Calculus 221 worksheet
Trig Limit and Sandwich Theorem
Example 1.
sin(x)
= 1. Use this limit along with the other “basic limits” to find the
Recall that lim
x→0
x
following:
1 − cos(x)
(1) lim
. [Hint: Multiply top and bottom by 1 + cos(x).]
x→0
x2
1 − cos(x)
.
(2) lim
x→0
x
tan(x)
(3) lim
.
x→0
x
Solution:
(1)
1 − cos(x)
1 − cos(x) 1 + cos(x)
= lim
·
2
x→0
x→0
x
x2
1 + cos(x)
2
1 − cos (x)
= lim 2
x→0 x (1 + cos(x))
sin2 (x)
= lim 2
x→0 x (1 + cos(x))
2 sin(x)
1
= lim
x→0
x
1 + cos(x)
1
1
=
=1·
1+1
2
lim
(2) Using (a),
1 − cos(x)
1 − cos(x)
= lim x ·
x→0
x→0
x
x2
1
=0· =0
2
lim
(3)
sin(x)
tan(x)
cos(x)
lim
= lim
x→0
x→0
x
x
sin(x)
1
= lim
·
x→0
x
cos(x)
1
=1· =1
1
Example 2. Evaluate the limit limπ tan(x) − sec(x) or show that it does not exist.
x→
2
1
2
Solution:
limπ tan(x) − sec(x) = limπ
x→
x→
2
2
= limπ
x→
2
= limπ
x→
2
= limπ
x→
2
= limπ
x→
2
1
sin(x)
−
cos(x) cos(x)
sin(x) − 1 sin(x) + 1
·
cos(x)
sin(x) + 1
sin2 (x) − 1
cos(x)(sin(x) + 1)
− cos2 (x)
cos(x)(sin(x) + 1)
− cos(x)
sin(x) + 1
0
= =0
2
sin(x)
or show that it does not exist.
x→0 1 − cos(x)
Example 3. Evaluate the limit lim
Solution:
lim
x→0
sin(x)
sin(x)
1 + cos(x)
= lim
·
x→0
1 − cos(x)
1 − cos(x) 1 + cos(x)
sin(x)(1 + cos(x))
= lim
x→0
1 − cos2 (x)
sin(x)(1 + cos(x))
= lim
x→0
sin2 (x)
1 + cos(x)
= lim
x→0
sin(x)
2
=‘ ’=∞
0
so the limit does not exist.
1
Example 4. Use the Sandwich Theorem to evaluate the limit lim x · sin
.
x→0
x
1
1
Solution: Since −1 ≤ sin
≤ 1 for all x, it follows that −|x| ≤ x sin
≤ |x| for all
x
x
x.
We know that lim |x| = 0 and lim −|x| = 0. Therefore, the Sandwich Theorem says that
x→0
x→0
1
lim x · sin
= 0.
x→0
x
3
Exercise
1 − cos(3x)
x→0
2x2
ex
2) lim
x→0 sin(πx)
1) lim
sin(sin(x))
x→0
sin(x)
3) lim
4) lim
sin(sin(x))
x
5) lim
1 − cos(5x)
sin2 (3x)
x→0
x→0
x − 4 sin(x)
.
x→∞
2x
6) Use the Sandwich Theorem to evaluate the limit lim
4
Answers
1)
9
4
2)
e
π
3) 1
4) 1
5)
25
18
6)
1
2