prooving limits and differentials

profileArun1990
420-chp4-part2.pptx

Continuous vs. Uniformly Continuous

A function is continuous at if

such that

is continuous on a set if is continuous at every

A function is uniformly continuous on if

such that

Theorem

If is uniformly continuous on , then is continuous on .

1

is uniformly continuous on is continuous on .

Whenever is compact,

is uniformly continuous on is continuous on .

Proof: Let . Choose /3. Thus,

and

Thus, is uniformly continuous on , which implies is continuous on

Example:

2

Uniformly Continuous examples

Proof: Let . Let . Choose . Thus,

and

Thus, is uniformly continuous on , which implies is continuous on

Example:

3

Uniformly Continuous examples

uniformly continuous

not

uniformly continuous

δ

δ

Prove that is not uniformly continuous on .

e.g. x^3 is NOT unif cont on R. P=n+1/n, Q=n (n suff. large)

e.g. 1/x^2 is NOT unif cont on (0,1]. P=1/n, Q=2/n (n suff. large)

4

Since the above must hold for every small another way to show non-uniform continuity is with sequences.

f is not

uniformly continuous

Example: 1/x is uniformly continuous on [a,b] for any 0<a<b, but not uniformly continuous on (0,1).

Consider the sequences x_n=1/2n and y_n=1/n. Notice that |1/x_n – 1/y_n| = n \ge 1=\varepsilon.

5

For last statement: try [-2,-1], [-1,0], and [0,2].

(x,y) points are (-2,-7), (-1,1), (0,-1), (2,13)

6

∀ε > 0 ∃δ > 0 such that ∀p, q ∈ E with | p − q |<δ ⇒| f ( p) − f (q) |<ε.

"e>0 $d>0 such that

"p,qÎE with |p-q|<dÞ|f(p)-f(q)|<e.

3

Prove that () is both continuous on [0,

1]

and uniformly continuous on [0,1].

fxx

=

3

Prove that () is both continuous on [,]

and uniformly continuous on [,].

fxxab

ab

=

To show f is not uniformly continuous on X , you should demonstrate that there is some ε> 0 such that no matter what > 0 you pick, you can find points p, q ∈ X such that | p − q | < and | f ( p) − f (q) | ≥ε.

∀ε> 0 ∃δ > 0 s.t. ∀p, q with | p − q |<δ ⇒ | f ( p) − f (q) | <ε

∀ε> 0( ) ∃δ > 0( ) ∀p, q( )(| p − q | <δ ⇒ | f ( p) − f (q) | <ε)

has negation

∃ε> 0( ) ∀δ > 0( ) ∃p, q( )(| p − q | <δ and | f ( p) − f (q) | ≥ε)

Prove f ( x) = x 3 + x 2 − 4 has a real zero.

Prove f(x)=x

3

+x

2

-4 has a real zero.

Prove f ( x) = x 3 + x 2 − 4 has a real zero in (0,2).

Prove f(x)=x

3

+x

2

-4 has a real zero in (0,2).

Prove f ( x) = 2x 3 + x 2 − 3x −1 has three real zeros.

Prove f(x)=2x

3

+x

2

-3x-1 has three real zeros.