18 014 Problem Set 4 Solutions
Total: 24 points
Problem : Establish the following limit formulas. You may assume the formula
limx!0 sinx x) = 1.
(a)
sin(5x)
lim
= 5.
x!0 sin(x)
(b)
lim
sin(5x)
x
x!0
(
Exam
Problem 1 Evaluate limx!0
1
x
1
log(x+1)
Solutions
.
Solution We begin by writing the problem as a single fraction, limx!0 log(x+1) x .
x log(x+1)
Observe that both numerator and denominator have
Exam
Solutions
October 29, 2010
Total: 60 points
Problem : Find the derivative of each of the following functions
g(x) = log(cos(x2 )
p
h(x) = e
x sin x
Solution For the rst problem, we invoke the c
G.l
Rational exponents - an application of the intermediate-value
theorem.
It is a consequence of the intersediate~va1ue theorem, that;
given a positive integer n and a real nulber a > 0,
there is
exa
10.1 The Geometric Series
CHAPTER 10
10.1
(page 373)
INFINITE SERIES
The Geometric Series
The advice in the text is: Learn the geometric series. This is the most important series and also the
simplest
\:;ii/>Integrabilitx g: bounded Eiecewisemonotonic functions.
The definition of "piecewise-monotonic" is given on p.
77 of the text.
Lemma. If f is bounded on [a,b] and monotonic on
(a,b), then f is
Continuity oi the sguare root function.
The following theorem shows that the square-root
function is continuous for x Z 0. We will give a different
proof,based on the intermediate-value theorem, sho
The Riemann condition.
The most useful criterion for determining whether f is
integrable on [a,b] is given in the following theorem. It is
called the Riemann condition for existence of the integral.
T
Drl
Progertiesgg integrals
In this section, we prove the four basic properties of the integral
that we shall need.
Theorem. (Progerties g: the integral)
(l) (Linearitx Erogertx.) g; f and g are integr
11.1 Vectors and Dot Products
(page 405)
CHAPTER 11
VECTORS AND MATRICES
Vectors and Dot Products
11.1
(page 405)
One side of the brain thinks of a vector as a pair of numbers (x, y) or a triple of nu
18 014 Problem Set 2 Solutions
Total: 24 points
P
Problem : Let f (x) = n=0 ck xk be a polynomial of degree n.
k
(d) If f (x) = 0 for n + 1 distinct real values of x, then every coe cient ck of f is
z
18 014 Problem Set 3 Solutions
Total: 12 points
Problem : Find all values of c for which
Rc
(a) R0 x(1 x)dx = 0.
c
(b) 0 |x(1 x)|dx = 0.
Solution (4 points)
(a) Computing, we get
Z
0
c
x(1
x) =
Z
c
(x
18 014 Problem Set 1 Solutions
Total: 24 points
Problem : If ab = 0, then a = 0 or b = 0.
Solution (4 points)
Suppose ab = 0 and b = 0. By axiom 6, there exists a real number y such that
6
by = 1. Hen
Problem : Find
integer x.)
R3
2
2x2 [ |x| ] dx. (Here, as usual, [x] denotes the largest
Solution Note that
Hence,
Z
3
8 2
> 2x for
>
<
0 for
2
2x b|x|c =
> 2x2 for
>
:
4x2 for
2
2
2x b|x|cdx =
Z
1
2