using the sandwich theorem to find. if we graph, it appears that

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