Transcript 1.2 Limits and Their Properties

1.2
Limits and Their Properties
• Suppose a football is placed on the 7 yard line.
After a penalty the referee places the ball half
the distance to the goal line (3.5 yards) and
then this is done again (1.75 yard line) and this
is done over and over and over an infinite
amount of times, the ball will never reach the
goal line, but it is approaching it. We then say
that the limit is zero (0).
Definition of a Limit
• As x approaches a, the limit of f(x) is written
•
lim f ( x )  L
xa
If all values of f(x) are close to L for values of x
that are sufficiently close, but not equal, to a.
Left Handed and Right Handed (side)
• Limits only exist if their right handed limit is equal to the
left hand limit.
– Theorem
if lim f ( x)  lim f ( x)  L , then lim f ( x)  L
x a 
x a
x a
• If they are different we state that the limit does not exist
(DNE)
• Notation
• Right hand limit
• Left hand limit
lim
xc
lim
xc
• If both sides of the limit are approaching the
same value, then the limit exists at that value.
• We can determine a limit graphically,
algebraically, and analytically
lim f ( x )
x0
lim f ( x )
x0
lim f ( x ) lim f ( x )
x 3
x  3
lim f ( x) lim f ( x)
x 1
lim f ( x )
x 3
x 1
lim f ( x )
x 1
f ( x)
lim f ( x ) lim
x4
x2
Wall Method
f(x) = sin (1/x)
lim f ( x ) 
x 3
f (3) 
Hwk. Pg. 55 9-24
1
f ( x) 
3
x2
1
lim
3
x 3 x  2
X=2.1
Y=
X=3.5
Y=
X=2.5
Y=
X=3.2
Y=
X=2.99
Y=
X=3.01
Y=
1
f ( x) 
3
x2
• What happens when x approaches 2
Limit principles
cc
• lim
xa
lim12  12
x 6
– The limit of a constant is that constant
n
lim  f ( x)  lim f ( x)   Ln
xa
 x

n
lim n f ( x)  n lim f ( x)  n L
xa
xa
The limit of a power is the power of the limit, and
the limit of a root is the root of the limit.
lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)  L  M
xa
xa
xa
assuming that lim f ( x)  L and lim g ( x)  M
xa
xa
lim( x  3x)  lim x  lim3x  4  6  10
2
x 2
2
x 2
x 2
lim  f ( x)  g ( x)   lim f ( x)  lim g ( x)  L  M
x a
x a
xa
lim x (4 x)  lim x  lim 4 x  9 12
2
x3
2
x3
x3
g ( x) M
g ( x) lim
xa
lim

 ; provided that L  0
x a f ( x )
lim f ( x) L
x a
lim  cf ( x)  c  lim f ( x)  c  L
xa
xa
lim 5 x  5  lim 5( x  1)  5  lim ( x  1)   5  5
x4
x4
x4
• So basically limits can be added, subtracted,
multiplied, divided, taken to a power and
roots can be taken, in addition to being scalar
multiplied.
Use Limit Principles to find
lim  x  3x  7 
2
x 4
lim  x  5 x  x  7 
4
x2
3
2
lim  5 x  12 
2
x4
Theorem on Limits of Rational
Functions
• For any rational function F, with a in the
domain of F, lim F ( x )  F ( a )
xa
This states that as long as you are approaching
a value that is in the domain of the function
then you can simply substitute a into F(x) and
evaluate to find the limit.
• When using the Theorem on Limits of Rational
Functions it is important to first determine
that values for x in which the function does
not exist. (Simplify all algebraic expressions,
and cancel denominators if possible) then
substitute to see if the limit exists.
lim  2 x  3x  4 x  1
4
x3
3
x 8
lim
x3 x  2
2
x  3x  18
lim
x3
x 3
2
• Best way of thinking about a limit is to ask the
question “what y value am I approaching as x
moves along the graph in a certain direction”
x 1
• Consider the following graph of f ( x) 
,x  0
x 1
3
x 1
lim
x 1 x  1
3
Algebraically
-factor numerator and denominator completely
-cancel like terms (quantities)
-evaluate remaining expression with x=c.
x 1
lim
x 1 x  1
3
Difference of a perfect cube
( x  1)( x 2  x  1)
lim
x 1
x 1
lim x  x  1
2
x 1
Evaluate with x = 1
1 11  3
2
x  7 x  12
lim
x 4
x4
2
( x  4)( x  3)
lim
x 4
x4
lim x  3
x4
Evaluate with x=4
4 3 1
This a problem because of 0 in the denominator, but how do we factor? Or
can we factor? Can we manipulate it so it looks different?
x  x  23  5
lim
x1
x 1
2
x 2  x  23  5 x 2  x  23  5
lim

x1
x 1
x 2  x  23  5
lim
x1
x 2  x  23  25
( x  1)( x 2  x  23  5
 lim
x1
( x  2)( x  1)
( x  1)( x 2  x  23  5
Continuity
• A function is CONTINUOUS OVER, or ON,
SOME INTERVAL, of the real line if its graph
can be traced without lifting the pencil from
the paper. If there is any point in an interval
where a “jump” or a “hole” occurs, then we
say that the function is not continuous over
the interval.
• All the previous graphs had discontinuity.
• A function is continuous on an interval if:
1. f (a) exists
2. lim f ( x) exists
theoutput at a exists
thelimit as x  a exists
3. lim f ( x)  f  a 
limit is thesameas theoutput
xa
xa
• Is the function given by f(x)=3x-2 continuous
at x=5? Why or why not?