18.014 Problem Set 10 Solutions
Sam Elder
November 18, 2015
Problem 1 (20 points). Compute the following limits.
1 − cos x
.
x→0
x2
Part 1.1. lim
Solution 1. The second Taylor polynomial for cos x gives us
cos x = 1 −
x2
+ o(x2 ).
2
Therefore,
1 − (1 − x2 /2 + o(x2 ))
1 − cos x
lim
=
lim
= lim
x→0
x→0
x→0
x2
x2
1 o(x2 )
+
2
x2
=
1
2
.
Solution 2. Since 1 − cos 0 = 0 and 02 = 0, we can apply L’Hôpital’s rule:
1 − cos x
sin x
= lim
.
2
x→0
x→0 2x
x
lim
Since sin 0 = 2 · 0 = 0, we can apply L’Hôpital’s rule again:
lim
x→0
Part 1.2. lim
x→0
sin x
cos x
cos 0
1
= lim
=
= .
x→0
2x
2
2
2
log(1 + x)
.
e2x − 1
Solution. Since log(1 + 0) = log 1 = 0 and e2·0 − 1 = e0 − 1 = 0, we can apply L’Hôpital’s rule:
lim
x→0
Part 1.3. lim
x→0
(1 + x)1/x
e
log(1 + x)
1/(1 + x)
1/(1 + 0)
1
= lim
=
=
.
x→0
e2x − 1
2e2x
2e2·0
2
1/x
.
Solution. We write this as an exponential of a fraction:
(1 + x)1/x
e
1/x
= exp log
(1 + x)1/x
e
1/x !
= exp
(1/x) log(1 + x) − 1
x
.
Since exp is continuous,
lim
x→0
(1 + x)1/x
e
1/x
(1/x) log(1 + x) − 1
= exp lim
x→0
x
1
log(1 + x) − x
= exp lim
x→0
x2
.
Since log(1 + 0) − 0 = 02 = 0, we can apply L’Hôpital’s rule:
lim
x→0
1/(1 + x) − 1
log(1 + x) − x
1 − (1 + x)
1
1
= lim
= lim
= lim −
=− .
2
x→0
x→0
x→0
x
2x
2x(x + 1)
2(x + 1)
2
Therefore, the original limit is
lim
x→0
Part 1.4. lim
x→0
(1 + x)1/x
e
1/x
1
= exp −
= e−1/2 .
2
log(cos(2x))
.
log(cos(3x))
Solution. Since log(cos(0)) = log(1) = 0, we can apply L’Hôpital’s rule:
−2 sin(2x)/ cos(2x)
2 tan(2x)
log(cos(2x))
= lim
= lim
.
log(cos(3x)) x→0 −3 sin(3x)/ cos(3x) x→0 3 tan(3x)
lim
x→0
Now since tan(0) = 0, we can apply L’Hôpital’s rule again:
2 tan(2x)
4 sec2 (2x)
4 sec2 (2 · 0)
4
= lim
=
=
.
x→0 3 tan(3x)
x→0 9 sec2 (9x)
9 sec2 (3 · 0)
9
lim
Problem 2. Let f (x) = tan−1 (x). Show that for any nonnegative integer n, the (2n+1)th Taylor polynomial
of f at 0 is
n
X
(−1)k 2k+1
x3
x5
x2n+1
T2n+1 (x) =
x
=x−
+
− · · · + (−1)n
.
2k + 1
3
5
2n + 1
k=0
Solution. We first investigate its derivative.
Lemma. For any nonnegative integer n, the (2n)th Taylor polynomial of f 0 (x) =
0
T2n
(x) =
n
X
1
at 0 is
1 + x2
(−1)k x2k .
k=0
Proof. We compute that
n
X
1
1 − (1 + x2 ) + (x2 + x4 ) − (x4 + x6 ) + · · · + (−1)n (x2n − x2n+2 )
(−1)n+1 x2n+2
−
(−1)k x2k =
=
2
2
1+x
1+x
1 + x2
k=0
n+1
2
x
→ 0 as x → 0, so this expression is o(x2n ). Therefore,
by telescoping. If we divide by x2n , we get (−1)
1+x2
this must be the Taylor polynomial, as desired.
Now the Taylor polynomial of an integral must match the integral of a Taylor polynomial, so
T2n+1 (x) = tan−1 (0) +
Z
0
x
0
T2n
(t) dt = 0 +
Z
0
n
xX
(−1)k t2k dt =
k=0
n
X
(−1)k
x
Z
t2k dt =
0
k=0
n
X
(−1)k 2k+1
x
,
2k + 1
k=0
as desired.
1
x→0 bx − sin x
Z
Problem 3 (7.13-15). Find constants a and b such that lim
Answer. (a, b) = (4, 1).
2
0
x
√
t2
dt = 1.
a+t
R x 2 dt
Solution. We notice that as x → 0, 0 √t a+t
→ 0, so we can apply L’Hôpital’s rule using the fundamental
theorem of calculus:
Z x
1
t2
1
1
x2
x2
√
√
lim
dt = lim
= √ lim
,
x→0 bx − sin x 0
x→0 b − cos x
a x→0 b − cos x
a+x
a+t
√
√
because a + x is continuous at x = 0, with value a.
If b 6= 1, then b − cos x → b − 1 6= 0, so the limit is 02 = 0. Since we want it to be 1, we must take b = 1.
There are three approaches to finish computing the limit.
The first uses o-notation: 1 − cos x = 1 − (1 − x2 /2 + o(x2 )) = x2 /2 + o(x2 ), so the limit is
x2
2
1
2
1
√ lim 2
= √ lim
=√ .
2
x→0
x→0
x /2 + o(x )
1 + o(1)
a
a
a
The second uses L’Hôpital’s rule twice:
x2
2x
2
2
= lim
= lim
=
= 2,
x→0 1 − cos x
x→0 sin x
x→0 cos x
cos 0
lim
where we check that 02 = 1 − cos 0 = 2 · 0 = sin 0 = 0, so L’Hôpital’s rule is valid.
The final approach is to notice that this is the reciprocal of the limit in the first part of the first problem
1
x2
1 − cos x
= , we have lim
= 2.
of this problem set. Since we showed there that lim
2
x→0 1 − cos x
x→0
x
2
2
In all approaches, the limit is √ . For this to equal 1, we need a = 4, as desired.
a
Problem 4. Suppose that f is defined in some interval containing 0 and satisfies f (x) = 1 + x + o(x) as
x → 0. Prove that
lim f (x)1/x = e.
x→0
Solution. We write
f (x)
1/x
= exp(log(f (x)
1/x
)) = exp
log f (x)
x
.
Since exp is continuous, we have
lim f (x)
1/x
x→0
log f (x)
= exp lim
x→0
x
.
We will need the following lemma about logarithms:
Lemma. As y → 0, log(1 + y) = y + o(y).
1
1
00
, and f 00 (y) = − (1+y)
Proof. If we let f (y) = log(1 + y), then f 0 (y) = 1+y
2 . Therefore, f (y) is continuous
on the interval (−1, 1) containing 0, so by Taylor’s theorem,
f (y) = f (0) + f 0 (0)y + o(y) = log(1 + 0) +
1
y + o(y) = y + o(y),
1+0
as desired.
Now since f (x) = 1 + x + o(x), by the properties of o(x) notation,
lim
x→0
log f (x)
log(1 + x + o(x))
x + o(x) + o(x + o(x))
= lim
= lim
x→0
x→0
x
x
x
x + o(x) + o(x) + o(x)
o(x)
= lim
= lim 1 +
=1
x→0
x→0
x
x
lim f (x)1/x = exp(1) = e,
x→0
as desired.
3