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Finally, using the continuity of both f and g applied to the positive numbers ϵ 1 = ϵ / ( 4 M 2 | g ( c ) | 2 ) and ϵ 2 = ϵ / ( 4 M 1 | g ( c ) | 2 ) , choose δ > 0 , with δ < min ( δ 1 , δ 2 , δ ' ) , and such that if | y - c | < δ and y S then | f ( y ) - f ( c ) | < ϵ 4 M 2 / | g ( c ) | 2 and | g ( c ) - g ( y ) | < ϵ 4 M 1 / | g ( c ) | 2 . Then, if | y - c | < δ and y S we have that

| f ( y ) g ( y ) - f ( c ) g ( c ) | < ϵ

as desired.

  1. Prove part (2) of the preceding theorem. (It's an ϵ / 2 argument.)
  2. Prove part (3) of the preceding theorem. (It's similar to the proof of part (5) only easier.)
  3. Prove part (4) of the preceding theorem.
  4. Prove part (6) of the preceding theorem.
  5. Suppose S is a subset of R . Verify the above theoremreplacing “ continuity” with left continuity and right continuity.
  6. If S is a subset of R , show that f is continuous at a point c S if and only if it is both right continuous and left continuous at c .

The composition of continuous functions is continuous.

Let S , T , and U be subsets of C , and let f : S T and g : T U be functions. Suppose f is continuous at a point c S and that g is continuous at the point f ( c ) T . Then the composition g f is continuous at c .

Let ϵ > 0 be given. Because g is continuous at the point f ( c ) , there exists an α > 0 such that | g ( t ) - g ( f ( c ) ) | < ϵ if | t - f ( c ) | < α . Now, using this positive number α , and using the fact that f is continuous at the point c , there exists a δ > 0 so that | f ( s ) - f ( c ) | < α if | s - c | < δ . Therefore, if | s - c | < δ , then | f ( s ) - f ( c ) | < α , and hence | g ( f ( s ) ) - g ( f ( c ) ) | = | g f ( s ) - g f ( c ) | < ϵ , which completes the proof.

  1. If f : C C is the function defined by f ( z ) = z , prove that f is continuous at each point of C .
  2. Use part (a) and Theorem 3.2 to conclude that every rational function is continuous on its domain.
  3. Prove that a step function h : [ a , b ] C is continuous everywhere on [ a , b ] except possibly at the points of the partition P that determines h .
  1. Let S be the set of nonnegative real numbers, and define f : S S by f ( x ) = x . Prove that f is continuous at each point of S . HINT: For c = 0 , use δ = ϵ 2 . For c 0 , use the identity
    y - c = ( y - c ) y + c y + c = y - c y + c y - c c .
  2. If f : C R is the function defined by f ( z ) = | z | , show that f is continuous at every point of its domain.

Using the previous theorems and exercises, explain why the following functions f are continuous on their domains. Describe the domains as well.

  1. f ( z ) = ( 1 - z 2 ) / ( 1 + z 2 ) .
  2. f ( z ) = | 1 + z + z 2 + z 3 - ( 1 / z ) | .
  3. f ( z ) = 1 + 1 - | z | 2 .
  1. If c and d are real numbers, show that max ( c , d ) = ( c + d ) / 2 + | c - d | / 2 .
  2. If f and g are functions from S into R , show that max ( f , g ) = ( f + g ) / 2 + | f - g | / 2 .
  3. If f and g are real-valued functions that are both continuous at a point c , show that max ( f , g ) and min ( f , g ) are both continuous at c .

Let N be the set of natural numbers, let P be the set of positive real numbers, and define f : N P by f ( n ) = 1 + n . Prove that f is continuous at each point of N . Show in fact that every function f : N C is continuous on this domain N .

HINT: Show that for any ϵ > 0 , the choice of δ = 1 will work.

  1. Negate the statement: “For every ϵ > 0 , | x | < ϵ . ' '
  2. Negate the statement: “For every ϵ > 0 , there exists an x for which | x | < ϵ . ' '
  3. Negate the statement that “ f is continuous at c . ' '

The next result establishes an equivalence between the basic ϵ , δ definition of continuity and a sequential formulation.In many cases, maybe most, this sequential version of continuity is easier to work with than the ϵ , δ version.

Let f : S C be a complex-valued function on S , and let c be a point in S . Then f is continuous at c if and only if the following condition holds: For every sequence { x n } of elements of S that converges to c , the sequence { f ( x n ) } converges to f ( c ) . Or, said a different way, if { x n } converges to c , then { f ( x n ) } converges to f ( c ) . And, said yet a third (somewhat less precise) way, the function f converts convergent sequences to convergent sequences.

Suppose first that f is continuous at c , and let { x n } be a sequence of elements of S that converges to c . Let ϵ > 0 be given. We must find a natural number N such that if n N then | f ( x n ) - f ( c ) | < ϵ . First, choose δ > 0 so that | f ( y ) - f ( c ) | < ϵ whenever y S and | y - c | < δ . Now, choose N so that | x n - c | < δ whenever n N . Then if n N , we have that | x n - c | < δ , whence | f ( x n ) - f ( c ) | < ϵ . This shows that the sequence { f ( x n ) } converges to f ( c ) , as desired.

We prove the converse by proving the contrapositive statement; i.e., we will show that if f is not continuous at c , then there does exist a sequence { x n } that converges to c but for which the sequence { f ( x n ) } does not converge to f ( c ) . Thus, suppose f is not continuous at c . Then there exists an ϵ 0 > 0 such that for every δ > 0 there is a y S such that | y - c | < δ but | f ( y ) - f ( c ) | ϵ 0 . To obtain a sequence, we apply this statement to δ 's of the form δ = 1 / n . Hence, for every natural number n there exists a point x n S such that | x n - c | < 1 / n but | f ( x n ) - f ( c ) | ϵ 0 . Clearly, the sequence { x n } converges to c since | x n - c | < 1 / n . On the other hand, the sequence { f ( x n ) } cannot be converging to f ( c ) , because | f ( x n ) - f ( c ) | is always ϵ 0 .

This completes the proof of the theorem.

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
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Sherica
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Sherica
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Tamia
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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what's the easiest and fastest way to the synthesize AgNP?
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China
Cied
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abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
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Ramkumar Reply
what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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