MAS111 Strand 1: Solutions to Week 4 Problems
Spring Semester 2014-2015
1. (a) Since −1 ≤ sin x1 ≤ 1, we have −x2 ≤ x2 sin x1 ≤ x2 for all x. Since
lim −x2 = lim x2 = 0,
x→0
x→0
it follows from the Sandwich Theorem that limx→0 x2 sin x1 = 0.
(b) We have limx→0 x2 sin x1 = limx→0 x limx→0 x sin x1 = 0.1 = 0.
2. (a) limx→0
sin(7x)
x
= limx→0
7 sin(7x)
7x
= 7 limx→0
sin(7x)
7x
= 7.1 = 7.
(b) We check that both numerator and denominator are 0 at x = 0, so
that we can apply l’Hopital’s rule: limx→0 sin(7x)
= limx→0 7 cos(7x)
= 7.
x
1
3. (a)
sin2 x
1 − cos x
=
x
x(1 + cos x)
⇔
x(1 + cos x)(1 − cos x)
= sin2 x
x
⇔
1 − cos2 x = sin2 x
⇔
cos2 x + sin2 x = 1.
So the required identity is equivalent to a standard one.
So
1 − cos x
sin2 x
= lim
x→0
x→0 x(1 + cos x)
x
sin x
sin x
= lim
lim
= 1.0 = 0.
x→0 x x→0 1 + cos x
lim
(b) We check that both numerator and denominator are 0 at x = 0, so
x
that we can apply l’Hopital’s rule: limx→0 1−cos
= limx→0 sin1 x = 0.
x
4. (a) We check that both numerator and denominator are 0 at x = 2. As
long as this continues to be the case, we keep differentiating top and
bottom.
x3 − 3x2 + 4
3x2 − 6x
6x − 6
6
3
=
lim
= lim
= = .
x→2 x3 − 2x2 − 4x + 8
x→2 3x2 − 4x − 4
x→2 6x − 4
8
4
lim
(b) We check that both numerator and denominator are 0 at x = 0. As
long as this continues to be the case, we keep differentiating top and
bottom.
lim
x→0
1 − cos x
sin x
cos x
1
= lim
= lim
= .
2
x→0
x→0
x
2x
2
2
1
(c) We check that both numerator and denominator are 0 at x = 0.
−ex
1 − ex
=
lim
= −1.
x→0 1 − 2x
x→0 x − x2
lim
ln x)
ln x
(d) This time, we write limx→0+ x ln x = limx→0+ 1/x
= limx→0+ −(−
=
1/x
ln x
, and we check that both numerator and denominator
− limx→0+ −1/x
go to infinity as x goes to zero. Then
lim x ln x = − lim
x→0+
x→0+
−1/x
− ln x
= − lim x = 0.
= − lim
1/x
x→0+
x→0+ −1/x2
Hence,
lim xx = lim+ ex ln x = e0 = 1.
x→0+
x→0
5. We check that both numerator and denominator go to infinity as x goes
to infinity. So we can apply l’Hopital’s Rule. However if we differentiate
2
1/2
top and bottom, in the first step we get (x +1)
and in the second step
x
we get back to where we started.
On the other hand, we have
lim
x→∞
x
1
1
= lim
=
= 1.
x→∞ (1 + 1/x2 )1/2
(x2 + 1)1/2
(1 + 0)1/2
6. We consider the three cases k < 0, k = 0 and k > 0 separately.
Firstly, if k < 0, then xk → 0 and ex → ∞ as x → ∞, so xk /ex → 0.
Secondly, if k = 0, then xk = 1 and xk /ex = 1/ex → 0 as x → ∞.
Thirdly, if k is a positive integer, then both xk → ∞ and ex → ∞ as
x → ∞, so we may apply l’Hopital’s Rule:
xk
kxk−1
= lim
.
x
x→∞ e
x→∞
ex
lim
Continuing to apply l’Hopital’s Rule, we find
xk
kxk−1
k(k − 1)xk−2
k!
=
lim
=
lim
= · · · = lim x = 0.
x→∞ ex
x→∞
x→∞
x→∞ e
ex
ex
lim
Finally, if k > 0, not necessarily an integer, then 0 < xk /ex < xdke /ex ,
where dke denotes the “ceiling of k”, the smallest integer not less than
k. Then the Sandwich Theorem gives the desired result, since we have
already seen that limx→0 xdke /ex = 0.
2
7.
The shaded area is
Z 1+ x1
1
1+ x1
1
1
dt = ln t 1
= ln 1 +
.
t
x
Using the smaller and larger rectangles, we see that
1
1
1
1
≤
ln
1
+
≤
(1).
x 1 + x1
x
x
Hence (since x > 0),
1
1+
1
x
1
≤ x ln 1 +
x
≤ 1.
1
= 1. So, by the Sandwich Theorem,
As x → ∞, x1 → 0, so 1+1 1 → 1+0
x
x
1
x ln 1 + x → 1 as x → ∞. Then taking exponentials, limx→∞ 1 + x1 =
e.
3