Math 220
September 13
I. Determine the values of a for which limx→a f (x) exists.


1 + sin(x) if x < 0
f (x) = cos(x)
if 0 ≤ x ≤ π


sin(x)
if x > π
II. Determine the infinite limit.
1.
lim−
x+3
x−4
lim+
x+3
x−4
x→4
2.
x→4
3.
3−x
x→4 (x − 4)2
lim
4. limx→4+ ln(x2 − 16)
5. limx→π+ cot(x)
6.
lim−
x→2
x2 + 5x + 6
x2 − 3x + 2
1
7.
x2 − 4
lim
x→2+ x2 + 2x − 8
III. Evaluate the limit, if it exists
1.
x2 − 4x
x→−1 x2 − 3x − 4
lim
2.
√
√
2+x− 2−x
lim
x→0
x
3.
√
3− x
lim
x→9 9x − x2
4.
(x + h)3 − x3
h→0
h
lim
5.
(x + h)−2 − x−2
h→0
h
lim
2
1
Solutions
I. Determine the values of a for which limx→a f (x) exists.


1 + sin(x) if x < 0
f (x) = cos(x)
if 0 ≤ x ≤ π


sin(x)
if x > π
Answer:
By inspection, we can see that limit exists for a < 0, 0 < a < π and a > π
lim 1 + sin(x) = 1 + sin(0) = 1 + 1 = 2
x→0−
lim cos(x) = cos(0) = 1
x→0+
lim f (x) = lim+ f (x)
x→0−
x→0
So we have the limit exists at a = 0.
lim cos(x) = cos(π) = −1
x→π −
lim sin(x) = sin(π) = 0
x→π +
lim f (x) 6= lim+ f (x)
x→π −
x→π
So the limit doesn’t exist at a = π.
The limit exists for a < π and a > π.
II. Determine the infinite limit.
3
1.
lim−
x→4
x+3
x−4
Answer:
x+3
lim−
x→4 x − 4
2.
lim+
x→4
7
0−
= −∞
x+3
x−4
Answer:
x+3
lim+
x→4 x − 4
3.
7
0+
=∞
3−x
x→4 (x − 4)2
lim
Answer:
x+3
lim−
x→4 (x − 4)2
x+3
lim+
x→4 (x − 4)2
7
0+
7
0+
=∞
=∞
x+3
=∞
x→4 (x − 4)2
lim
4
4.
lim+ ln(x2 − 16)
x→4
Answer:
lim+ ln(x2 − 16)
ln(0+ ) = −∞
x→4
5.
lim cot(x)
x→π +
Answer:
cos(x)
lim+ cot(x) = lim+
x→π
x→π sin(x)
6.
lim−
x→2
−1
0−
=∞
x2 + 5x + 6
x2 − 3x + 2
Answer:
(x + 3)(x + 2)
x2 + 5x + 6
= lim−
lim− 2
x→2 (x − 2)(x − 1)
x→2 x − 3x + 2
7.
lim+
x→2
20
0−
= −∞
x2 − 4
x2 + 2x − 8
Answer:
lim+
x→2
x2 − 4
(x + 2)(x − 2)
(x + 2)
4
2
= lim+
= lim+
= =
2
x + 2x − 8 x→2 (x − 2)(x + 4) x→2 (x + 4)
6
3
5
III. Evaluate the limit, if it exists
1.
x2 − 4x
x→−1 x2 − 3x − 4
lim
Answer:
x2 − 4x
x(x − 4)
x
1
= lim
= lim
=
2
x→−1 x − 3x − 4
x→−1 (x − 4)(x + 1)
x→−1 (x + 1)
2
lim
2.
√
√
2+x− 2−x
lim
x→0
x
Answer:
√
lim
x→0
3.
2+x−
x
√
2−x
√
√
√
√
2+x− 2−x 2+x+ 2−x
√
√
= lim
x→0
x
2+x+ 2−x
(2 + x) − (2 − x)
√
= lim √
x→0 x( 2 + x +
2 − x)
2x
√
= lim √
x→0 x( 2 + x +
2 − x)
2
√
= lim √
x→0 ( 2 + x +
2 − x)
2
= √
√
( 4 + 2 − 2)
2
=
2
=1
√
3− x
lim
x→9 9x − x2
6
Answer:
√
√ √ 3− x
3− x 3+ x
√
lim
= lim
x→9 9x − x2
x→9 9x − x2
3+ x
9−x
√
= lim
x→9 (9x − x2 )(3 +
x)
9−x
√
= lim
x→9 x(9 − x)(3 +
x)
1
√
= lim
x→9 x(3 +
x)
1
√
=
9(3 + 9)
1
=
9(6)
1
=
54
4.
(x + h)3 − x3
h→0
h
lim
Answer:
(x + h)3 − x3
(x3 + 3x2 h + 3xh2 + h3 ) − x3
= lim
h→0
h→0
h
h
3x2 h + 3xh2 + h3
= lim
h→0
h
2
= lim 3x + 3xh + h2
lim
h→0
2
= 3x
5.
(x + h)−2 − x−2
h→0
h
lim
7
Answer:
(x + h)−2 − x−2
lim
= lim
h→0
h→0
h
1
1 1
−
(x + h)2 x2 h
x2
1
(x + h)2
= lim
− 2
2
2
2
h→0
x (x + h)
x (x + h) h
2
2
x − (x + h)
1
= lim
2
2
h→0
x (x + h)
h
2
2
x − (x + 2xh + h2 ) 1
= lim
h→0
x2 (x + h)2
h
2
1
−2xh − h
= lim
2
2
h→0
x (x + h)
h
(−2x − h)h 1
= lim
h→0
x2 (x + h)2 h
(−2x − h)
= lim
h→0
x2 (x + h)2
−2x
= 2 2
xx
−2x
= 4
x
−2
= 3
x
8