Math 1A: Homework 3 Solutions
July 9
1. Evaluate the following limits.
√
(a) limx→2
x2 +12−4
.
x−2
We have
√
lim
x→2
=
=
=
=
=
=
(b) limt→1
√
x2 + 12 − 4
x2 + 12 + 4
×√
x−2
x2 + 12 + 4
(x2 + 12) − 42
√
lim
x→2 (x − 2)( x2 + 12 + 4)
x2 − 4
√
lim
x→2 (x − 2)( x2 + 12 + 4)
(x − 2)(x + 2)
√
lim
x→2 (x − 2)( x2 + 12 + 4)
x+2
lim √
2
x→2
x + 12 + 4
2+2
√
4 + 12 + 4
1
4
= .
8
2
x2 + 12 − 4
= lim
x→2
x−2
√
t3 −1
.
t4 −1
We have
t3 − 1
(t − 1)(t2 + t + 1)
=
lim
t→1 (t − 1)(t3 + t2 + t + 1)
t→1 t4 − 1
t2 + t + 1
= lim 3
t→1 t + t2 + t + 1
1+1+1
3
=
= .
1+1+1+1
4
lim
(c) limx→−8 (x + 7)2015 .
We have
lim (x + 7)2015 = (−8 + 7)2015
x→−8
= (−1)2015
= −1.
1
(d) limx→0
6x3 −7x
.
9x3 +2x2 −6x
We have
x(6x2 − 7)
6x3 − 7x
=
lim
x→0 x(9x2 + 2x − 6)
x→0 9x3 + 2x2 − 6x
6x2 − 7)
= lim 2
x→0 9x + 2x − 6
0 − 7)
=
0+0−6
7
=
.
6
lim
(e) limx→−∞
sin(x)
.
x2
Observe that
−1 ≤ sin(x) ≤ 1
sin(x)
1
1
≤ 2
(x 6= 0)
⇒ − 2 ≤
2
x
x
x
1
sin(x)
1
⇒ lim − 2 ≤ lim
≤ lim 2
2
x→−∞
x→−∞
x→−∞ x
x
x
sin(x)
⇒ 0 ≤ lim
≤ 0.
x→−∞
x2
We conclude that limx→−∞
(f)
sin(x)
x2
= 0.
x
limx→∞ exe+10 .
We have
(ex )/ex
x→∞ (ex + 10)/ex
1
= lim
x→∞ 1 + 10e−x
1
=
= 1.
1+0
ex
=
x→∞ ex + 10
lim
(g) limx→∞
lim
9x4 −13x3 +8x2 +2x−8
.
4x4 +25x2 −6x
We have
9x4 − 13x3 + 8x2 + 2x − 8
=
x→∞
4x4 + 25x2 − 6x
(9x4 − 13x3 + 8x2 + 2x − 8)/x4
x→∞
(4x4 + 25x2 − 6x)/x4
9 − 13/x + 8/x2 + 2/x3 − 8/x4
= lim
x→∞
4 + 25/x2 − 6/x3
9
=
.
4
lim
2
lim
2. Let f (x) =
√
x. Use the definition of the derivative to show that f 0 (x) =
1
√
.
2 x
We have for x > 0
f (x + h) − f (x)
h→0
h
√
√
x+h− x
lim
h→0
h
√
√
√
√
x+h− x
x+h+ x
lim
×√
√
h→0
h
x+h+ x
(x + h) − x
lim √
√
h→0 h( x + h +
x)
h
lim √
√
h→0 h( x + h +
x)
1
lim √
√
h→0
x+h+ x
1
√
√
x+ x
1
√ .
2 x
f 0 (x) = lim
=
=
=
=
=
=
=
3. Using the definition of infinite limits, prove the following.
(a) limx→3−
2x
x−3
= −∞.
Let M < 0 be given. Choose δ =
have
6
2−M
6
−
x−3
6
2+
3−x
6 − 6 − 2x
3−x
2x
x−3
−
6
.
2−M
6
Suppose − 2−M
< x − 3 < 0. Then, we
< x−3
> 2−M
((2 − M ) > 0, (x − 3) < 0 so the inequality flips)
< M
< M
< M.
Since we can choose a δ > 0 for any M < 0 such that
2x
−δ < x − 3 < 0, we conclude that limx→3− x−3
= −∞.
(b) limx→0
x4 +1
x2
= ∞.
3
2x
x−3
< M whenever
Let M > 0 be given. Choose δ =
√1 .
M
Suppose 0 < |x − 0| < δ. We then have
1
|x| < √
M
1
x2 <
M
1
> M
x2
1
x2 + 2 > M
x
4
x +1
> M.
x2
Since we can choose a δ > 0 for any M > 0 such that
4
0 < |x − 0| < δ, we conclude that limx→0 x x+1
= ∞.
2
x4 +1
x2
> M whenever
4. (a) Show that the equation x3 − 1 = 15x has at least three solutions in (−4, 4).
Let f (x) = x3 − 1 − 15x. Note that f is continuous on R since it is a polynomial.
Observe that:
f (−4)
f (−3)
f (0)
f (4)
=
=
=
=
−64 − 1 + 60 = −5
−27 − 1 + 45 = 17
0 − 1 + 0 = −1
64 − 1 − 60 = 3.
By the Intermediate Value Theorem, it follows that f (x) = 0 has solutions in
(−4, −3), (−3, 0) and (0, 4). Thus, x3 − 1 = 15x has at least three solutions in
(−4, 4).
(b) If f (x) = x3 − 12x2 + 4 cos(x), show that there exist α, β and γ such that f (α) =
−1729, f (β) = π 3 and f (γ) = eγ .
Note first that since 3 < π < 4, we have 27 < π 3 < 64. Next, note that f is a
continuous function on R since it is the sum of a polynomial and a cosine. Observe
that:
f (−10) =
=
f (0) =
f (100) =
−1000 − 12(100) + 4 cos(20)
−2200 + 4 cos(−10) < −2195 (∵ 4 cos(−10) ≤ 4 < 5)
4 cos(0) = 4
106 − 12(104 ) + 4 cos(10) = 88(104 ) + 4 cos(10) > 80(104 ).
Note then that
f (−10) < −2195 < −1729 < f (0) < 27 < π 3 < 64 < 80(10)4 < f (100).
By the Intermediate Value Theorem, it follows that there exist α in (−10, 0) and
β in (0, 100) such that f (α) = −1729 and f (β) = π 3 .
4
Finally, define g(x) = f (x) − ex = x3 − 12x2 + 4 cos(x) − ex . Observe that g is
again continuous on R and that
g(−10) = −2200 + 4 cos(−10) − e−10 < −2194 (∵ e−10 < 1)
g(0) = 4 − e0 = 3.
Since g(−10) < 0 < g(0), it follows from the Intermediate Value Theorem that
there exists γ in (−10, 0) such that g(γ) = 0 ⇒ f (γ) = eγ .
5. Let f : [0, 1] → [0, 1] be continuous. Prove that there exists some c in [0, 1] such that
f (c) = c.
Let g(x) = f (x) − x. We need to show that g(c) = 0 for some c in [0, 1]. Since g is the
difference of two continuous functions, it is also continuous.
Note that g(0) = f (0) − 0 ≥ 0 since f (x) ≥ 0. Also note that g(1) = f (1) − 1 ≤ 0
since f (x) ≤ 1. Observe that if either g(0) or g(1) is equal to zero, we can take c to
be 0 or 1 accordingly. If g(0), g(1) are both non-zero, it follows that g(1) < 0 < g(0).
By the Intermediate Value Theorem, there exists some c in (0, 1) such that g(c) = 0.
6. (a) Using limx→0 sin(x)
= 1, show that limx→0
x
2
1 − 2 sin (x/2)).
1−cos(x)
x
= 0. (Hint: use cos(x) =
We have
1 − cos(x)
1 − (1 − 2 sin2 (x/2))
= lim
x→0
x→0
x
x
2 sin2 (x/2)
= lim
x→0
x
sin(x/2)
= lim
(sin(x/2))
x→0
x/2
i
sin(x/2) h
= lim
lim (sin(x/2))
x→0
x→0
x/2
= [1] [0] = 0.
lim
(b) Let g(x) = sin x. Use the definition of the derivative to show that g 0 (x) = cos(x)
We have
g(x + h) − g(x)
h→0
h
sin(x + h) − sin(x)
lim
h→0
h
sin(x) cos(h) + sin(h) cos(x) − sin(x)
lim
h→0
h
cos(h) − 1
sin(h)
sin(x) lim
+ cos(x) lim
h→0
h→0
h
h
sin(x) [0] + cos(x) [1]
(∵ (a))
cos(x).
g 0 (x) = lim
=
=
=
=
=
5
7. Find the equation of the tangent line to the curve of each of the following functions at
the given points.
(a) f (x) =
x3/2 +x2
cos(x)
at (0, 0).
We have
1/2
cos(x)( 3x2
+ 2x) − (x3/2 + x2 )(− sin(x))
cos2 (x)
cos(0)(0) − (0)(− sin(0))
f 0 (0) =
= 0.
cos2 (0)
0
f (x) =
Thus, the equation of the tangent line through (0, 0) is
y − 0 = 0(x − 0) ⇒ y = 0.
(b) f (x) = sin3 (x) + cos(x) at (π, −1).
We have
f 0 (x) = 3 sin2 (x) cos(x) − sin(x)
f 0 (π) = 3 sin2 (π) cos(π) − sin(π) = 0.
Thus, the equation of the tangent line through (π, −1) is
y − (−1) = 0(x − π) ⇒ y = −1.
(c) f (x) = e4x sin(x) at (−π, 0). We have
f 0 (x) = 4e4x sin(x) + e4x cos(x)
f 0 (−π) = 4e−4π sin(−π) + e−4π cos(−π)
= 0 + e−4π (1) = e−4π .
Thus, the equation of the tangent line through (−π, 0) is
y − (0) = e−4π (x − (−π)) ⇒ y = e−4π (x + π).
(d) f (x) = x2 e−2x −
We have
cos2 (x)
x2 +1
at (0, −1).
(x2 + 1)(2 cos(x))(− sin(x)) − (cos2 (x))(2x)
f (x) = x (−2e ) + (2x)e
−
(x2 + 1)2
(0 + 1)(2 cos(0))(− sin(0)) − (cos2 (0))(0)
0
0
0
f (0) = (0)(−2e ) + (0)e −
= 0.
(0 + 1)2
0
2
−2x
−2x
Thus, the equation of the tangent line through (0, −1) is
y − (−1) = 0(x − 0) ⇒ y = −1.
6