MATH 101 HOMEWORK 2 – SOLUTIONS
1. Evaluate the limits:
n + 2n2 − n3
n−2 + 2n−1 − 1
1
=− .
=
lim
3
−3
n→∞
n→∞
1 + 2n
n +2
2
(a) lim
1 − cos(2/n)
1 − cos(2x)
2 sin(2x)
4 cos(2x)
4
= lim
= lim
= lim
=
n→∞ 1 − cos(3/n)
x→0 1 − cos(3x)
x→0 3 sin(3x)
x→0 9 cos(3x)
9
(b) lim
(we used l’Hopital’s rule twice.)
2. Evaluate the sums:
(a)
92 + 122 + 152 + 182 + . . . + (3n − 3)2 + (3n)2 =
n
X
(3j)2 = 9
n
X
j=3
j2
j=3
n
X
n(n + 1)(2n + 1) .
= 9 − 12 − 22 +
j2 = 9 − 5 +
6
j=1
(b)
n X
i
X
i=1
=
i
n n
n
X
X
X
2i − 1 X i+1 i
2i+j =
2i−1 =
2i+1 ·
2 (2 − 1)
2i+1
=
2
−
1
j=1
i=1
j=1
i=1
i=1
n
X
i=1
22i+1 −
n
X
2i+1 = 8
i=1
n
X
4i−1 − 4
i=1
=
n
X
2i−1 = 8 ·
i=1
2n − 1
4n − 1
−4·
4−1
2−1
8 n
(4 − 1) − 4(2n − 1).
3
R5
3. Write the upper and lower Riemann sums approximating 2 x−3 dx, corresponding to
the partition of [2, 5] into n intervals of equal length. (Do not try to evaluate the sums.)
To partition [2, 5] into n intervals of equal length, we take ∆xj = 3/n, so that the partition
points are
3
6
3n
2, 2 + , 2 + , . . . , 2 +
= 5,
n
n
n
i.e. xj = 2 + 3j/n. The function x−3 is decreasing, so its maximum value on [xj−1 , xj ]
is attained at xj−1 , and the minimum value – at xj . So the upper and lower sums are,
respectively,
n
n
X
X
3
3
3(j − 1) −3
3j −3
, L=
.
U=
2+
2+
n
n
n
n
j=1
j=1
1
4. Identify the limit lim
n→∞
n
X
3n − 2i
i=1
n2
as an area of a planar region, and use this to evaluate
it.
n
X
1
2i 3−
. The sum inside the limit can be identified
n→∞
n
n
i=1
Z 1
with the Riemann sums approximating
(3 − 2x)dx, with ∆xj = 1/n and xj = j/n,
We first rewrite it as lim
0
j = 0, 1, . . . , n. Thus the limit should be equal to the area of the trapezoid with vertices
(0, 0), (0, 3), (1, 1) and (1, 0). This area equals 2: to see this, use the formula for the area
of a trapezoid, or else observe that it is a union of a square of sidelength 1 and a right
triangle with leg lengths 1 and 2.
2