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
MATH
when
MATH

Theorem. Suppose that
MATH
and and suppose also that
MATH
exists. Then
MATH
exists, and
MATH

PROOF. The hypothesis that MATH exists contains two implicit assumptions:

(1) there is an interval MATH such that $f^{\prime }(x)$ and $g^{\prime }(x)$ exist for all $x$ in MATH except, perhaps, for $x=\alpha $,

(2) in this interval MATH with, once again, the possible exception of $x=\alpha .$

On the other hand, $f$ and $g$ are not even assumed to be defined at $\alpha $. If we define MATH (changing the previous values of $f(\alpha )$ and $g(\alpha )$, if necessary), then $f$ and $g$ are continuous at $\alpha $. If MATH, then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to $f$ and $g$ on the interval $[\alpha ,x]$ (and a similar statement holds for MATH). First applying the Mean Value Theorem to $g$, we see that $g(x)\ne 0$, for if $g(x)=0$ there would be some $x_{1}$ in $(\alpha ,x)$ with MATH, contradicting (2). Now applying the Cauchy Mean Value Theorem to $f$ and $g$, we see that there is a number $b$. in $(\alpha ,x)$ such that
MATH
or
MATH
Now $b$ approaches $a$ as $x$ approaches $\alpha $, because $b$, is in $(\alpha ,x)$; since MATH exists, it follows that
MATH

Uniform Continuity

Definition. The function $f$ is uniformly continuous on an interval $A$ if for every $\epsilon >0$ there is some $\delta >0$ such that, for all $x$ and $y$ in $A,$


MATH

Lemma. Let $a<b<c$ and let $f$ be continuous on the interval $[a,c].$ Let $\epsilon >0$, and suppose that statements (i) and (ii) hold:

(i) if $x$ and $y$ are in $[a,b]\;$and $\left| x-y\right| $ $<$ $\delta _1$, then MATH $<\epsilon $,

(ii) if $x$ and $y$ are in $[b,c]\;$and $\left| x-y\right| $ $<$ $\delta _2$, then MATH

Then there is a $\delta $$>$ 0 such that
MATH

Proof. Since $f$ is continuous at $b$, there is a $\delta _3$ $>$ 0 such that
MATH

It follows that

(iii) if MATH and MATH then MATH

Choose $\delta $ to be the minimum of MATH and $\delta _3$. We claim that this $\delta $ works. In fact, suppose that $x$ and $y$ are any two points in $[a,c]$ with MATH. If $x$ and $y$ are both in $[a,b],$ then MATHby (i); and if $x$ and $y$ are both in $[b,c],$ then MATHby (ii). The only other possibility is that

$x<b<y$ or $y<b<x.$

In either case, since $\left| x-y\right| $ $<\delta $ , we also have $\left| x-b\right| $ $<\delta $ and $\left| y-b\right| $ $<\delta .$ So MATHby (iii).

THEOREM If $f$ is continuous on $[a,b]$, then $f$ is uniformly continuous on $[a,b].$

PROOF: For $\epsilon >0$ let's say that $f\;$is $\epsilon -$good on $[a,b]$ if there is some $\delta >0$ such that, for all $y$ and $z$ in $[a,b],$if $\left| y-z\right| $ $<\delta $, then MATH. Then we're trying to prove that $f$ is $\epsilon $-good on $[a,b]$ for all $\epsilon >0.$

Consider any particular $\epsilon >0$. Let


MATH
Then $A\ne \phi \;$(since a is in $A$), and $A$ is bounded above (by $b$), so $A$ has a least upper bound $\alpha $. We really should write $\alpha _\epsilon $, since $A$ and $\alpha $ might depend on $\epsilon $. But we won't since we intend to prove that $\alpha =b$, no matter what $\epsilon \;$is.

Suppose that we had $\alpha <b$. Since $f$ is continuous at $\alpha $, there is some $\delta _0$$>0$ such that, if MATH $<\delta _0$, then MATH. Consequently, if MATH $<\delta _0$ and MATH, then MATH So $f$ is surely $\epsilon $-good on the interval MATH. On the other hand, since $\alpha $ is the least upper bound of $A$, it is also clear that $f$ is $\epsilon $-good on MATH. Then the above Lemma implies that $f$ is $\epsilon $-good on MATH, so $\alpha +\delta _0$ is in A, contradicting the fact that $\alpha $ is an upper bound.

To complete the proof we just have to show that $\alpha =b$ is actually in $A $. The argument for this is practically the same: Since $f$ is continuous at $b$, there is some $\delta _{0}>0$ such that, if MATH, then MATHSo $f$ is $\epsilon $-good on $[b-\delta _{0},b]$. But $f$ is also $\epsilon $-good on $[a,b-\delta _{0}]$, so the Lemma implies that $f$ is $\epsilon $-good on $[a,b]$.

Logarithm

Definition. If $x>0$, define
MATH

It follows from the Fundamental Theorem of Calculus that $\ln x$ is differentiable and
MATH
For each fixed $y>0$, define $f(x)=\ln (xy)$. Then MATH. Therefore there is a constant $c$ so that
MATH
Setting $x=1$, we see that
MATH
Therefore
MATH

Corollary. If $n$ is a natural number and $x>0$, then
MATH

Corollary. If $x,y>0$, then
MATH

Exponential Function

Since MATH for all $x$, $\ln $ is an increasing function on $(0,+\infty )$. It is unbounded from above:
MATH
and it is unbounded from below:
MATH
Therefore the inverse function $\ln ^{-1}$ of $\ln $ exists. The exponential function $\exp $ is defined as $\ln ^{-1}$. Thus $\exp (x)$ is the unique positive number such that
MATH
holds for all real number $y.$

Derivative of Exp

Since MATH holds for all $x>0$, it follows that, writing $y=\ln \ x$ and $y_0=\ln \ x_0,$
MATH
Therefore $\exp $ is differentiable at any $y_0$ with MATH
MATH

Write MATH, MATHThen MATH, MATH. Then
MATH
so
MATH

Definition of $e$

Definition: $e=\exp (1).$

Thus $e$ is the unique number satisfying
MATH

As
MATH
and
MATH
it follows that
MATH

Since MATH, it follows that
MATH
holds for all integers $n$ and
MATH
holds for all integers $m,n$ with $m\neq 0$, i.e.,
MATH
holds for all rational numbers.

Definition. For any number $x$, $e^x=\exp (x).$

Note that if $x$ is rational, then
MATH

Definition. If $a>0$, then for any real number $x,$
MATH

In case $e=a$, this definition is consistent with the earlier definition.

Properties of $a^x$

(1) If $a>0$, then $(a^b)^c=a^{(bc)}$ for all $b,c.$

(2) If $a>0$, then $a^1=a$, MATH for all $x,y.$

Uniqueness of the solution to MATH

Theorem. If $f$ is differentiable and
MATH
then there is a number $c$ such that
MATH

Proof. Consider
MATH
Then
MATH
Therefore $g$ must be constant $c.$

Properties of Integrals

Corollary: If $f$ is integrable on $[a,b]$, then $|f|$ is integrable on $[a,b]$ and
MATH

Proof: MATH

unless there exists a constant $c$ such that $g=cf.$

Proof. Write
MATH

The algebraic properties of integrals show that

(i) $<f,f>>0$ unless $f\equiv 0.$

(ii) $<f,g>=<g,f>$

(iii) MATH

(iii') MATH

We need to show that
MATH

Case (1) $g\equiv 0$. The result holds since both sides are zero.

Case (2)$<g,g>>0$. Then
MATH

MATH

MATH
Therefore
MATH

If the equality holds then
MATH
consequently
MATH

If $f$ and $g$ are integrable on $[a,b]$, then
MATH

Proof.
MATH

MATH

MATH

Continuity of Indefinite Integrals

Recall that the first fundamental theorem of calculus asserts that if $f$ is integrable on $[a,b]$ and if $f$ is continuous at $c$ then the function defined by $F(x)=\int_a^xf$ is differentiable at $c$ and MATH. Note that the function $f$ is not assumed to be continuous throughout $[a,b]$. The following result, however, does hold true.

Theorem. If $f$ is integrable on $[a,b]$ and $F$ is defined by
MATH
then $F$ is continuous on $[a,b].$

Proof. Fix $c\in [a,b]$. Since $f$ is integrable on $[a,b]$ it is, by definition, bounded on $[a,b]$. Choose $M>0$ so that $|f(x)|\leq M$ for all $x\in [a,b]$. Let $\epsilon >0$ be given. Set MATH. We now show that MATH for $|h|<\delta .$

Case (1). $0<h<\delta $. Then
MATH

Since
MATH
it follows that
MATH
Therefore,
MATH

Case (2). $-\delta <h<0$. Then
MATH
Since
MATH
it follows that
MATH
Therefore,
MATH
Thus MATH in both cases.

Mean Value Theorem for Integrals

Theorem. Suppose that $f$ is continuous on $[a,b]$ and that $g$ is integrable and nonnegative on $[a,b]$. Then there exists $\xi $ in $[a,b]$ such that
MATH

Corollary. Suppose that $f$ is continuous on $[a,b]$. Then there exists $\xi $ in $[a,b]$ such that
MATH

Proof. Since $f$ is continuous on $[a,b]$, there exists $\alpha $ in $[a,b]$ such that
MATH
and there exists $\beta $ in $[a,b]$ such that
MATH
It follows from the inequality
MATH
that
MATH

If $g\equiv 0$ then the result clearly holds. If $g$ is not identically $0$ then $\int_a^bg(x)dx>0$ and so
MATH
Now the function $h$ given by
MATH
is positive for $t=\beta $ and negative for $t=\alpha $. It follows from the Intermediate Value Theorem that there exists $\xi $ (between $\alpha $ and $\beta )$ with $h(\xi )=0.$

Riemann Sums

Let $a<b$. Let $P$ be a partition of the interval $[a,b]$ consisting of
MATH

For a bounded function $f$ defined on $[a,b]$ and for each $i$ fix a point $t_i$ in the interval $[x_{i-1},x_i]$. Then
MATH
Any sum of the form
MATH
is called a Riemann sum of $f$ for $P.$

Theorem. Suppose that $f$ is continuous on $[a,b]$. Then for every $\epsilon >0$ there is some $\delta >0$ such that, if $P$ be a partition of the interval $[a,b]$ consisting of
MATH
with MATH for all $i$, then
MATH
for any Riemann sum formed by choosing $t_i$ in $[x_{i-1},x_i].$

Proof. Given $\epsilon >0$, choose $\delta >0$ so that for all $x,y$ in $[a,b]$
MATH
This is possible since $f$ is uniformly continuous on $[a,b]$. Now for any partition $P$ satisfies the stated condition we have
MATH
But we have also
MATH
and
MATH
The result follows from the above three inequalities.

Length of a Parametric Curve

Let a curve be given parametrically by
MATH
For any partition $P$
MATH
of $[a,b]$ the total length of the polygonal line segments connecting
MATH
where $x_i=x(t_i)$ and $y_i=y(t_i)$ for all $i$, is given by
MATH

MATH
We now show that for each given $\epsilon >0$ there exists a $\delta >0$ such that if all $t_i-t_{i-1}$ in the partition $P\ $are $<\delta $, then
MATH

(1) For the given $\epsilon >0$, there exists $\delta _1>0$ so that
MATH
whenever $P$ is a partition
MATH
with MATH for all $i$.

(2) Since the function MATH is uniformly continuous for all $s > 0$, it follows that there exists $\epsilon _1$ such that


MATH

(3) Since the function $x^{\prime }$ is continuous on $[a,b]$, it is uniformly continuous there. Thus there exists $\delta _2>0$ such that
MATH
Let MATH. Consider a partition $P$ consisting of
MATH
with MATH for all $i$. The Mean Value Theorem shows that there exist $\xi _i$ and $\eta _i$ in $(t_{i-1},t_i)$ such that
MATH
Since MATH, it follows that
MATH
and so
MATH

MATH

The desired result follows since
MATH

Length of a Parametric Curve

If $x^{\prime }$ and $y^{\prime }$ are continuous on $[a,b]$, the length of the parametric curve
MATH
is defined to be the definite integral
MATH

Length of the Graph of a Continuously Differentiable Function

If $f^{\prime }$ is continuous on $[a,b]$, then the length of its graph is the same as the length of the parametric curve
MATH
Therefore it is given by
MATH

Length of a Polar Curve

If the curve is given by the polar equation
MATH
It is just a particular instance of the parametric equations
MATH
Since
MATH
and
MATH
it follows that
MATH
so the length of its graph is given by
MATH

Indefinite Integral of a Quadratic Rational Function

MATH

Example. Find MATH

Set MATH. Then MATH and MATH. Thus
MATH
Since MATH, MATH, we have
MATH

MATH

Method of Completing the Square


MATH

Example. MATH

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