(3− 1

x− 2x2

)limx→∞

(5+

4x+1x2

)

=limx→∞

3− limx→∞

1x− limx→∞

2x2

limx→∞

5+ limx→∞

4x+ limx→∞

1x2

=3− 0− 05+ 0+ 0

=35

The horizontal asymptote is y =35.





f (x) =

√2x2 + 13x − 5 .

limx→∞

√2x2 + 13x − 5 = lim

x→∞

√2x2 + 1√x2

3x − 5x

,√x2 = x for x > 0

= limx→∞

√2x2

x2+1x2

3xx− 5x

= limx→∞

√2+

1x2

3− 5x

=limx→∞

√2+

1x2

limx→∞

(3− 5

x

) =√limx→∞

2+ limx→∞

1x2

limx→∞

3− limx→∞

5x

=

√23




Example - Finding the Asymptotes - continue

limx→−∞

√2x2 + 13x − 5 = lim

x→−∞

−√2+

1x2(

3− 5x

) ,√x2 = −x for x < 0

=− limx→∞

√2+

1x2

limx→−∞

(3− 5

x

) = −√23





4 2 2 4

4

2

2

4

x

y

The horizontal asymptotes are: y = ±√23.

The vertical asymptote is when 3x − 5 = 0, that is, x = 53.





f (x) =√x2 + 1− x

limx→∞

(√x2 + 1− x

)= lim

x→∞

(√x2 + 1− x

)·

(√x2 + 1+ x

)(√

x2 + 1+ x)

= limx→∞

(x2 + 1

)− x2√

x2 + 1+ x= limx→∞

1√x2 + 1+ x

= limx→∞

1x√

x2 + 1+ x√x2

= limx→∞

1x√

x2

x2+1x2+ 1

= limx→∞

1x√

1+1x2+ 1

=0√

1+ 0+ 1= 0





4 2 0 2 4

5

10

x

y

The horizontal asymptote is y = 0.




Example

limx→∞

x3 = ∞ and limx→−∞

x3 = −∞.

4 2 2 4

100

50

50

100

x

y




Example

limx→∞

(x2 − x

). Note that the properties of limits cannot be applied to

infinite limits since ∞ is not a number. So,

limx→∞

(x2 − x

)= limx→∞

x (x − 1) = ∞




Example

limx→∞

x2 + x3− x .

limx→∞

x2 + x3− x = lim

x→∞

x2

x+xx

3x− xx

= limx→∞

x + 13x− 1

=∞−1 = −∞




Continuous Functions at a Point

Definition 5

A function f is continuous at a if

limx→a

f (x) = f (a) (10)

y

x

y = f(x)

a

f(a)

f (a) is defined (a is in the domain of f )limx→a

f (x) exists.

limx→a

f (x) = f (a)




Example

y

x1 3 50 2 4 6

Discontinuities at 1, 3, and 5.

at a = 1, f is undefined

at a = 3, f is defined but limx→3

f (x) DNE;

at a = 5, f is defined and limx→5

f (x) exists, but limx→5

f (x) 6= f (5) .




Example

f (x) =x2 − x − 2x − 2 is discontinuous at 2 because f (2) is undefined.

1 1 2 3 4 51

1

2

3

4

5

x

y




Example

g (x) =

{ 1x2

if x 6= 01 if x = 0

is defined at 0 but limx→0

g (x) = limx→0

1x2

does not exist. This discontinuity is called infinite discontinuity.

4 2 2 41

1

2

3

4

5

x

y




Example

h (x) =

x2 − x − 2x − 2 if x 6= 2

1 if x = 2is defined at 2 and lim

x→2h (x) = 3,

but limx→2

h (x) 6= h (2) . This discontinuity is called removablediscontinuity.

1 1 2 3 4 51

1

2

3

4

5

x

y




Example

k (x) = bxc has discontinuities at all of the integers because limx→n

k (x)

does not exist if n is an integer. These discontinuities are called jumpdiscontinuities.

1 1 2 3 4 51

1

2

3

4

5

x

y




Theorem 6

If f and g are continuous at x = a and c is a constant, then thefollowing functions are also continuous at a.

(a) f ± g(b) cf

(c) fg

(d)fgif g (a) 6= 0




Theorem 7

The following functions are continuous at every number in their domains.

(a) Polynomial functions.

(b) Rational functions.

(c) Power and root functions

(d) Trigonometric Functions




Example

f (x) = x100 − 2x37 + 75 is a polynomial function. So it iscontinuous everywhere: (−∞,∞)

g (x) =x2 + 2x + 17x2 − 1 is a rational function, and continuous on its

domain {x | x 6= ±1} = (−∞,−1) ∪ (−1, 1) ∪ (1,∞) .




Example

h (x) =√x +

x + 1x − 1 −

x + 1x2 + 1

Let h1 (x) =√x ; h2 (x) =

x + 1x − 1 ; and h3 (x) =

x + 1x2 + 1

.

h1 (x) is a root function and continuous on [0,∞).h2 (x) is a rational function and continuous on (−∞, 1) ∪ (1,∞) ,and

h3 (x) is also a rational function and continuous everywhere on R.

So, h (x) is continuous on [0, 1) ∪ (1,∞) .




Example

f (x) =sin x

2+ cos x

Let f1 (x) = sin x , and let f2 (x) = 2+ cos x .

f1 (x) and f2 (x) are trigonometric functions. So, they arecontinuous. Note that cos x ≥ −1. So, f2 (x) = 2 cos x is alwayspositive.

Hence, f (x) =f1 (x)f2 (x)

=sin x

2+ cos xis continuous everywhere on R.




Theorem 8

If g is continuous at a and f is continuous at g (a) , then(f ◦ g) (x) = f (g (x)) is continuous at a.




Example

f (x) = sin(x2)

Let F (x) = sin x , and let G (x) = x2.

F and G are continuous on R.

So, f (x) = F (G (x)) = sin(x2)is continuous on R.



Benginning Calculus Lecture notes 2 - limits and continuity

Education

limits of functions

x y lim x

b b b b b b b b b b

c c c c c c c c c c

limit of f x

existence of limits

calculus limits

idea of limits