1.1 A Preview of Calculus and 1.2 Finding Limits Graphically and Numerically
Finding Limits Graphically and Numerically An Introduction to Limits Limits that Fail to Exist A...
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Transcript of Finding Limits Graphically and Numerically An Introduction to Limits Limits that Fail to Exist A...
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