Finding Limits Graphically and Numerically

•An Introduction to Limits•Limits that Fail to Exist•A Formal Definition of a Limit

An Introduction to Limits

• Graph:

• What can we expect at x = 1?

• Approach x=1 from the left.• Approach x=1 from the right.• Are we approaching a specific value from both

sides? What is that number?• Do Now: evaluate f(1.1)

f (x) =x3 −1x−1

Numericallyx 0.75 0.90 0.99 0.999 1 1.001 1.01 1.10 1.25

f(x) ?

f (x) =x3 −1x−1

Fill in chart for all values of x:

Numericallyx 0.75 0.90 0.99 0.999 1 1.001 1.01 1.10 1.25

f(x) 2.31 2.71 2.97 2.997 ? 3.003 3.0301 3.31 3.81

limx→1

x3 −1x−1

=3

Notation

Lxfcx

)(limThe limit of f(x) as x approaches c is L.

Exploration

2

232

2lim

x

xx

x

x 1.75 1.90 1.99 1.999 2 2.001 2.01 2.10 2.25

f(x)

Exploration

x→ 2lim

x2 −3x+2x−2

=1

x 1.75 1.90 1.99 1.999 2 2.001 2.01 2.10 2.25

f(x) .75 .9 .99 .999 Und.

1.001 1.01 1.1 1.25

Example 1: Estimating a Limit Numerically

11lim0 x

x

x

Where is it undefined?What is the limit?

Example 1: Estimating a Limit Numerically

11lim0 x

x

x

Where is it undefined? 0What is the limit? 2

x -.1 -.01 -.001 0 .001 .01 .1

f(x) 1.95 1.995 1.9995 Und.

2.0005

2.005 2.05

Estimating a Limit Numerically

• It is important to realize that the existence or nonexistence of f(x) at x = c has no bearing on the existence of the limit of f(x) as x approaches c.

• The value of f(c) may be the same as the limit as x approaches c, or it may not be.

Finding the limit by substitution

• Always try evaluating a function at c first:• Examples:

1)limx→ 2

x4

2) limx→−3

(3x+2)

3)limx→ 7

5xx+2⎛

⎝⎜

⎞

⎠⎟

Finding the limit by substitution

• Always try evaluating a function at c first:• Simple and boring!

1)limx→ 2

x4 =16

2) limx→−3

(3x+2)=−7

3)limx→ 7

5xx+2⎛

⎝⎜

⎞

⎠⎟=

359

Substitution needing analytical approach:

Factor and simplify:

limx→−5

x+5x2 −25

Indeterminate forms occur when substitution in the limit results in 0/0. In such cases either factor or rationalize the expressions.

Ex.25

5lim

25x

x

x

Notice form0

0

5

5lim

5 5x

x

x x

Factor and cancel common factors

5

1 1lim

5 10x x

Indeterminate Forms

Using Algebraic Methods• When substitution renders an indeterminate

value, try factoring and simplifying:• Hint for # 3: use synthetic division to factor numerator (see if

x+2 is a factor) limx→−1

x2 −1x+1

limx→−1

2x2 −x−3x+1

limx→−2

x3 +8x+2

Using Algebraic Methods• Now try to substitute in “c”

limx→−1

x2 −1x+1

=(x+1)(x−1)

x+1=x−1

limx→−1

2x2 −x−3x+1

=(2x−3)(x+1)

x+1=2x−3

limx→−2

x3 +8x+2

=(x2 −2x+ 4)(x+2)

x+2=x2 −2x+ 4

Using Algebraic Methods

• Substitution works for the simplified version.

limx→−1

x2 −1x+1

=x−1=−2

limx→−1

2x2 −x−3x+1

=2x−3=−5

limx→−2

x3 +8x+2

=x2 −2x+ 4 =12

More complicated algebraic methods

• Involving radicals:

9

3a) lim

9x

x

x

9

( 3)

( 3)

( 3) = lim

( 9)x

x

x

x

x

9

9 lim

( 9)( 3)x

x

x x

9

1 1 lim

63x x

Other Algebraic Methods:

• 1) Try simplifying a complex fraction • 2) Try rationalizing (the numerator):

1)limx→ 0

12 + x

−12

2x

2)limx→ 0

x+ 3 − 3x

Other Algebraic Methods:

• Try simplifying a complex fraction or rationalizing (a numerator or denominator):

1)limx→ 0

12 + x

−12

2x= −14x+8

=−18

2)limx→ 0

x+ 3 − 3x

= 1x+ 3 + 3

= 12 3

Do Now: graph the piecewise function:

Find2

3 if 2lim ( ) where ( )

1 if 2x

x xf x f x

x

-2

62 2

lim ( ) = lim 3x x

f x x

Note: f (-2) = 1

is not involved

=−3(−2) =6

Using a graph to find the limit:

Ex 2: Finding the limit as x → 2

2,0

2,1)(

x

xxf

1. Numerical Approach – Construct a table of values.2. Graphical Approach – Draw a graph by hand or using technology.

Ex 2: Finding the limit as x → 2

f (x) =1,x≠20,x=2

⎧⎨⎩

al Approach – Use algebra or calculus.

x

−.50.511.522.5

y

1

1

1

1

1

0

1

limx→ 2

f(x) =1

2

2

4( 4)a. lim

2x

x

x

0

1, if 0b. lim ( ), where ( )

1, if 0x

xg x g x

x

20

1c. lim ( ), where f ( )

xf x x

x

Use your calculator to evaluate the limits

2

2

4( 4)a. lim

2x

x

x

0

1, if 0b. lim ( ), where ( )

1, if 0x

xg x g x

x

20

1c. lim ( ), where f ( )

xf x x

x

Answer : 16

Answer : no limit

Answer : no limit

3) Use your calculator to evaluate the limits

2 if 3( )

2 if 3

x xf x

x x

ExamplesDo Now: Graph the function:

Limits that Fail to Exist-this one approaches a different value from

the left and the right

x→ 0lim

xx

1. Numerical Approach – Construct a table of values.2. Graphical Approach – Draw a graph by hand or using technology.3. Analytical Approach – Use algebra or calculus.

Ex 4: Unbounded Behavior

20

1lim xx

1. Numerical Approach – Construct a table of values.2. Graphical Approach – Draw a graph by hand or using technology.3. Analytical Approach – Use algebra or calculus.

Ex 5: Oscillating Behavior

xx

1sinlim

0

1. Numerical Approach – Construct a table of values.2. Graphical Approach – Draw a graph by hand or using technology.3. Analytical Approach – Use algebra or calculus.

x 1 .5 .1 .01 .001 .0001 As x approaches 0?

x

1sin

Ex 5: Oscillating Behavior

xx

1sinlim

0

1. Numerical Approach – Construct a table of values.2. Graphical Approach – Draw a graph by hand or using technology.3. Analytical Approach – Use algebra or calculus.

x 1 .5 .1 .01 .001 .0001 As x approaches 0?

.84 .91 -.54 -.51 .827 -.31 0? No! It doesn’t exist!

x

1sin

Common Types of Behavior Associated with the Nonexistence of a Limit

1. f(x) approaches a different number from the right side of c than it approaches from the left side.

2. f(x) increases or decreases without bound as x approaches c.

3. f(x) oscillates between two fixed values as x approaches c.

A Formal Definition of a Limit

• Lim x→c f(x) = L

• If for every number ε > 0

• There is a number δ > 0

• Such that |f(x) – L| < ε

• Whenever 0 < |x – c| < δ

Using the formal definition.

• Prove: lim x→3 (4x – 5) = 7Lim x→c f(x) = L

If for every number ε > 0There is a number δ > 0Such that |f(x) – L| < ε

Whenever 0 < |x – c| < δ

The right-hand limit of f (x), as x approaches a, equals L

written:

if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to the right of a.

lim

x→ a+f (x) =L

a

L( )y f x

One-Sided Limit One-Sided Limits

The left-hand limit of f (x), as x approaches a, equals M

written:

if we can make the value f (x) arbitrarily close to L by taking x to be sufficiently close to the left of a.

lim ( )x a

f x M

a

M

( )y f x

2 if 3( )

2 if 3

x xf x

x x

1. Given

3lim ( )x

f x

Find

Find 3

lim ( )x

f x

ExamplesExamples of One-Sided Limit

2 if 3( )

2 if 3

x xf x

x x

1. Given

3lim ( )x

f x

3 3lim ( ) lim 2 6x x

f x x

2

3 3lim ( ) lim 9x x

f x x

Find

Find 3

lim ( )x

f x

ExamplesExamples of One-Sided Limit

So and therefore, does not exist! limx→ 3+

f (x) ≠ limx→ 3−

f (x) limx→ 3

f(x)

2. Let f (x)

x1, if x 0−x−1, if x≤0.

⎧⎨⎩ Find the limits:

0lim( 1)x

x

0

a) lim ( )x

f x

0b) lim ( )

xf x

=lim

x→ 0−(−x−1)

1c) lim ( )

xf x

1lim( 1)x

x

1d) lim ( )

xf x

1lim( 1)x

x

More Examples

2. Let f (x)

x1, if x 0−x−1, if x≤0.

⎧⎨⎩ Find the limits:

0lim( 1)x

x

0 1 1 0

a) lim ( )x

f x

0b) lim ( )

xf x

=lim

x→ 0−(−x−1) 0 1 1

1c) lim ( )

xf x

1lim( 1)x

x

1 1 2

1d) lim ( )

xf x

1lim( 1)x

x

1 1 2

More Examples

lim ( ) if and only if lim ( ) and lim ( ) .x a x a x a

f x L f x L f x L

For the function

1 1 1lim ( ) 2 because lim ( ) 2 and lim ( ) 2.x x x

f x f x f x

But

0 0 0lim ( ) does not exist because lim ( ) 1 and lim ( ) 1.x x x

f x f x f x

This theorem is used to show a limit does not exist.

A Theorem

Limits at infinity

• 3 cases: when the degree is:• “top heavy”- goes to negative or positive

infinity• “bottom heavy”- goes to zero• “equal” – put terms over each other and

reduce. What does this mean?

Limits at Infinity

For all n > 0,1 1

lim lim 0n nx xx x

provided that is defined.

1

xn

Ex. 2

2

3 5 1lim

2 4x

x x

x

Divide by 2x

Limits at Infinity

For all n > 0,1 1

lim lim 0n nx xx x

provided that is defined.1nx

Ex.2

2

3 5 1lim

2 4x

x x

x

limx→∞

3x2 x2 5xx2

1x2

2x2

−4x2

x2

3 0 0 3

0 4 4

Divide by 2x

2

2

5 1lim 3 lim lim

2lim lim 4

x x x

x x

x x

x

More Examples

1. limx→∞

2x3 −3x2 +2x3 −x2 −100x+1

⎛

⎝⎜

⎞

⎠⎟

More Examples

3 2

3 2

2 3 21. lim

100 1x

x x

x x x

3 2

3 3 3

3 2

3 3 3 3

2 3 2

lim100 1x

x xx x x

x x xx x x x

3

2 3

3 22

lim1 100 1

1x

x x

x x x

22

1

2

3 2

4 5 212. lim

7 5 10 1x

x x

x x x

2 2 43. lim

12 31x

x x

x

0

2

3 2

4 5 212. lim

7 5 10 1x

x x

x x x

2 3

2 3

4 5 21

lim5 10 1

7x

x x x

x x x

0

7

2 2 43. lim

12 31x

x x

x

2 2 4

lim12 31x

x xx x x

xx x

42

lim31

12x

xx

x

2

12

Limits at infinity

• When the numerator has a larger degree than the denominator…

limx→∞

2x4

1+ 3x limx→−∞

2x4

1+ 3x

Limits at infinity

• When the numerator has a larger degree than the denominator…

limx→∞

2x4

1+ 3x=∞ lim

x→−∞

2x4

1+ 3x=−∞

51

Limits at infinity

lim 0nx

a

x If n is a positive integer, the , where a is some

constant.

• Property:

The denominator has a higher degree

• Find the limit

3

5

7 3 2lim

4 3p

p p

p

The denominator has a higher degree

• Find the limit

3

5

7 3 2lim

4 3p

p p

p

=0

When the degrees are equal…

• Reduce the equal terms

limx→∞

3x5 −2x4x+ 7x5

When the degrees are equal…

• Reduce the equal terms

limx→∞

3x5 −2x4x+ 7x5

=37

56

Example

Evaluate the limit

2

2

2lim

3 1t

t

t t

57

Example

Evaluate the limit

2

2

2lim

3 1t

t

t t

1

3

Continuity

A function f is continuous at the point x = a if the following are true:

This one fails iii !

) ( ) is definedi f a) lim ( ) exists

x aii f x

a

f(a)) lim ( ) ( )

x aiii f x f a

A function f is continuous at the point x = a if the following are true:

) ( ) is definedi f a) lim ( ) exists

x aii f x

) lim ( ) ( )x a

iii f x f a

a

f(a)

At which value(s) of x is the given function discontinuous?

1. ( ) 2f x x 2

92. ( )

3

xg x

x

Continuous everywhere Continuous everywhere

except at 3x

( 3) is undefinedg

lim( 2) 2 x a

x a

and so lim ( ) ( )x a

f x f a

-4 -2 2 4

-2

2

4

6

-6 -4 -2 2 4

-10

-8

-6

-4

-2

2

4

Examples

2, if 13. ( )

1, if 1

x xh x

x

1lim ( )x

h x

and

Thus h is not cont. at x=1.

11

lim ( )x

h x

3

h is continuous everywhere else

1, if 04. ( )

1, if 0

xF x

x

0lim ( )x

F x

1 and

0lim ( )x

F x

1

Thus F is not cont. at 0.x

F is continuous everywhere else

0o

Continuous Functions

A polynomial function y = P(x) is continuous at every point x.

A rational function is continuous at every point x in its domain.

( )( ) ( )p xR x q x

If f and g are continuous at x = a, then

f ±g, fg, and fg g(a) ≠0( ) are continuous

at x=a