Limit Squeeze Theorem
lim f ( x)  A
xa
If f ( x)  g ( x)  h( x) and lim h( x)  A
xa
g ( x)  A
Then, lim
x a
f ( x)
Ex1 : If 2  x 2  f ( x)  2  x 2 , find lim
x0


lim 2  x 2  2
x0


lim 2  x 2  2
x0
 By Squeeze Theorem
lim f ( x)  2
x 0
2
4
Ex2 : Draw f ( x)  x and h( x)  x
Suppose that
and
x 4  g ( x)  x 2
if
x 1
x 2  g ( x)  x 4
if
x 1
We are saying that g(x) is a function that sits between f(x) and h(x).
g ( x)
a) Find, xlim
 1
2
x
 1 and
A. xlim
1
lim  x 2  1
 lim x 2  1
lim  x 4  1
 lim x 4  1
x1
4
x
 1 and
B. xlim
1
x1
x1
x1
 From (A) and (B), according to the Squeeze Theorem
lim g ( x )  1
x  1
g ( x)
b) Find, lim
x0
lim x 4  0
x0
lim x 2  0
x0
g ( x)  0
 By the Squeeze Theorem, lim
x 0
g ( x)
c) Find the lim
x1
1
2
lim
x
sin
 0
Ex2: Show that x0
 x
1
1
2
2
lim
x
sin

lim
x

lim
sin


  because…
We cannot use x0
x0
 x  x0
 x
1
lim sin   Does not exist
x 0
 x
1

1

sin
  1
However, we know that
 x
1
2
2
2
2

x

x
sin
 x
(if we multiply by x ) →
 x


 
 x 2  0 and lim x 2  0
Since we know lim
x0
x0
 
x 2 sin    0
 By the Squeeze Theorem, lim
x0
x
1
 
(What the graph might look like) →
In other words, if a function lies between 2 functions and the limit
of the two outside functions is equal, then the limit of the inside
function has the same value.