18.014 Exam 3 Solutions
Sam Elder
November 25, 2015
Problem 1. Find real numbers a, b, c such that
2
1 + cx
lim
−
= 3.
x→0 x
a sin x + b cos x
Answer. We have a = 12 , b = 0, and c = − 23 .
Solution. We start by writing this difference with a common denominator:
2
1 + cx
2a sin x + 2b cos x − x − cx2
−
=
.
x a sin x + b cos x
ax sin x + bx cos x
Both the numerator and denominator are continuous. At x = 0, they take on values of 2a sin 0 + 2b cos 0 −
0 − c · 02 = 2b and a · 0 sin 0 + b · 0 cos 0 = 0, respectively. Therefore, we must have b = 0 or the limit does
not converge.
With b = 0, we can compute
2a sin x − x − cx2
(2a − 1)x − cx2 + o(x2 )
(2a − 1) − cx + o(x)
=
=
.
ax sin x
ax2 + o(x3 )
ax + o(x2 )
Therefore, if a 6= 21 , the limit of this new numerator is nonzero, while the limit of the denominator is zero,
and therefore, the limit of the ratio is undefined. Taking a = 12 , we are left with
−c + o(1)
−cx + o(x)
=
→ −2c
x/2 + o(x2 )
1/2 + o(x)
as x → 0. Therefore, if this equals 3, c = − 32 .1
Problem 2.
Part 2.1. Compute the integral
Z
2
xex dx.
Solution. We make the substitution u = x2 , so du = 2x dx and
Z
xe
x2
1
dx =
2
Z
eu du =
1 u
1 2
e + C = eu + C .
2
2
Part 2.2. Compute the integral
Z
1
2
x3 ex dx.
0
1 This problem can also be solved by L’Hôpital’s Rule at each moment a factor of x is cancelled from both numerator and
denominator. The proof is otherwise identical.
1
Solution 1. We start with the substitution u = x2 , so du = 2x dx:
Z
Z 1
1 1 u
3 x2
ue du.
x e dx =
2 0
0
Now we apply integration by parts with v = eu :
Z
Z
1 1 u
1
1
1 1 u
1
1
ue du = (1 · e1 − 0 · e0 ) −
e du = e1 − (e1 − e0 ) =
.
2 0
2
2 0
2
2
2
2
2
Solution 2. We start by integrating by parts: v = ex , so dv = 2xex dx and u = 21 x2 , so du = x dx. Then
Z 1
Z 1
2
1 2 12
3 x2
2 02
x e = (1 e − 0 e ) −
xex dx.
2
0
0
Now we apply the result from the previous part:
Z 1
2
2
1 2
xex dx = (e1 − e0 ).
2
0
Putting it all together, we get
1
Z
2
x3 ex =
0
1
1
1
e − (e − 1) = .
2
2
2
Problem 3. True-false problems.
Part 3.1. If p(x) is a polynomial of degree 3, then there exist real numbers A1 , A2 , A3 such that
Z
A2
p(x) dx
= A1 x +
+ A3 log(x − 2) + C.
(x − 1)2 (x − 2)
x−1
Solution. This is false. One term is missing on the right side.
Suppose that ax3 is the leading term of p(x). Then p(x) − a(x − 1)2 (x − 2) is a polynomial with degree
at most 2. Therefore, by what we’ve learned about partial fractions,
p(x)
p(x) − a(x − 1)2 (x − 2)
b1
b2
b3
=
a
+
=a+
+
+
.
(x − 1)2 (x − 2)
(x − 1)2 (x − 2)
x − 1 (x − 1)2
x−2
Hence
Z
p(x)
b2
= ax + b1 log(x − 1) +
+ b3 log(x − 2) + C.
(x − 1)2 (x − 2)
x−1
The problem statement is missing the b1 log(x − 1) term. For an example of a possible cubic p(x) which
would produce a nonzero value for b1 , take p(x) = (x − 1)2 (x − 2) + (x − 1)(x − 2). Then
p(x)
1
=1+
.
2
(x − 1) (x − 2)
x−1
Part 3.2. If f : R → R is twice-differentiable and satisfies f (0) = 0 and f 00 (x) = −f (x) for all x ∈ R, then
there exists a ∈ R such that f (x) = a sin x for all x ∈ R.
Solution. This is true. First suppose f satisfies these criteria with f 0 (0) 6= 0. Then g(x) = ff0(x)
(0) satisfies
g(0) = 0, g 00 (x) = −g(x) for all x, and g 0 (0) = 1. Therefore, g(x) = sin x and f (x) = f 0 (0) sin x, as desired.
If f 0 (0) = 0, we repeat the argument from showing that sine is unique to show that f (x) = 0 identically.
Indeed, call h(x) = f (x)2 + f 0 (x)2 . Then
h0 (x) = 2f (x)f 0 (x) + 2f 0 (x)f 00 (x) = 2f 0 (x)(f (x) + f 00 (x)) = 0,
so h(x) is constant. But h(0) = f (0)2 + f 0 (0)2 = 02 + 02 = 0. Therefore, f (x) = f 0 (x) = 0 for all x ∈ R, as
desired.
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Part 3.3. If f (x) = o(x2 ) as x → 0, then f (x) = o(x sin x) as x → 0.
f (x)
Solution. This is true. If f (x) = o(x2 ) as x → 0, by definition, lim 2 = 0. Now we’ve established
x→0 x
sin x
many times in this class that lim
= 1. Since this is nonzero, we can divide these two limits and get
x→0 x
f (x)
0
lim
= = 0.
x→0 x sin x
1
Part 3.4.
xx
2 = +∞.
x→+∞ ex
lim
Solution. This is false. We rewrite xx = ex log x , so
xx
ex log x
=
= exp(x log x − x2 ).
2
ex
ex2
Now we note that x2 > x log x for all x, and therefore, the fraction in question is never greater than e0 = 1,
and hence cannot go to +∞. (In fact, the exponent goes to −∞, so the fraction will go to 0 as x → ∞, by
continuity of exp.)
Part 3.5. If f (x) is a polynomial of degree n, then the nth Taylor polynomial of f at the point 1 is equal
to f .
Solution. This is true. The nth Taylor polynomial of f is defined as the unique degree ≤ n polynomial g
such that f (k) (1) = g (k) (1) for k = 0, 1, . . . , n. Since f itself satisfies these criteria, we must have g = f and
the statement is true.
Part 3.6. Let f : R → R be the inverse function to the polynomial p(x) = 13 x3 + x. Then f is differentiable
and
1
f 0 (x) =
for all x ∈ R.
1 + f (x)2
Solution. This is true. Since p0 (x) = x2 + 1, the inverse function theorem states that
f 0 (x) =
1
p0 (f (x))
=
1
,
1 + f (x)2
as desired.
Problem 4. Suppose that f : R → R is infinitely differentiable and satisfies f (n) (0) = 1 for every integer
n ≥ 0, where f (n) is hte nth derivative of f . Prove that
f (x) − ex
=0
x→0
xn
lim
for any integer n ≥ 0.
Solution 1. Since f (x) and ex are both infinitely differentiable, their nth-degree Taylor expansions are both
f (0) + f 0 (0)x +
f 00 (0)x2
f (n) xn
x2
xn
+ ··· +
+ o(xn ) = 1 + x +
+ ··· +
+ o(xn ).
2!
n!
2!
n!
Therefore,
f (x) − ex
1
x2
xn
x2
xn
n
n
=
lim
1
+
x
+
+
·
·
·
+
+
o(x
)
−
1
+
x
+
+
·
·
·
+
+
o(x
)
x→0
x→0 xn
xn
2!
n!
2!
n!
o(xn )
= lim
= 0,
x→0 xn
lim
by the definition of o(xn ).
3
Solution 2. We apply L’Hôpital’s Rule n times. First we must check that this is valid. After applying it k
times, where 0 ≤ k < n, the fraction is
(n − k)! f (k) (x) − ex
,
n!
xn−k
which evaluates to f (k) (0) − e0 = 1 − 1 = 0 in the numerator and 0n−k = 0 in the denominator, as long as
k < n. So each time we can apply L’Hôpital’s Rule, and after the nth time, we are left with
f (k) − ex
1−1
=
= 0,
x→0
n!
n!
lim
as desired.
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