Math 201
Activity 2 Key
Section 3
Spring 2013
sin x
x−→0 x
Squeeze Theorem and lim
y
6
M N
sin θ
θ
O
P
cos θ
Q
-
x
-
Assume1 that OM = OQ = 1
d
The arc M
Q is part of the circle with radius 1, centered at O.
d
Q, prove that
1. Considering the segment M P and the arc M
0 ≤ sin θ < θ.
d
Solution. The arc M
Q is greater than the linear segment M Q. Being the hypotenuse
of the right 4M P Q, the linear segment M Q must be larger than the side M P . By
d
trigonometry we have that the arc M
Q equals θ(1) = θ, whereas M P = sin θ. Since
0 < θ, we have 0 ≤ sin θ < θ, as required.
2. Use the Squeeze Theorem to prove that
lim sin θ = 0.
θ−→0+
1
diagsins.tex
Solution. If θ −→ 0+ , then by squeezing 0 ≤ sin θ < θ, we obtain the result.
3. In words, sketch a possible proof for
lim sin θ = 0.
θ−→0−
Solution. We could draw a diagram like the above but with a negative θ, prove that
θ < sin θ ≤ 0 and squeeze.
4. Now prove that
lim cos θ = 1.
θ−→0
[Hint: Find an identity involving sin θ and cos θ, solve for cos θ,p
and use (3).]
2
2
1 − sin2 θ. Thus
Solution. From sin
p θ + cos θ = 1 it follows that cos θ =
lim cos θ = lim 1 − sin2 θ = 1, by 2 and 3.
θ−→0
θ−→0
5. Show that the area of the triangle 4ON Q equals 12 tan θ.
Solution. The area of the triangle 4ON Q equals (1/2)(OQ)(N Q) = (1/2)N Q. On
the other hand,
NQ
NQ
tan θ =
=
= N Q.
1
OQ
Thus the area of the triangle 4ON Q equals (1/2) tan θ, as required.
6. Considering (1), (5), and the area of the circular sector OM Q, prove that
sin θ < θ < tan θ.
Solution. From (1) we obtained the first inequality. For the second one, notice that
the area of the circular sector OM Q equals (1/2)θ(1)2 = (1/2)θ and is smaller
than the area of the 4ON Q, which by (5) equals (1/2) tan θ. Thus (1/2)θ <
(1/2) tan θ =⇒ θ < tan θ, as required.
7. Thus show that
θ
1
1<
<
.
sin θ
cos θ
Solution. sin θ < θ < tan θ can be rewritten as
sin θ < θ <
sin θ
cos θ
Now, since θ > 0 we have that sin θ > 0. Thus dividing (A) by sin θ, yields
1<
as required.
θ
1
<
sin θ
cos θ
(A)
8. Hence
sin θ
< 1.
θ
Solution. All the terms involved in (7) are positive, so we can exchange denominator
with numerators on each inequality.
9. Now prove that
sin θ
= 1.
lim+
θ−→0
θ
Solution. By (4) lim+ cos θ = 1. Also lim+ 1 = 1. Thus squeezing inequality 8,
cos θ <
θ−→0
θ−→0
yields the result.
10. In words, sketch a possible proof for
lim−
θ−→0
sin θ
= 1.
θ
Solution. We could consider a diagram with negative θ. Then the areas would be
negative and the inequalities involved would be reversed.