CALCULUS I HOMEWORK 2
1. Find the limit, if it exists. If the limit does not exist, explain why.
(a) Compute
lim
x→1
x2 + x − 2
.
x2 − 1
Solution: Factor
x2 + x − 2
(x − 1)(x + 2)
x+2
=
=
2
x −1
(x − 1)(x + 1)
x+1
Hence
lim
x→1
x2 + x − 2
x+2
= lim
= 3/2.
2
x→1 x + 1
x −1
(b) Compute
lim
x→1
1 + 5x2 + 3x3
2 + x2
2
.
Solution: The function is continuous at x = 1, hence
1 + 5x2 + 3x3 2
1+5+3 2
lim
=
= 9.
x→1
2 + x2
2+1
(c) Compute
lim
x→1
1 − |x|3
.
1 − x3
Solution: Near x = 1, we have |x| = x. Hence
lim
x→1
1 − |x|3
1 − x3
=
lim
= 1.
x→1 1 − x3
1 − x3
(d) Compute
lim
t→0
t2
√
1
1
− 2.
1 + t3 t
1
2
CALCULUS I HOMEWORK 2
Solution: We rewrite
1
1
√
− 2
t2 1 + t3 t
1
√
−1
1 + t3
√
1
1 + t3
1
√
√
−
t2
1 + t3
1 + t3
√
1 1 − 1 + t3
√
.
t2
1 + t3
1
t2
=
=
=
Multiplying by the conjugate
1
1
√
− 2
2
3
t
t 1+t
√
√
1 1 − 1 + t3
1 + 1 + t3
√
√
= 2
t
1 + t3
1 + 1 + t3
3
1
1 − (1 + t )
√
= 2 √
t ( 1 + t3 )(1 + 1 + t3 )
t
√
√
=
.
( 1 + t3 )(1 + 1 + t3 )
Hence
lim
t→0
1
1
√
− 2
2
3
t
t 1+t
= lim
t→0
t
√
√
3
( 1 + t )(1 + 1 + t3 )
(e) Compute
lim
x→0
(x − 6)2 − 36
x
Solution: Expanding the numerator
(x − 6)2 − 36
(x2 − 12x + 36 − 36
=
= x − 12.
x
x
Hence
lim
x→0
(x − 6)2 − 36
= lim (x − 12) = −12.
x→0
x
(f) Compute
lim
x→0
|x|
x
Solution: For any x > 0, we have
|x|
x
= = 1.
x
x
For any x < 0, we have
|x|
−x
=
= −1.
x
x
Hence the limit does not exist.
= 0.
CALCULUS I HOMEWORK 2
3
2. Find the limit, if it exists. If the limit does not exist, explain why.
(a) Compute
x
lim √
x→∞
x2 + x + 1
Solution: We will factor out x from the denominator. For any x > 0, then
x
x
x
√
=q
= q
x2 + x + 1
x2 (1 + 1 ) + 1
x( 1 + 1 + 1 )
x
Hence
lim √
x→∞
.
x
x
1
= lim q
x2 + x + 1 x→∞ 1 + 1 +
x
x
=1
1
x
(b) Compute
p
lim ( x2 + 4x − x)
x→∞
Solution: First, multiply through by the conjugate.
√
p
p
(x2 + 4x) − x2
x2 + 4x + x
2
2
= √
.
x + 4x − x = ( x + 4x − x) √
x2 + 4x + x
x2 + 4x + x
Therefore, for all x > 0 the previous expression is equal to
4x
4x
√
.
= q
x2 + 4x + x
x( 1 + 4 1 + 1)
x
Hence
lim (
x→∞
p
x2 + 4x − x) = lim
x→∞
4
4
q
=
= 2.
1+1
( 1 + 4 x1 + 1
3. The Heaviside function H(x) is defined as
1
if x ≥ 0
H(x) =
0
otherwise
Find the following limits or explain why the limit does not exist.
(a) Compute
lim H(x)
x→0
Solution: If x > 0, we have
H(x) = 1,
while if x < 0, we have
H(x) = 0.
It follows that the limit does not exist.
4
CALCULUS I HOMEWORK 2
(b) Compute
lim x H(x)
x→0
Solution: We will use the squeeze theorem. Since |H(x)| ≤ 1 for all x, we
have
lim |xH(x)| ≤ lim |x| = 0.
x→0
x→0
Hence
lim x H(x) = 0.
x→0
Note: although I chose to solve this problem with the squeeze theorem, you
may also solve this problem by using left and right hand limits.
(c) Compute
lim
x→∞
1
H(x)
x
Solution: For any x > 0, we have H(x) = 1. Therefore
1
1
lim H(x) = lim
= 0.
x→∞ x
x→∞ x
(d) Compute
1
H(x − 1)
x→0 x
lim
Solution: For any x < 12 , we have H(x − 1) = 0. Therefore, for any x 6= 0
such that x < 12 ,
1
1
H(x − 1) = 0 = 0.
x
x
It follows that
1
lim H(x − 1) = 0.
x→0 x
(e) Compute
lim e−x sin x H(x + π)
x→−π
Solution: We use the squeeze theorem.
|e−x sin x H(x + π)| ≤ e−x | sin x| · 1.
Hence
lim |e−x sin x H(x + π)| ≤ lim e−x | sin x| = 0.
x→−π
x→−π
Therefore
lim e−x sin x H(x + π) = 0.
x→−π
Again, this problem may also be solved using left and right hand limits.
CALCULUS I HOMEWORK 2
5
(f) Compute
1
lim H(sin x) −
x→∞
2
Solution: For any integer n, if 2πn ≤ x ≤ 2πn + π, then
1
1
1
=1− = .
H(sin x) −
2
2
2
For any integer n, if 2πn + π < x < 2πn + 2π, then
1
1
1
H(sin x) −
=0− =− .
2
2
2
In other words, the function keeps oscillating between
that the limit does not exist.
1
2
and − 21 . It follows
(g) Compute
lim e
x→∞
−x
1
H(sin x) −
2
Solution: We use the squeeze theorem.
−x
1 1
e
H(sin x) −
≤ e−x · .
2
2
Therefore
−x
1 e−x
lim e
H(sin x) −
≤
lim
= 0.
x→∞
2 x→∞ 2
Hence
1
−x
lim e
H(sin x) −
= 0.
x→∞
2