Section 15-1
Limits
 If
there is a number L such that the value of
f(x) gets closer and closer to L as x gets
closer to a number a, then L is called the
limit of f(x) as x approaches a.
 In
symbols:
L = l i m f(x)
x  a

Consider the graph of the function y = f(x). Find each
pair of values.

f(-2) and
lim

f(2) and
lim

x  2
x 2
Sometimes f(a) and lim f(x) are the same, but at other times they are
x a
different. This has to do with whether a function is continuous or
discontinuous. Examples of continuous functions include polynomials as
well as sin x, cos x.

f(x) is continuous at a if and only if
lim f(x) = f(a)

x a
 Evaluate
each limit.
 1. lim 2x - 6x -10x + 10 = -18
3
2
x 2
2. lim
x 
s in 3 x
x
= 0




The definition of a limit has x approaching a. The x
values do not actually reach a.
Look at this example:
lim
x 2
x 2  2x
x2 4
lim
x ( x  2)
( x  2(
)x  2)
x 2
=
=
lim
x 2
x
x 2
=½
B ecau se x≠ 2 , w e can lo o k at w h at h ap p en s to th is
function as x approaches 2. We see that as x approaches
2, the function approaches 1/2

Sometimes algebra is not sufficient to find the limit. For example, the
problem l i m s ixn x can not be found by simply replacing x with 0.
x 
0

The function is not continuous at x = 0. You can use the calculator to
compute values of the function s i n x for x values that get closer
and closer to 0 from both sides. x

Notice: As x gets closer and closer to 0 from both sides s i n x gets
closer to 1. So l i m s ixn x = 1
x
x 
x
1
s in x
x
.841470984808
-1
s in x
x
.841470984808
.1
.998334166468
-.1
.998334166468
.01
.999983333417
-.01
.999983333417
.001
.999999833333
-.001
.999999833333
-.0001
.999999998333
x


0
.0001 .999999998333
#21-22,24
HW#32:
Section 15.1
Pg 946 #12-18,
#21-22,24
HW #33:
Section 15.1
Pg 946 #19,20,23,
25-37 odd