Proof that $f(x)=ax+b$ is continuous [closed]

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I'm looking to prove $f(x)=ax+b$ is continuous using the epsilon-delta method. Now I know it’s already been proven to be uniformly continuous and proving that means it has to be continuous anyway, but I’m still interested in how one would prove basic continuity. I would like to use the epsilon-delta method, and am having a little difficulty understanding it. So far I have,

$f(x)=ax+b$ for some $epsilon>0$, there exists a $delta$>0 s.t.
$|x-c|<delta$ if $|f(x)-f(c)|<epsilon$

which means
$|a(x-c)|<epsilon$

Not really sure where to take the proof from here and would appreciate any advice, this is my first post also so apologies if the latex is wrong or anything! Again specifying that the reason I think this question hasnt been answered here is I'm not looking for uniform continuity.

continuity

share|cite|improve this question

edited Jul 25 at 13:50

asked Jul 25 at 12:24

Robbie Meaney

246

closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, Mostafa Ayaz, max_zorn Jul 26 at 0:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You actually want $|x-c|<delta implies |f(x)-f(c)|<epsilon$, so implies not if, and you want it to be true for all $c$. You should try to find a value of $delta$ (possibly as a function of $epsilon$ and $c$) which works for this case.
  – Henry
  Jul 25 at 12:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Possible duplicate of Prove that any polynomial function is continuous
  – Dietrich Burde
  Jul 25 at 12:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$|a(x-c)|<epsiloniff |x-c|<fracepsilon$$
  – Yves Daoust
  Jul 25 at 12:45
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

I'm looking to prove $f(x)=ax+b$ is continuous using the epsilon-delta method. Now I know it’s already been proven to be uniformly continuous and proving that means it has to be continuous anyway, but I’m still interested in how one would prove basic continuity. I would like to use the epsilon-delta method, and am having a little difficulty understanding it. So far I have,

$f(x)=ax+b$ for some $epsilon>0$, there exists a $delta$>0 s.t.
$|x-c|<delta$ if $|f(x)-f(c)|<epsilon$

which means
$|a(x-c)|<epsilon$

Not really sure where to take the proof from here and would appreciate any advice, this is my first post also so apologies if the latex is wrong or anything! Again specifying that the reason I think this question hasnt been answered here is I'm not looking for uniform continuity.

continuity

share|cite|improve this question

edited Jul 25 at 13:50

asked Jul 25 at 12:24

Robbie Meaney

246

closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, Mostafa Ayaz, max_zorn Jul 26 at 0:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You actually want $|x-c|<delta implies |f(x)-f(c)|<epsilon$, so implies not if, and you want it to be true for all $c$. You should try to find a value of $delta$ (possibly as a function of $epsilon$ and $c$) which works for this case.
  – Henry
  Jul 25 at 12:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Possible duplicate of Prove that any polynomial function is continuous
  – Dietrich Burde
  Jul 25 at 12:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$|a(x-c)|<epsiloniff |x-c|<fracepsilon$$
  – Yves Daoust
  Jul 25 at 12:45
  
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

I'm looking to prove $f(x)=ax+b$ is continuous using the epsilon-delta method. Now I know it’s already been proven to be uniformly continuous and proving that means it has to be continuous anyway, but I’m still interested in how one would prove basic continuity. I would like to use the epsilon-delta method, and am having a little difficulty understanding it. So far I have,

$f(x)=ax+b$ for some $epsilon>0$, there exists a $delta$>0 s.t.
$|x-c|<delta$ if $|f(x)-f(c)|<epsilon$

which means
$|a(x-c)|<epsilon$

Not really sure where to take the proof from here and would appreciate any advice, this is my first post also so apologies if the latex is wrong or anything! Again specifying that the reason I think this question hasnt been answered here is I'm not looking for uniform continuity.

continuity

share|cite|improve this question

edited Jul 25 at 13:50

asked Jul 25 at 12:24

Robbie Meaney

246

I'm looking to prove $f(x)=ax+b$ is continuous using the epsilon-delta method. Now I know it’s already been proven to be uniformly continuous and proving that means it has to be continuous anyway, but I’m still interested in how one would prove basic continuity. I would like to use the epsilon-delta method, and am having a little difficulty understanding it. So far I have,

$f(x)=ax+b$ for some $epsilon>0$, there exists a $delta$>0 s.t.
$|x-c|<delta$ if $|f(x)-f(c)|<epsilon$

which means
$|a(x-c)|<epsilon$

Not really sure where to take the proof from here and would appreciate any advice, this is my first post also so apologies if the latex is wrong or anything! Again specifying that the reason I think this question hasnt been answered here is I'm not looking for uniform continuity.

continuity

share|cite|improve this question

edited Jul 25 at 13:50

asked Jul 25 at 12:24

Robbie Meaney

246

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 13:50

edited Jul 25 at 13:50

edited Jul 25 at 13:50

asked Jul 25 at 12:24

Robbie Meaney

246

asked Jul 25 at 12:24

Robbie Meaney

246

asked Jul 25 at 12:24

Robbie Meaney

246

246

closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, Mostafa Ayaz, max_zorn Jul 26 at 0:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Dietrich Burde, José Carlos Santos, Mostafa Ayaz, max_zorn Jul 26 at 0:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Mostafa Ayaz, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You actually want $|x-c|<delta implies |f(x)-f(c)|<epsilon$, so implies not if, and you want it to be true for all $c$. You should try to find a value of $delta$ (possibly as a function of $epsilon$ and $c$) which works for this case.
  – Henry
  Jul 25 at 12:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Possible duplicate of Prove that any polynomial function is continuous
  – Dietrich Burde
  Jul 25 at 12:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$|a(x-c)|<epsiloniff |x-c|<fracepsilon$$
  – Yves Daoust
  Jul 25 at 12:45
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You actually want $|x-c|<delta implies |f(x)-f(c)|<epsilon$, so implies not if, and you want it to be true for all $c$. You should try to find a value of $delta$ (possibly as a function of $epsilon$ and $c$) which works for this case.
  – Henry
  Jul 25 at 12:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Possible duplicate of Prove that any polynomial function is continuous
  – Dietrich Burde
  Jul 25 at 12:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$|a(x-c)|<epsiloniff |x-c|<fracepsilon$$
  – Yves Daoust
  Jul 25 at 12:45
  
  

  
  
  
  

You actually want $|x-c|<delta implies |f(x)-f(c)|<epsilon$, so implies not if, and you want it to be true for all $c$. You should try to find a value of $delta$ (possibly as a function of $epsilon$ and $c$) which works for this case.
– Henry
Jul 25 at 12:28

You actually want $|x-c|<delta implies |f(x)-f(c)|<epsilon$, so implies not if, and you want it to be true for all $c$. You should try to find a value of $delta$ (possibly as a function of $epsilon$ and $c$) which works for this case.
– Henry
Jul 25 at 12:28

Possible duplicate of Prove that any polynomial function is continuous
– Dietrich Burde
Jul 25 at 12:28

Possible duplicate of Prove that any polynomial function is continuous
– Dietrich Burde
Jul 25 at 12:28

$$|a(x-c)|<epsiloniff |x-c|<fracepsilon$$
– Yves Daoust
Jul 25 at 12:45

$$|a(x-c)|<epsiloniff |x-c|<fracepsilon$$
– Yves Daoust
Jul 25 at 12:45

add a comment | 

3 Answers
3

active

oldest

votes

up vote
0
down vote



accepted

To show, that $f:mathbbRtomathbbR$, $xmapsto ax+b$ is continuous, we have to show, that $forallepsilon>0,existsdelta>0:forall xinmathbbR:|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon$

Simply said, we have to find a $delta$ for every $epsilon$.

Keep in mind, that we have $|x-y|<delta$ as an assumption.

You can find your $delta$ like this. First of all:

$|f(x)-f(y)|=|ax+b-(ay+b)|=|ax+b-ay-b|=|a(x-y)|=|a||x-y|$

We know, that $|x-y|<delta$ so we can stipulate like this:

$|a||x-y|<|a|delta$

We want to find for every $epsilon$ a $delta$ such that, when $|x-y|<delta$ we have $|f(x)-f(y)|<epsilon$, this means that $delta$ can be dependent from $epsilon$.

We choose $delta:=fracepsilon$ for $aneq 0$.
For $a=0$ we have to find a seperate $delta$. But this is easy, since if $a=0$, we just get $0$ and we can choose $delta=1$ for example. It does not matter.

Note, that this is all a scatch and not the proof yet. We just searched our $delta$.

Now for the proof:

Let $epsilon >0$ be arbitrary. If $a=0$ choose $delta=1$ and we get, that $|f(x)-f(y)|=|0cdot x+b-0cdot y-b|=0<epsilon$.

For $aneq 0$ choose $delta=fracepsilon>0$.

Then:

$|f(x)-f(y)|=|ax+b-(ay+b)|=dotso =|a||x-y|<|a|delta =|a|cdot fracepsilon=epsilon$

And we are done.

share|cite|improve this answer

edited Jul 25 at 12:51

answered Jul 25 at 12:46

Cornman

2,40021127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for taking the time to answer this, I'm currently working through Rudins Principles of Maths Analysis and you have been a huge help!
  – Robbie Meaney
  Jul 25 at 12:56
  

  
  
  
  

add a comment | 

up vote
2
down vote

Let $e>0$ be given. Let $d=frace2($. Then, $|x-c|<d$ implies
$$
|f(x)-f(c)|=|a|cdot|x-c|leq(1+|a|)d<e
$$
whether $a=0$ or not. Can you complete the argument now?

share|cite|improve this answer

edited Jul 26 at 16:29

answered Jul 25 at 12:30

yurnero

6,8401824

add a comment | 

up vote
1
down vote

Your statement about continuity is really wrong.

In order to prove that $f$ is continuous at $c$, you need

for every $varepsilon>0$, there exists $delta>0$ such that, for every $x$ with $|x-c|<delta$, it is true that $|f(x)-f(c)|<varepsilon$.

Can you spot the difference with what you stated?

In order to prove continuity for your function, let $varepsilon>0$ be given. Then we want to solve the inequality $|f(x)-f(c)|<varepsilon$ and see whether the solution set has the above property. Since
$$
|f(x)-f(c)|=|ax+b-ac-b|=|a||x-c|
$$
we see that

1. if $a=0$, then every $delta>0$ fits;


2. if $ane0$, the inequality $|a||x-c|<varepsilon$ is equivalent to
   $$|x-c|<fracvarepsilon,$$ so we can take $delta=varepsilon/|a|$.

share|cite|improve this answer

answered Jul 25 at 12:45

egreg

164k1180187

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

To show, that $f:mathbbRtomathbbR$, $xmapsto ax+b$ is continuous, we have to show, that $forallepsilon>0,existsdelta>0:forall xinmathbbR:|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon$

Simply said, we have to find a $delta$ for every $epsilon$.

Keep in mind, that we have $|x-y|<delta$ as an assumption.

You can find your $delta$ like this. First of all:

$|f(x)-f(y)|=|ax+b-(ay+b)|=|ax+b-ay-b|=|a(x-y)|=|a||x-y|$

We know, that $|x-y|<delta$ so we can stipulate like this:

$|a||x-y|<|a|delta$

We want to find for every $epsilon$ a $delta$ such that, when $|x-y|<delta$ we have $|f(x)-f(y)|<epsilon$, this means that $delta$ can be dependent from $epsilon$.

We choose $delta:=fracepsilon$ for $aneq 0$.
For $a=0$ we have to find a seperate $delta$. But this is easy, since if $a=0$, we just get $0$ and we can choose $delta=1$ for example. It does not matter.

Note, that this is all a scatch and not the proof yet. We just searched our $delta$.

Now for the proof:

Let $epsilon >0$ be arbitrary. If $a=0$ choose $delta=1$ and we get, that $|f(x)-f(y)|=|0cdot x+b-0cdot y-b|=0<epsilon$.

For $aneq 0$ choose $delta=fracepsilon>0$.

Then:

$|f(x)-f(y)|=|ax+b-(ay+b)|=dotso =|a||x-y|<|a|delta =|a|cdot fracepsilon=epsilon$

And we are done.

share|cite|improve this answer

edited Jul 25 at 12:51

answered Jul 25 at 12:46

Cornman

2,40021127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for taking the time to answer this, I'm currently working through Rudins Principles of Maths Analysis and you have been a huge help!
  – Robbie Meaney
  Jul 25 at 12:56
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

To show, that $f:mathbbRtomathbbR$, $xmapsto ax+b$ is continuous, we have to show, that $forallepsilon>0,existsdelta>0:forall xinmathbbR:|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon$

Simply said, we have to find a $delta$ for every $epsilon$.

Keep in mind, that we have $|x-y|<delta$ as an assumption.

You can find your $delta$ like this. First of all:

$|f(x)-f(y)|=|ax+b-(ay+b)|=|ax+b-ay-b|=|a(x-y)|=|a||x-y|$

We know, that $|x-y|<delta$ so we can stipulate like this:

$|a||x-y|<|a|delta$

We want to find for every $epsilon$ a $delta$ such that, when $|x-y|<delta$ we have $|f(x)-f(y)|<epsilon$, this means that $delta$ can be dependent from $epsilon$.

We choose $delta:=fracepsilon$ for $aneq 0$.
For $a=0$ we have to find a seperate $delta$. But this is easy, since if $a=0$, we just get $0$ and we can choose $delta=1$ for example. It does not matter.

Note, that this is all a scatch and not the proof yet. We just searched our $delta$.

Now for the proof:

Let $epsilon >0$ be arbitrary. If $a=0$ choose $delta=1$ and we get, that $|f(x)-f(y)|=|0cdot x+b-0cdot y-b|=0<epsilon$.

For $aneq 0$ choose $delta=fracepsilon>0$.

Then:

$|f(x)-f(y)|=|ax+b-(ay+b)|=dotso =|a||x-y|<|a|delta =|a|cdot fracepsilon=epsilon$

And we are done.

share|cite|improve this answer

edited Jul 25 at 12:51

answered Jul 25 at 12:46

Cornman

2,40021127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for taking the time to answer this, I'm currently working through Rudins Principles of Maths Analysis and you have been a huge help!
  – Robbie Meaney
  Jul 25 at 12:56
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

To show, that $f:mathbbRtomathbbR$, $xmapsto ax+b$ is continuous, we have to show, that $forallepsilon>0,existsdelta>0:forall xinmathbbR:|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon$

Simply said, we have to find a $delta$ for every $epsilon$.

Keep in mind, that we have $|x-y|<delta$ as an assumption.

You can find your $delta$ like this. First of all:

$|f(x)-f(y)|=|ax+b-(ay+b)|=|ax+b-ay-b|=|a(x-y)|=|a||x-y|$

We know, that $|x-y|<delta$ so we can stipulate like this:

$|a||x-y|<|a|delta$

We want to find for every $epsilon$ a $delta$ such that, when $|x-y|<delta$ we have $|f(x)-f(y)|<epsilon$, this means that $delta$ can be dependent from $epsilon$.

We choose $delta:=fracepsilon$ for $aneq 0$.
For $a=0$ we have to find a seperate $delta$. But this is easy, since if $a=0$, we just get $0$ and we can choose $delta=1$ for example. It does not matter.

Note, that this is all a scatch and not the proof yet. We just searched our $delta$.

Now for the proof:

Let $epsilon >0$ be arbitrary. If $a=0$ choose $delta=1$ and we get, that $|f(x)-f(y)|=|0cdot x+b-0cdot y-b|=0<epsilon$.

For $aneq 0$ choose $delta=fracepsilon>0$.

Then:

$|f(x)-f(y)|=|ax+b-(ay+b)|=dotso =|a||x-y|<|a|delta =|a|cdot fracepsilon=epsilon$

And we are done.

share|cite|improve this answer

edited Jul 25 at 12:51

answered Jul 25 at 12:46

Cornman

2,40021127

To show, that $f:mathbbRtomathbbR$, $xmapsto ax+b$ is continuous, we have to show, that $forallepsilon>0,existsdelta>0:forall xinmathbbR:|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon$

Simply said, we have to find a $delta$ for every $epsilon$.

Keep in mind, that we have $|x-y|<delta$ as an assumption.

You can find your $delta$ like this. First of all:

$|f(x)-f(y)|=|ax+b-(ay+b)|=|ax+b-ay-b|=|a(x-y)|=|a||x-y|$

We know, that $|x-y|<delta$ so we can stipulate like this:

$|a||x-y|<|a|delta$

We want to find for every $epsilon$ a $delta$ such that, when $|x-y|<delta$ we have $|f(x)-f(y)|<epsilon$, this means that $delta$ can be dependent from $epsilon$.

We choose $delta:=fracepsilon$ for $aneq 0$.
For $a=0$ we have to find a seperate $delta$. But this is easy, since if $a=0$, we just get $0$ and we can choose $delta=1$ for example. It does not matter.

Note, that this is all a scatch and not the proof yet. We just searched our $delta$.

Now for the proof:

Let $epsilon >0$ be arbitrary. If $a=0$ choose $delta=1$ and we get, that $|f(x)-f(y)|=|0cdot x+b-0cdot y-b|=0<epsilon$.

For $aneq 0$ choose $delta=fracepsilon>0$.

Then:

$|f(x)-f(y)|=|ax+b-(ay+b)|=dotso =|a||x-y|<|a|delta =|a|cdot fracepsilon=epsilon$

And we are done.

share|cite|improve this answer

edited Jul 25 at 12:51

answered Jul 25 at 12:46

Cornman

2,40021127

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 12:51

edited Jul 25 at 12:51

edited Jul 25 at 12:51

answered Jul 25 at 12:46

Cornman

2,40021127

answered Jul 25 at 12:46

Cornman

2,40021127

answered Jul 25 at 12:46

Cornman

2,40021127

2,40021127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for taking the time to answer this, I'm currently working through Rudins Principles of Maths Analysis and you have been a huge help!
  – Robbie Meaney
  Jul 25 at 12:56
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for taking the time to answer this, I'm currently working through Rudins Principles of Maths Analysis and you have been a huge help!
  – Robbie Meaney
  Jul 25 at 12:56
  

  
  
  
  

1

1

Thank you for taking the time to answer this, I'm currently working through Rudins Principles of Maths Analysis and you have been a huge help!
– Robbie Meaney
Jul 25 at 12:56

Thank you for taking the time to answer this, I'm currently working through Rudins Principles of Maths Analysis and you have been a huge help!
– Robbie Meaney
Jul 25 at 12:56

add a comment | 

up vote
2
down vote

Let $e>0$ be given. Let $d=frace2($. Then, $|x-c|<d$ implies
$$
|f(x)-f(c)|=|a|cdot|x-c|leq(1+|a|)d<e
$$
whether $a=0$ or not. Can you complete the argument now?

share|cite|improve this answer

edited Jul 26 at 16:29

answered Jul 25 at 12:30

yurnero

6,8401824

add a comment | 

up vote
2
down vote

Let $e>0$ be given. Let $d=frace2($. Then, $|x-c|<d$ implies
$$
|f(x)-f(c)|=|a|cdot|x-c|leq(1+|a|)d<e
$$
whether $a=0$ or not. Can you complete the argument now?

share|cite|improve this answer

edited Jul 26 at 16:29

answered Jul 25 at 12:30

yurnero

6,8401824

add a comment | 

up vote
2
down vote

up vote
2
down vote

Let $e>0$ be given. Let $d=frace2($. Then, $|x-c|<d$ implies
$$
|f(x)-f(c)|=|a|cdot|x-c|leq(1+|a|)d<e
$$
whether $a=0$ or not. Can you complete the argument now?

share|cite|improve this answer

edited Jul 26 at 16:29

answered Jul 25 at 12:30

yurnero

6,8401824

Let $e>0$ be given. Let $d=frace2($. Then, $|x-c|<d$ implies
$$
|f(x)-f(c)|=|a|cdot|x-c|leq(1+|a|)d<e
$$
whether $a=0$ or not. Can you complete the argument now?

share|cite|improve this answer

edited Jul 26 at 16:29

answered Jul 25 at 12:30

yurnero

6,8401824

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 16:29

edited Jul 26 at 16:29

edited Jul 26 at 16:29

answered Jul 25 at 12:30

yurnero

6,8401824

answered Jul 25 at 12:30

yurnero

6,8401824

answered Jul 25 at 12:30

yurnero

6,8401824

6,8401824

add a comment | 

add a comment | 

up vote
1
down vote

Your statement about continuity is really wrong.

In order to prove that $f$ is continuous at $c$, you need

for every $varepsilon>0$, there exists $delta>0$ such that, for every $x$ with $|x-c|<delta$, it is true that $|f(x)-f(c)|<varepsilon$.

Can you spot the difference with what you stated?

In order to prove continuity for your function, let $varepsilon>0$ be given. Then we want to solve the inequality $|f(x)-f(c)|<varepsilon$ and see whether the solution set has the above property. Since
$$
|f(x)-f(c)|=|ax+b-ac-b|=|a||x-c|
$$
we see that

1. if $a=0$, then every $delta>0$ fits;


2. if $ane0$, the inequality $|a||x-c|<varepsilon$ is equivalent to
   $$|x-c|<fracvarepsilon,$$ so we can take $delta=varepsilon/|a|$.

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answered Jul 25 at 12:45

egreg

164k1180187

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up vote
1
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Your statement about continuity is really wrong.

In order to prove that $f$ is continuous at $c$, you need

for every $varepsilon>0$, there exists $delta>0$ such that, for every $x$ with $|x-c|<delta$, it is true that $|f(x)-f(c)|<varepsilon$.

Can you spot the difference with what you stated?

In order to prove continuity for your function, let $varepsilon>0$ be given. Then we want to solve the inequality $|f(x)-f(c)|<varepsilon$ and see whether the solution set has the above property. Since
$$
|f(x)-f(c)|=|ax+b-ac-b|=|a||x-c|
$$
we see that

1. if $a=0$, then every $delta>0$ fits;


2. if $ane0$, the inequality $|a||x-c|<varepsilon$ is equivalent to
   $$|x-c|<fracvarepsilon,$$ so we can take $delta=varepsilon/|a|$.

share|cite|improve this answer

answered Jul 25 at 12:45

egreg

164k1180187

add a comment | 

up vote
1
down vote

up vote
1
down vote

Your statement about continuity is really wrong.

In order to prove that $f$ is continuous at $c$, you need

for every $varepsilon>0$, there exists $delta>0$ such that, for every $x$ with $|x-c|<delta$, it is true that $|f(x)-f(c)|<varepsilon$.

Can you spot the difference with what you stated?

In order to prove continuity for your function, let $varepsilon>0$ be given. Then we want to solve the inequality $|f(x)-f(c)|<varepsilon$ and see whether the solution set has the above property. Since
$$
|f(x)-f(c)|=|ax+b-ac-b|=|a||x-c|
$$
we see that

1. if $a=0$, then every $delta>0$ fits;


2. if $ane0$, the inequality $|a||x-c|<varepsilon$ is equivalent to
   $$|x-c|<fracvarepsilon,$$ so we can take $delta=varepsilon/|a|$.

share|cite|improve this answer

answered Jul 25 at 12:45

egreg

164k1180187

Your statement about continuity is really wrong.

In order to prove that $f$ is continuous at $c$, you need

for every $varepsilon>0$, there exists $delta>0$ such that, for every $x$ with $|x-c|<delta$, it is true that $|f(x)-f(c)|<varepsilon$.

Can you spot the difference with what you stated?

In order to prove continuity for your function, let $varepsilon>0$ be given. Then we want to solve the inequality $|f(x)-f(c)|<varepsilon$ and see whether the solution set has the above property. Since
$$
|f(x)-f(c)|=|ax+b-ac-b|=|a||x-c|
$$
we see that

1. if $a=0$, then every $delta>0$ fits;


2. if $ane0$, the inequality $|a||x-c|<varepsilon$ is equivalent to
   $$|x-c|<fracvarepsilon,$$ so we can take $delta=varepsilon/|a|$.

share|cite|improve this answer

answered Jul 25 at 12:45

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 12:45

egreg

164k1180187

answered Jul 25 at 12:45

egreg

164k1180187

answered Jul 25 at 12:45

egreg

164k1180187

164k1180187

add a comment | 

add a comment |