A REVIEW OF CALCULUS and AN OVERVIEW OF
ANALYSIS
Analysis is the study of the limit concept and its ramifications. These include the
following four important terms that they are defined by means of a limit –the
continuity of a function, the derivative of a function, the definite integral of a
function defined on a closed interval, and the convergence of infinite sequences and
series.
We now list several main theorems of calculus. Most will be familiar, at lieat
by name, to the student who has finished a course in calculus.
Result 1 (Differentiability implies continuity)
If f (c ) exists, then f (x) is continuous at c . The converse of this result is not
true─consider f ( x)  x at c  0.
Result 2 ( Continuity implies integrability)
If f (x) is continuous on a, b then
b
 f ( x)dx
exists.
a
Result 3 ( Mean Value Theorem for derivatives)
If f (x) exist on a, b , then there is a value c in (a, b) so that
f (c) 
f (b)  f (a )
.
ba
Result 3 ( Mean Value Theorem for integrals)
See the text book
Some Problems From Calculus
The following examples are intended to demonstrate the range of topics in calculus
and to provide motivation to learn the theory underlaying the techniques used in
these solutions.
Example 1 Use mathematical induction to prove that 2 n  n! for all n  4.
Solution. Exercise
Example 2 Find lim x0 x sin
Solution. lim x0 x sin
1
1
and lim x  x sin .
x
x
1
1
 0 because for x  0, sin  1, and so
x
x
0

x sin
1
0
x

x sin
1
x
=
x sin

x.
1
x
We then appeal to the Squeeze Theorem for limits (see Theorem 10.4), which
asserts that as x approaches 0, the quantity x sin
1
must approach 0 because it is
x
“squeezed” between 0 and x , both of which also approach 0. We observe that the
Limit Theorem (Theorem 10.2) cannot be used here since lim x 0 sin
1
does not exist
x
because of oscillation.
1
 4
 x sin
Example 3 Find f (0) if f ( x)  
x
 0
for x  0
for x  0
Solution The usual approach of a calculus studente to this type of problem is to
calculate f (x) and then set x  0 . While this approach is often succesful, we show
that it will not work for this problem. By use of the Product rule, Chain rule, and
the derivative folrmulas for sin x and cos x , we find that for x  0 ,
1
1
 x 2 cos
x
x
f (x)
=
4 x 3 sin
f (x)
=
12 x 2 sin
1
1
1
 6 x cos  sin .
x
x
x
While we are able to get f (0)  lim x0 f ( x)  0 by the Squeeze Theorem approach
used in Exaple 2, that method does not apply to f (0) because lim x 0 sin
1
does not
x
exist, and so lim x0 f ( x) does not exist either. This proves that f  is not
continuous at 0, but this does give information as to whether or not f (0) exists.
We now apply the derivative definition to find
f ( x)  f (0)
1
1

 lim  4 x 2 sin  x cos   0.
x

0
x0
x
x

f (0)  lim
x 0
Example 4 Evaluate
1
x
0
p
dx, where p  R.
Solution In order to evaluate this integral, we need to distinguish two possibilities.
If p  0, then
1
x
p
0
dx, is a proper integral because x p is continuous on [0,1]. Since
we know an antiderivative, we use the Fundamental Theorem to write
1
x p 1
1
x
dx


0
p 1 0 p 1
1
p
For values of p  0 , x p has an infinite discontinuity at 0, so
1
x
0
p
dx is an improper
integral. We evaluate it by considering three subcases.
Case 1:  1  p  0
1
x
dx =
0
=
and so the integral converges to
p
 1
t p 1 
1

 
lim

t 0   p  1
p  1 p  1

1
.
p 1
Case 2: p  1
1
1
p
 x dx  lim ln x ln x  lim (0  ln t )  
0
t 0
t
t 0
and so the integral diverges to  in this case.
Case 3: p  1
 1
t p 1 
p


=
x
dx
lim

0
t 0   p  1
p  1 

1
1
x p 1
lim  x dx  lim
t 0
t 0 p  1
t
t
1
p
since p  1  0, and so the integral again diverges to  .
THEOREM (Mean Value Theorem)
If f is continuous on [a,b] and differentiable on (a,b), then there is at least one
value c  (a, b) such that
f (c) 
f (b)  f (a )
.
ba
Example 5 Use the Mean Value Theorem to find upper and lower bounds for 67 .
Solution By applying the Mean Value Theorem to f ( x)  x on the interval
[64,67], we have
67  64
1
for some value c for which 64 < c < 67 <

67  64
2 c
81.This leads to the following string of inequalities.
1
18
1
18
1
8
6
<
<
1
2 c
67  8
3
<
<
<
67
<
8.16666 <
67
<
1
16
1
16
3
8
16
8.18785
