# L'Hôpital's Rule

The Cauchy Mean Value Theorem is the basic tool needed to prove a theorem which facilitates evaluation of limits of the form

when

Theorem. Suppose that

and and suppose also that

exists. Then

exists, and

PROOF. The hypothesis that exists contains two implicit assumptions:

(1) there is an interval such that and exist for all in except, perhaps, for ,

(2) in this interval with, once again, the possible exception of

On the other hand, and are not even assumed to be defined at . If we define (changing the previous values of and , if necessary), then and are continuous at . If , then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to and on the interval (and a similar statement holds for ). First applying the Mean Value Theorem to , we see that , for if there would be some in with , contradicting (2). Now applying the Cauchy Mean Value Theorem to and , we see that there is a number . in such that

or

Now approaches as approaches , because , is in ; since exists, it follows that

# Uniform Continuity

Definition. The function is uniformly continuous on an interval if for every there is some such that, for all and in

Lemma. Let and let be continuous on the interval Let , and suppose that statements (i) and (ii) hold:

(i) if and are in and , then ,

(ii) if and are in and , then

Then there is a 0 such that

Proof. Since is continuous at , there is a 0 such that

It follows that

(iii) if and then

Choose to be the minimum of and . We claim that this works. In fact, suppose that and are any two points in with . If and are both in then by (i); and if and are both in then by (ii). The only other possibility is that

or

In either case, since , we also have and So by (iii).

THEOREM If is continuous on , then is uniformly continuous on

PROOF: For let's say that is good on if there is some such that, for all and in if , then . Then we're trying to prove that is -good on for all

Consider any particular . Let

Then (since a is in ), and is bounded above (by ), so has a least upper bound . We really should write , since and might depend on . But we won't since we intend to prove that , no matter what is.

Suppose that we had . Since is continuous at , there is some such that, if , then . Consequently, if and , then So is surely -good on the interval . On the other hand, since is the least upper bound of , it is also clear that is -good on . Then the above Lemma implies that is -good on , so is in A, contradicting the fact that is an upper bound.

To complete the proof we just have to show that is actually in . The argument for this is practically the same: Since is continuous at , there is some such that, if , then So is -good on . But is also -good on , so the Lemma implies that is -good on .

# Logarithm

Definition. If , define

It follows from the Fundamental Theorem of Calculus that is differentiable and

For each fixed , define . Then . Therefore there is a constant so that

Setting , we see that

Therefore

Corollary. If is a natural number and , then

Corollary. If , then

# Exponential Function

Since for all , is an increasing function on . It is unbounded from above:

and it is unbounded from below:

Therefore the inverse function of exists. The exponential function is defined as . Thus is the unique positive number such that

holds for all real number

# Derivative of Exp

Since holds for all , it follows that, writing and

Therefore is differentiable at any with

Write , Then , . Then

so

# Definition of

Definition:

Thus is the unique number satisfying

As

and

it follows that

Since , it follows that

holds for all integers and

holds for all integers with , i.e.,

holds for all rational numbers.

Definition. For any number ,

Note that if is rational, then

Definition. If , then for any real number

In case , this definition is consistent with the earlier definition.

# Properties of

(1) If , then for all

(2) If , then , for all

# Uniqueness of the solution to

Theorem. If is differentiable and

then there is a number such that

Proof. Consider

Then

Therefore must be constant

# Properties of Integrals

• Hereditary: If is integrable on , then is integrable on for any

• Order Preserving: If and are integrable on and , then

Corollary: If is integrable on , then is integrable on and

Proof:

• Linearity: If and are integrable on , then

• Cauchy-Schwarz Inequality

Theorem. If and are integrable on , then

unless there exists a constant such that

Proof. Write

The algebraic properties of integrals show that

(i) unless

(ii)

(iii)

(iii')

We need to show that

Case (1) . The result holds since both sides are zero.

Case (2). Then

Therefore

If the equality holds then

consequently

• Triangular Inequality

If and are integrable on , then

Proof.

# Continuity of Indefinite Integrals

Recall that the first fundamental theorem of calculus asserts that if is integrable on and if is continuous at then the function defined by is differentiable at and . Note that the function is not assumed to be continuous throughout . The following result, however, does hold true.

Theorem. If is integrable on and is defined by

then is continuous on

Proof. Fix . Since is integrable on it is, by definition, bounded on . Choose so that for all . Let be given. Set . We now show that for

Case (1). . Then

Since

it follows that

Therefore,

Case (2). . Then

Since

it follows that

Therefore,

Thus in both cases.

# Mean Value Theorem for Integrals

Theorem. Suppose that is continuous on and that is integrable and nonnegative on . Then there exists in such that

Corollary. Suppose that is continuous on . Then there exists in such that

Proof. Since is continuous on , there exists in such that

and there exists in such that

It follows from the inequality

that

If then the result clearly holds. If is not identically then and so

Now the function given by

is positive for and negative for . It follows from the Intermediate Value Theorem that there exists (between and with

# Riemann Sums

Let . Let be a partition of the interval consisting of

For a bounded function defined on and for each fix a point in the interval . Then

Any sum of the form

is called a Riemann sum of for

Theorem. Suppose that is continuous on . Then for every there is some such that, if be a partition of the interval consisting of

with for all , then

for any Riemann sum formed by choosing in

Proof. Given , choose so that for all in

This is possible since is uniformly continuous on . Now for any partition satisfies the stated condition we have

But we have also

and

The result follows from the above three inequalities.

# Length of a Parametric Curve

Let a curve be given parametrically by

For any partition

of the total length of the polygonal line segments connecting

where and for all , is given by

We now show that for each given there exists a such that if all in the partition are , then

(1) For the given , there exists so that

whenever is a partition

with for all .

(2) Since the function is uniformly continuous for all , it follows that there exists such that

(3) Since the function is continuous on , it is uniformly continuous there. Thus there exists such that

Let . Consider a partition consisting of

with for all . The Mean Value Theorem shows that there exist and in such that

Since , it follows that

and so

The desired result follows since

# Length of a Parametric Curve

If and are continuous on , the length of the parametric curve

is defined to be the definite integral

# Length of the Graph of a Continuously Differentiable Function

If is continuous on , then the length of its graph is the same as the length of the parametric curve

Therefore it is given by

# Length of a Polar Curve

If the curve is given by the polar equation

It is just a particular instance of the parametric equations

Since

and

it follows that

so the length of its graph is given by

# Indefinite Integral of a Quadratic Rational Function

Example. Find

Set . Then and . Thus

Since , , we have

# Method of Completing the Square

Example.

This document created by Scientific WorkPlace 4.0.