Download calc 2.2 - Souderton Math

2.2 Computing Limits
We will find limits algebraically
Theorem 2.2.1
• Let a and k be real numbers.
(a ) lim k  k
xa
Theorem 2.2.1
• Let a and k be real numbers.
(a ) lim k  k
xa
(b) lim x  a
xa
Theorem 2.2.1
• Let a and k be real numbers.
(a ) lim k  k
xa
(b) lim x  a
xa
(c) lim
x 0
1
 
x
Theorem 2.2.1
• Let a and k be real numbers.
(a ) lim k  k
xa
(b) lim x  a
xa
(c) lim
x 0
1
 
x
1
(d ) lim  
x 0 x
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
xa
lim g ( x)  L2
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
(a) lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)  L1  L2
x a
x a
x a
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
(a) lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)  L1  L2
x a
x a
x a
• The limit of a sum is the sum of the limits.
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
(b) lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)  L1  L2
x a
x a
x a
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
(b) lim [ f ( x)  g ( x)]  lim f ( x)  lim g ( x)  L1  L2
x a
x a
x a
• The limit of a difference is the difference of the
limits.
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
(c) lim [ f ( x) g ( x)]  lim f ( x)  lim g ( x)  L1 L2
x a
x a
x a
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
(c) lim [ f ( x) g ( x)]  lim f ( x)  lim g ( x)  L1 L2
x a
x a
x a
• The limit of a product is the product of the limits
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
f ( x) L
f ( x) lim
x a
(d ) lim

 1 , L2  0
x a g ( x)
lim g ( x) L2
x a
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
f ( x) L
f ( x) lim
x a
(d ) lim

 1 , L2  0
x a g ( x)
lim g ( x) L2
x a
• The limit of a quotient is the quotient of the limits
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and L2,
respectively. Then:
(e) lim
x a
n
f ( x)  n lim f ( x)  n L1
x a
Theorem 2.2.2
• Let a be a real number, and suppose that
lim f ( x )  L1
lim g ( x)  L2
xa
xa
• That is, the limits exist and have values L1 and
L2, respectively. Then:
(e) lim
x a
n
f ( x)  n lim f ( x)  n L1
x a
• The limit of an nth root is the nth root of the
limit.
Constants
• A constant factor can be moved through a limit
symbol.
lim (kg( x))  lim k  lim g ( x)  k lim g ( x)
xa
xa
x a
xa
Example 5
• Find
lim ( x 2  4 x  3).
x 5
Example 5
lim ( x  4 x  3).
2
• Find
x 5
 
lim ( x  4 x  3)  lim x  4 lim x  lim 3
2
x 5
2
x 5
x 5
x 5
Example 5
lim ( x  4 x  3).
2
• Find
x 5
 
lim ( x  4 x  3)  lim x  4 lim x  lim 3
2
x 5
2
x 5
x 5
x 5
52  4(5)  3  8
Theorem 2.2.3
• For any polynomial
p( x)  c0  c1 x  ...  cn x n
• and any real number a,
lim p( x)  c0  c1 x  ...  cn x n  p(a)
xa
Rational Functions
•
We will have three situations when dealing with
rational functions.
1. The denominator is not zero.
Rational Functions
•
We will have three situations when dealing with
rational functions.
1. The denominator is not zero.
2. The denominator and the numerator are both zero.
Rational Functions
•
We will have three situations when dealing with
rational functions.
1. The denominator is not zero.
2. The denominator and the numerator are both zero.
3. The denominator is zero and the numerator is not.
Denominator not Zero
• When the denominator is not zero, we evaluate the
numerator and denominator and for this problem
use theorem 2.2.2d: the limit of a quotient is the
quotient of the limit.
5x3  4
lim
x2 x  3
Denominator not Zero
• When the denominator is not zero, we evaluate the
numerator and denominator and for this problem
use theorem 2.2.2d: the limit of a quotient is the
quotient of the limit.
5 x 3  4 5(2)3  4
lim

 44
x 2 x  3
23
Numerator and Denominator both
Zero.
• We know that when the numerator and denominator
are both zero, there is a hole in the graph. Even
though the value doesn’t exist, the limit does.
Numerator and Denominator both
Zero.
• We know that when the numerator and denominator
are both zero, there is a hole in the graph.
• Factor, cancel, evaluate.
x2  4
lim
x2 x  2
Numerator and Denominator both
Zero.
• We know that when the numerator and denominator
are both zero, there is a hole in the graph.
• Factor, cancel, evaluate.
x  4 ( x  2)( x  2)
lim

 x24
x 2 x  2
x2
2
Numerator and Denominator both
Zero.
• We know that when the numerator and denominator
are both zero, there is a hole in the graph.
• Factor, cancel, evaluate.
x 1
lim
x 1
x 1
Numerator and Denominator both
Zero.
• We know that when the numerator and denominator
are both zero, there is a hole in the graph.
• Factor, cancel, evaluate.
x  1 ( x  1)( x  1)
lim

 x 1  2
x 1
x 1
x 1
Numerator and Denominator both
Zero.
• We know that when the numerator and denominator
are both zero, there is a hole in the graph.
• Rationalize the denominator, cancel, evaluate.
x 1
x  1 ( x  1)( x  1)
lim


2
x 1
( x  1)
x 1 x  1
Denominator is Zero, the
Numerator is not
• We know that when the denominator is zero and
the numerator is not, there is a vertical asymptote.
Denominator is Zero, the
Numerator is not
• We know that when the denominator is zero and
the numerator is not, there is a vertical asymptote.
• Sign analysis- Determine what the graph is doing
around the asymptote.
Denominator is Zero, the
Numerator is not
• We know that when the denominator is zero and
the numerator is not, there is a vertical asymptote.
• Sign analysis
2 x
2 x
2 x
lim
or lim
or lim
x 4 ( x  4)( x  2)
x 4  ( x  4)( x  2)
x 4 ( x  4)( x  2)
___+___|_____-_____|___+___|___-___
2
2
4
Denominator is Zero, the
Numerator is not
• We know that when the denominator is zero and
the numerator is not, there is a vertical asymptote.
• Sign analysis
2 x
2 x
2 x
lim
or lim
or lim
x 4 ( x  4)( x  2)
x 4  ( x  4)( x  2)
x 4 ( x  4)( x  2)


DNE
___+___|_____-_____|___+___|___-___
2
2
4
Denominator is Zero, the
Numerator is not
• We know that when the denominator is zero and
the numerator is not, there is a vertical asymptote.
• Sign analysis
2 x
2 x
2 x
lim
or lim
or lim
x 4 ( x  4)( x  2)
x 4  ( x  4)( x  2)
x 4 ( x  4)( x  2)


DNE
• Remember, all three answers mean that the limit
does not exist. The first two answers tell us exactly
what the function is doing and why the limit DNE.
Limits of Piecewise Functions
• Let
f ( x) 
1
x2
x  2
x2  5
x  13
2 x3
x3
• Find
(a ) lim f ( x)
x  2
(b) lim f ( x)
x 0
(c ) lim f ( x )
x3
Limits of Piecewise Functions
• Let
f ( x) 
1
x2
x2  5
x  13
x  2
2 x3
x3
• Find
(a ) lim f ( x)
x  2
lim  
x  2
1
 
x2
lim f ( x)  DNE
x  2
lim   x 2  5  1
x2
Limits of Piecewise Functions
• Let
f ( x) 
1
x2
x  2
x2  5
x  13
2 x3
x3
• Find
(a ) lim f ( x)
x  2
(b) lim f ( x)
x 0
lim  02  5  5
x 0
Limits of Piecewise Functions
• Let
f ( x) 
1
x2
x  2
x2  5
x  13
2 x3
x3
• Find
(a ) lim f ( x)
x  2
(b) lim f ( x)
(c ) lim f ( x )
x 0
lim  x 2  5  4
x 3
lim  x  13  4
x 3
x3
lim f ( x)  4
x 3
Homework
• Pages 121-122
• 1-31 odd