Aim: What are limits?
Do Now:
John is given 24 inches of wire and is
asked to form a rectangle whose area is
as large as possible. What are the
dimensions of this rectangle?
40
P = 2w + 2l
24 = 2w + 2l
12 - w = l
A = lw
30
20
A = (12 – w)(w)
A = 12w – w2
lim (12 x  x 2 )  36
10
x 6
l – 6, w - 6
Aim: Limits
Course: Calculus
Definition of Limit
If f(x) becomes arbitrarily close to a unique
is the valuec of
y as
x side,
numberWhat
L as x approaches
from
either
approaches/equals
somec is
value
the limit
of f(x) as x approaches
L. This
is written as
g x  = 3x-2
lim f ( x )  L
x c
example:
Estimate the limit
lim ( 3 x  2) = 4
x2
let g(x) = 3x - 2
5
A: (2.00, 4.00)
4
A
3
2
continuous
1
2
evaluate for values of x as x approaches 2
x 1.9 1.99 1.999 2.0 2.001 2.01 2.1
g(x) 3.7 3.97 3.997 4
Aim: Limits
4.003 4.03 4.3
Course: Calculus
Model Problems
Use a table to estimate numerically the limit
x
lim
x 0
x 1 1
x
Let h( x ) 
=2
x 1 1
evaluate for values of x as x approaches 0
0 0.0001 0.001
0.01
x -0.01 -0.001 -0.0001 0
h(x)
1.995
4
h x  =
1.9995 1.99995 2
E 2.00005 2.0005
x
 x+1 0.5-1
3
2
1
discontinuous:
f is undefined at x = 0
2
Aim: Limits
4
Course: Calculus
2.005
NOTE!
lim
x 0
4
h x  =
x
=2
x 1 1
x
 x+1 0.5-1
3
2
1
discontinuous:
h is undefined at x = 0
2
4
Even though the function h is not
defined for x = 0, function h still has a
limit as x approaches 0.
This is situation is common
Aim: Limits
Course: Calculus
Model Problem - Graphically
x  x  x 1
Find lim
=2
x 1
x 1
3
q x  =
2
x 3-x 2+x-1
4
x-1
3
2
discontinuous:
q is undefined at x = 1
1
-2
2
Because of discontinuity,
unable to evaluate when x = 1
Aim: Limits
Course: Calculus
Conditions where Limits don’t Exist
The limit of f(x) as x  c does not exist if
any of the following conditions is true.
1. f(x) approaches a different number from
the right side of c than from the left side
of c.
x
lim
x0 x
r x  =
2
x
r( x) 
x
x
x
1
r(x)  1
-2
2
-1
r(x)  -1
Aim: Limits
Course: Calculus
Conditions where Limits don’t Exist
The limit of f(x) as x  c does not exist if
any of the following conditions is true.
2. f(x) increases or decreases without bound
as x approaches c.
1
s( x )  2
x
1
lim 2
x0 x
s x  =
1
3
x2
2
s(x)  
as x approaches 0
1
-2
2
s(x)  
as x approaches 0
Aim: Limits
Course: Calculus
Conditions where Limits don’t Exist
The limit of f(x) as x  c does not exist if
any of the following conditions is true.
3. f(x) oscillates between two fixed values as
x approaches c.
1
1
lim sin  s( x ) 
x 0
x
 x
t x  = sin

1
x
1
-2
2
-1
Aim: Limits
Course: Calculus
Aim: What are limits?
Do Now:
Find the limit algebraically,
numerically, and graphically for
x2  x  2
lim
x 1
x 1
if it exists.
Aim: Limits
Course: Calculus
Do Now
f x  =
x 2+x+2
10
x+1
8
6
4
2
-10
-5
5
10
-2
-4
-6
-8
Aim: Limits
Course: Calculus
Properties of Limits
Let b and c be real numbers and let n be a
positive integer.
1. lim b  b
xc
2. lim x  c
xc
3. lim x  c
n
n
x c
4. lim n x  n c
for n even and c  0
x c
Limits for these “well behaved” functions
can be found by direct substitution i.e.
the limit of f(x) as x  c is f(c)
Aim: Limits
Course: Calculus
Operations with Limits
Let b and c be real numbers, let n be a
positive integer, and let f and g be
functions with the following limits.
lim f ( x )  L
lim g ( x )  K
x c
xc
1. Scalar multiple lim [bf ( x )]  bL
xc
2. Sum or
difference
3. Product
4. Quotient
5. Power
Aim: Limits
lim [ f ( x )  g( x )]  L  K
x c
lim [ f ( x ) g ( x )]  LK
x c
f ( x)
L
]
K 0
lim [
x c g( x )
K
n
n
[
f
(
x
)]

L
lim
x c
Course: Calculus
Model Problems
Find the limit of the following as x
approaches 2
1, x  2
f ( x)  
 0, x  2
1. lim b  b
xc
1.5
1
lim f ( x )  1
x2
0.5
-1
1
2
-0.5
Aim: Limits
Course: Calculus
Model Problem

Find the lim  x 2  1
x 5

x  1 
lim [ f ( x ) g ( x )]  LK
3. Product
x c


lim x 2  1  lim x  1
x 5
x 5
(25 + 1) (2) = 52
Aim: Limits
Course: Calculus
Model Problems
Find each of the following limits
lim x
2
x 4
lim 5
= 42 = 16
=5
x 4
2
(
x
 2 x  5)
lim
x 4
x  2x  5
lim
x 4
x 1
= 19
2
Aim: Limits
= 19/3
Course: Calculus