MATH 101 HOMEWORK 2 – SOLUTIONS
1. Evaluate the following limits:
n5 − 7n
1 − 7n−4
1
=
lim
=
−
.
n→∞ n2 + n3 − 3n5
n→∞ n−3 + n−2 − 3
3
(a) lim
(b) lim sin(nπ) = lim 0 = 0, where we used that sin(nπ) = 0 if n is an integer.
n→∞
n→∞
In this problem, n was meant to denote integers – however, this was not stated explicitly.
Some students may have understood n to denote all real numbers, and consequently said
that the limit does not exist. Such solutions will also be considered correct.
2. Evaluate:
n
X
(a)
32n−i == 32n−1 + 32n−2 + . . . + 3n+1 + 3n = 3n (3n−1 + 3n−2 + . . . + 3 + 1)
i=1
=3
n
n
X
3i−1 = 3n ·
i=1
3n (3n − 1)
3n − 1
=
.
3−1
2
i
n X
n X
i
i
n
n
X
X
X
X
X
i(i + 1)
3
1
2
(b)
(i + j) =
(
i+
j) =
(i +
( i2 + i)
)=
2
2
2
i=1 j=1
i=1 j=1
j=1
i=1
i=1
n(n + 1)2
3 n(n + 1)(2n + 1) 1 n(n + 1)
3X 2 1X
+
=
.
i +
i=
=
2 i=1
2 i=1
2
6
2
2
2
n
n
R3 2
3. Write the upper and lower Riemann sums approximating 1 ex dx, corresponding to the
partition of [1, 3] into n intervals of equal length. Do not attempt to evaluate these sums!
2
We have xi = 1 + (2i/n), ∆xi = 2/n. The function ex is increasing, hence li = xi−1 and
ui = xi . Therefore
n
n
X
X
2 (1+ 2(i−1) )2
2 (1+ 2i )2
n
n
L=
e
e
, U=
.
n
n
i=1
i=1
4. Prove that the area of the region enclosed by the lines y = 0, x = 1, x = 3, and the graph
2
of y = ex is at least e(1 + e3 ). In this problem, you are not allowed to use a calculator.
The area in question is bounded from below by the lower Riemann sum with n = 2,
x0 = 1, x1 = 2, x2 = 3:
L = e · 1 + e4 · 1 = e(1 + e3 )
as required.
1