mean value theorem

Left Hand Limit

We say that the limit of

f (x)

x

approaches

c

from the left equals

L

when

\forall ϵ \in ℝ^{> 0}, \exists δ \in ℝ^{> 0} such that \forall x \in dom (f), c - δ < x < c \Rightarrow | f (x) - f (c) | < ϵ

Right Hand limit

We say that the limit of

f (x)

x

approaches

c

from the right equals

L

when

\forall ϵ \in ℝ^{> 0}, \exists δ \in ℝ^{> 0} such that \forall x \in dom (f), c < x < c + δ \Rightarrow | f (x) - f (c) | < ϵ

Limit Exists iff the Left and Right limits Exist and Agree

Suppose that

f (x)

is defined on an open interval containing

c

, then

{lim}_{x \to c} f (x)

is defined if and only if

{lim}_{x \to c^{+}} f (x)

and

{lim}_{x \to c^{-}} f (x)

are both defined and equal.

Function goes to Infinity at a Point From the Right

Suppose that

S \subseteq R

and that

f : S \to R

, is a function then if

a \in S

then we say that

f

goes to positive infinity at a point at

a

if for every

M \in R

there exists an

δ \in R^{+}

such that for all

x \in S

a < x < a + δ ⟹ f (x) > M

and we write

{lim}_{x \to a^{+}} f (x) = \infty

Function goes to Infinity at a Point From the Right Alternate Characterization

The regular definition of going to infinity from the right yields a new characterization:

{lim}_{x \to a^{+}} f (x) = \infty ⟹ \forall δ, M^{'} \in R^{+}, \exists x^{'} \in S st (a < x^{'} < a + δ^{'} \land f (x^{'}) > M^{'})

Suppose that ${lim}_{x \to a^{+}} f (x) = \infty$ , now let $δ^{'}, M^{'} \in R^{+}$ setting $M = M^{'}$ we obtain some $δ$ such that for any $x \in (a, a + δ)$ we have that $f (x) > M$ note that we have to find an $x^{'} \in (a, a + δ^{'})$ therefore note that for any $y \in (a, min (δ, δ^{'}))$ we have $f (y) > M$ as we know that $(a, a + min (δ, δ^{'})) \subseteq (a, a + δ)$ , therefore take $x^{'} = a + \frac{min (δ, δ^{'})}{2}$ so then we have that $f (x^{'}) > M$

Note that the above is not an if and only if as $\frac{1}{x} \sin (\frac{1}{x})$ on the domain $R^{+}$ satisfies the characterization, but does not satisfy the right hand limit as it always dips back down to zero infinitely often as we move toward zero.

local maximum

Let

f : I \to ℝ

where

I

is some interval and let

c \in I

, we say that

f

has a local maximum at

c

, when there exists some

δ \in ℝ^{> 0}

such that for all

x \in d o m (f)

, if

| x - c | < δ

then

f (x) \leq f (c)

local minimum

Let

f : I \to ℝ

where

I

is some interval and let

c \in I

we say that

f

has a local maximum at

c

, when there exists some

δ \in ℝ^{>} . .

such that for all

x \in d o m (f)

| (x - c |

then

f . j

local extremum

c

is a local minimum or maximum of

f

then it's said to be a local extremum

local extreme value

suppose that

f

has a local extremum at

c

and

c

is an interior point of

i

, then

f^{^{'}} (c) = 0

or does not exist

Without loss of generality assume that $f$ has a local maximum at $c$ , we want to show that the limit $f^{^{'}} (c) = {lim}_{x \to c} \frac{f (x) - f (c)}{x - c}$ exists and is $0$ , or doesn't exist.

Given a limit it either exists or it doesn't, assuming this limit exists, we need to show that it equals zero. So let's assume that the limit $f^{^{'}} (c)$ exists.

Since it exists the we can consider $x \to c^{+}$ , in this case $x - c > 0$ , since we know that $f$ has a local maximum at $c$ then there exists some $δ \in ℝ^{> 0}$ such that for all $p \in dom (f)$ , if $| p - c | < δ$ then $f (p) \leq f (c)$

c

is a local extrema of

f : I \to ℝ

then

c

is not an endpoint of the interval

I

mean value

Suppose that

a, b \in ℝ

a < b

and suppose

f : [a, b] \to ℝ

, if

f

is continuous on

[a, b]

and differentiable on

(a, b)

, then there is some

c \in (a, b)

such that

f^{'} (c) = \frac{f (b) - f (a)}{b - a}

🏗️ ΘρϵηΠατπ🚧 (under construction)

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)