• What is “calculus”?
• What do you learn in a calculus class?
• How do algebra and calculus differ?
 You will be able to answer all of these
questions after you finish the course.
10.1 Introduction to Limits
• One of the basic concepts to the study of
calculus is the concept of limit.
• This concept will help to describe the behavior of
f(x) when x is approaching a particular value c.
• In this section, we will review and learn more
about functions, graphs, and limits
Example 1a: f(x) = 2x - 1
Discuss the behavior of of f(x) when x is close to 2
using graph
• Graph f(x) = 2x - 1
• When x is closer and
closer to 2,
F(x) is closer to 3
Therefore:
The limit of f(x) as x
approaches 2 is 3
lim(2x-1) = 3 = f(2)
X2
Example 1b: f(x) = 2x - 1
Discuss the behavior of the values of f(x) when x is close to 2
using table
• Find: lim (x+2) and lim (3x+1)
X0
X -1
Do you get 2 and -2? If not, try again
x2  4
f ( x) 
x2
Example 2:
Discuss the behavior of f(x) when x is closer to 2
If x = 0, f(x) is undefined. If you graph, you will see a hole
there.
x
1.5
1.9
1.99
1.999
2
2.001
2.01
2.1
2.5
f (x)
3.5
3.9
3.99
3.999
?
4.001
4.01
4.1
4.5
Therefore, when x is closer and closer to 2, f(x) is closer to 4
lim f(x) = 4 = f(2)
X2
or
f ( x) 
x2
x2
Example 2:
Discuss the behavior of the values of f(x) when x is closer to 2.
Does the limit exist?
x
0
1
1.9
1.99
2
f (x)
-1
-1
-1
-1
?
2.001 2.01 2.1
* This function is not defined when x = 2.
* The limit does not exist because the limit on
the left and the limit on the right are not the
same.
Lim f(x) = -1 represents the limit on the left of 2
X2 -
Lim f(x) = 1 represents the limit on the right of 2
X2 +
1
1
1
2.5
1
■ We write
and call K the limit from the left (or left-hand limit) if
f (x) is close to K whenever x is close to c, but to the left
of c on the real number line.
■ We write
and call L the limit from the right (or right-hand limit) if
f (x) is close to L whenever x is close to c, but to the right
of c on the real number line.
■ In order for a limit to exist, the limit from the left and
the limit from the right must exist and be equal.
Example 3
(A) Discuss the behavior of
f(x) for x near 0
Lim f(x) = 0
X0 -
Lim f(x) = 0
X 0 +
Lim f(x) = 0
X0
F(0) = 0
Example 3 - continue
(B) Discuss the behavior of
f(x) for x near 1
Lim f(x) = 1
X1 -
Lim f(x) = 2
X 1 +
Lim f(x) = does not exist
X1
F(1) = not defined
Example 3 - continue
(C) Discuss the behavior of
f(x) for x near 3
Lim f(x) =
X3
-
Lim f(x) =
3
3
X 3 +
Lim f(x) =
3
X3
F(3) = not defined
Example 4
Lim (x2 + 5x)
X-2
= lim x2 + lim 5x
X-2
Property 1
X-2
= (lim x) (lim x) + 5 (lim x)
X-2
=
=
(-2)
X-2
Property 4 and 3
X-2
(-2) + 5 (-2)
4 – 10 = -6
Try lim (x4 + 3x – 2)
X-1
If you don’t get -4, try again
Example 5
Lim
3x 2  6
X2
=
lim(3 x  6)
2
Property 8
X2
=
Try lim
X-1
2x  2
2
3(2)( 2)  6  12  6  6
If you don’t get 2, try again
Example 6
lim
x  2
x
2
x 1
2
lim x
=
Note that this is a
rational function with a
nonzero denominator at
x = -2
2
x 2
lim ( x  1)
2
x 2
x
Try lim 4
x 1 x  2
=
(2)(2)
4

(2)(2)  1 5
If you don’t get 1/3, try again
Example 7
2 x  3
f ( x)  
 x  12
If x < 5
If x > 5
A) Lim f(x) =
X5-
lim (2x+3) = 2(5)+ 3 = 13
X5-
B) Lim f(x) =
X5+
lim (-x+12)=-5+12=7
X5+
C) Lim f(x) =
X5
Does not exist because the left hand the right hand limits are not equal
D) F(5) =
is not defined
Example 8: Use algebraic and/or graphical
techniques to analyze each of the following
indeterminate forms
A)
x 1
lim 2
x 1 x  1
B)
( x  1)
lim 2
x 1 x  1
C)
x2 1
lim
x 1 ( x  1) 2
2
See next page for step by
step instruction
Example 8 - Solutions
Note: when you find the limits of the above problems, you must factor first
and then simplify before you substitute the number for x
Different Quotient (pre-cal)
f ( x  h)  f ( x )
h
Indeterminate form
Examples:
• Find the following limit for the following functions
9)
10)
11)
f ( 4  h )  f ( 4)
f ( x )  7  2 x; lim
h 0
h
f (1  h)  f (1)
f ( x )  x  1 ; lim
h 0
h
f ( x) 
f (3  h)  f (3)
x ; lim
h 0
h
See the next pages for step by step instruction
Because the limit on the left and the litmit
on the right are not the same.
Therefore, this limit does not exist
Rationalize the numerator