using the sandwich theorem to find. if we graph, it appears that
TRANSCRIPT
Using the Sandwich theorem to find 0
sinlimx
x
x
If we graph , it appears thatsin x
yx
0
sinlim 1x
x
x
If we graph , it appears thatsin x
yx
0
sinlim 1x
x
x
We might try to prove this using the sandwich theorem as follows:
sin 1 and sin 1x x
0 0 0
1 sin 1 lim lim lim
x x x
x
x x x
Unfortunately, neither of these new limits are defined, since the left and right hand limits do not match.
We will have to be more creative. Just see if you can follow this proof. Don’t worry that you wouldn’t have thought of it.
Unfortunately, neither of these new limits are defined, since the left and right hand limits do not match.
(1,0)
1
Unit Circle
cos
sin
P(x,y)
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
tan1
AT
tanAT
1, tan
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
1, tan
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
1, tan
Area AOP
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
1, tan
Area AOP Area sector AOP
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
1, tan
Area AOP Area sector AOP Area OAT
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
1, tan
11 sin
2
Area AOP Area sector AOP Area OAT
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
1, tan
11 sin
2
Area sector AOP
2
2r
2
2
Area AOP Area sector AOP Area OAT
(1,0)
1
Unit Circle
cos
sin
P(x,y)
T
AO
1, tan
11 sin
2
2
11 tan
2
Area AOP Area sector AOP Area OAT
11 sin
2
2
11 tan
2
sin tan multiply by two
sinsin
cos
11
sin cos
divide by sin
sin1 cos
Take the reciprocals, which reverses the inequalities.
sincos 1
Switch ends.
11 sin
2
2
11 tan
2
sin tan
sinsin
cos
11
sin cos
sin1 cos
sincos 1
0 0 0
sinlim cos lim lim1
0
sin1 lim 1
By the sandwich theorem:
0
sinlim 1
0
sinlim 1x
x
x
0
1 coslim 0x
x
x
0
sin8limx
x
x0
3sin 5limx
x
x0
5sin 3lim
2x
x
x