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
