Suppose $f$ is defined and differentiable proof. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Derivatives and continuity of one variable functionsSuppose that $f: mathbb R^q to mathbb R^p$ is a linear map. Prove that $f$ is differentiable and that $f'(x) = f$ for every $x in mathbb R^q$Propositions concerning a differentiable function.Differentiable function and limitProve that $f(x)$ is defined and differentiable for $x>0$Suppose $f$ is differentiable at zero and $f(0) = 0$. Show that $f(x) = xg(x)$Calculating limit of the form of 1×inf - infSuppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)ne 0,forall xin(0,1)$Is the piecewise-defined function differentiablebounded differentiable functions
Product of Mrówka space and one point compactification discrete space.
What is "gratricide"?
Is there a kind of relay that only consumes power when switching?
How do living politicians protect their readily obtainable signatures from misuse?
How to unroll a parameter pack from right to left
Why do we bend a book to keep it straight?
If Windows 7 doesn't support WSL, then what does Linux subsystem option mean?
How do I find out the mythology and history of my Fortress?
Do wooden building fires get hotter than 600°C?
Putting class ranking in CV, but against dept guidelines
What is the meaning of 'breadth' in breadth first search?
How to dry out epoxy resin faster than usual?
What initially awakened the Balrog?
Belief In God or Knowledge Of God. Which is better?
How to draw/optimize this graph with tikz
Why weren't discrete x86 CPUs ever used in game hardware?
Parallel Computing Problem
Strange behavior of Object.defineProperty() in JavaScript
Sentence order: Where to put もう
Can anything be seen from the center of the Boötes void? How dark would it be?
How does the math work when buying airline miles?
Is it fair for a professor to grade us on the possession of past papers?
MLE of the unknown radius
What is the appropriate index architecture when forced to implement IsDeleted (soft deletes)?
Suppose $f$ is defined and differentiable proof.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Derivatives and continuity of one variable functionsSuppose that $f: mathbb R^q to mathbb R^p$ is a linear map. Prove that $f$ is differentiable and that $f'(x) = f$ for every $x in mathbb R^q$Propositions concerning a differentiable function.Differentiable function and limitProve that $f(x)$ is defined and differentiable for $x>0$Suppose $f$ is differentiable at zero and $f(0) = 0$. Show that $f(x) = xg(x)$Calculating limit of the form of 1×inf - infSuppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)ne 0,forall xin(0,1)$Is the piecewise-defined function differentiablebounded differentiable functions
$begingroup$
Suppose $f$ is defined and differentiable for every $x>0$, and $lim_xto infty f'(x)=0$.
Put $g(x)=f(x+1)-f(x)$. Prove that $lim_xto infty g(x)=0$
calculus analysis
edited Mar 27 at 20:31
Bernard
124k742117
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
$endgroup$
add a comment |
$begingroup$
Suppose $f$ is defined and differentiable for every $x>0$, and $lim_xto infty f'(x)=0$.
Put $g(x)=f(x+1)-f(x)$. Prove that $lim_xto infty g(x)=0$
calculus analysis
edited Mar 27 at 20:31
Bernard
124k742117
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
$endgroup$
$begingroup$
To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
$endgroup$
– avs
Mar 27 at 20:15
$begingroup$
You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
$endgroup$
– avs
Mar 28 at 5:54
add a comment |
$begingroup$
Suppose $f$ is defined and differentiable for every $x>0$, and $lim_xto infty f'(x)=0$.
Put $g(x)=f(x+1)-f(x)$. Prove that $lim_xto infty g(x)=0$
calculus analysis
edited Mar 27 at 20:31
Bernard
124k742117
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
$endgroup$
Suppose $f$ is defined and differentiable for every $x>0$, and $lim_xto infty f'(x)=0$.
Put $g(x)=f(x+1)-f(x)$. Prove that $lim_xto infty g(x)=0$
calculus analysis
calculus analysis
edited Mar 27 at 20:31
Bernard
124k742117
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
edited Mar 27 at 20:31
Bernard
124k742117
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
edited Mar 27 at 20:31
Bernard
124k742117
edited Mar 27 at 20:31
Bernard
124k742117
edited Mar 27 at 20:31
Bernard
124k742117
124k742117
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
asked Mar 27 at 20:11
BernhardRiemannBernhardRiemann
12
12
$begingroup$
To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
$endgroup$
– avs
Mar 27 at 20:15
$begingroup$
You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
$endgroup$
– avs
Mar 28 at 5:54
add a comment |
$begingroup$
To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
$endgroup$
– avs
Mar 27 at 20:15
$begingroup$
You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
$endgroup$
– avs
Mar 28 at 5:54
$begingroup$
To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
$endgroup$
– avs
Mar 27 at 20:15
$begingroup$
To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
$endgroup$
– avs
Mar 27 at 20:15
$begingroup$
You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
$endgroup$
– avs
Mar 28 at 5:54
$begingroup$
You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
$endgroup$
– avs
Mar 28 at 5:54
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.
answered Mar 27 at 20:16
avsavs
4,280515
$endgroup$
$begingroup$
i am confused because our professor said the derivative use the definition
$endgroup$
– BernhardRiemann
Mar 27 at 20:20
$begingroup$
sorry, not understanding your comment.
$endgroup$
– avs
Mar 27 at 21:33
$begingroup$
For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
$endgroup$
– avs
Mar 28 at 16:19
add a comment |
$begingroup$
Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.
So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.
Using the mean-value theorem I think you can show that the same M works for |g(x)| too.
answered Mar 27 at 20:47
bluebirdbluebird
214
$endgroup$
$begingroup$
Yes , our professor mentioned about exactly but I did not understand anything
$endgroup$
– BernhardRiemann
Mar 27 at 21:23
add a comment |
$begingroup$
By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.
If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
$endgroup$
add a comment |
$begingroup$
MVT:
$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.
We have $lim_x rightarrow infty f'(x)=0:$
Let $epsilon >0$ be given.
There is a $M >0$, real, s.t. for $x >M$:
$|f'(x)| < epsilon.$
For $x >M$:
Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have
$|g(x)| =|f'(t_x)| < epsilon,$ i .e
$lim_x rightarrow inftyg(x)=0.$
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
The proof is simple:
$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$
For large $x$, $x+1 approx x$, so, by the previous result,
$lim_x rightarrow infty g(x)=0$
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
$endgroup$
$begingroup$
One has to also prove that $lim_xrightarrow infty f(x)$ exists!
$endgroup$
– folouer of kaklas
Mar 27 at 21:30
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3165068%2fsuppose-f-is-defined-and-differentiable-proof%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.
answered Mar 27 at 20:16
avsavs
4,280515
$endgroup$
$begingroup$
i am confused because our professor said the derivative use the definition
$endgroup$
– BernhardRiemann
Mar 27 at 20:20
$begingroup$
sorry, not understanding your comment.
$endgroup$
– avs
Mar 27 at 21:33
$begingroup$
For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
$endgroup$
– avs
Mar 28 at 16:19
add a comment |
$begingroup$
Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.
answered Mar 27 at 20:16
avsavs
4,280515
$endgroup$
$begingroup$
i am confused because our professor said the derivative use the definition
$endgroup$
– BernhardRiemann
Mar 27 at 20:20
$begingroup$
sorry, not understanding your comment.
$endgroup$
– avs
Mar 27 at 21:33
$begingroup$
For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
$endgroup$
– avs
Mar 28 at 16:19
add a comment |
$begingroup$
Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.
answered Mar 27 at 20:16
avsavs
4,280515
$endgroup$
Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.
answered Mar 27 at 20:16
avsavs
4,280515
answered Mar 27 at 20:16
avsavs
4,280515
answered Mar 27 at 20:16
avsavs
4,280515
answered Mar 27 at 20:16
avsavs
4,280515
4,280515
$begingroup$
i am confused because our professor said the derivative use the definition
$endgroup$
– BernhardRiemann
Mar 27 at 20:20
$begingroup$
sorry, not understanding your comment.
$endgroup$
– avs
Mar 27 at 21:33
$begingroup$
For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
$endgroup$
– avs
Mar 28 at 16:19
add a comment |
$begingroup$
i am confused because our professor said the derivative use the definition
$endgroup$
– BernhardRiemann
Mar 27 at 20:20
$begingroup$
sorry, not understanding your comment.
$endgroup$
– avs
Mar 27 at 21:33
$begingroup$
For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
$endgroup$
– avs
Mar 28 at 16:19
$begingroup$
i am confused because our professor said the derivative use the definition
$endgroup$
– BernhardRiemann
Mar 27 at 20:20
$begingroup$
i am confused because our professor said the derivative use the definition
$endgroup$
– BernhardRiemann
Mar 27 at 20:20
$begingroup$
sorry, not understanding your comment.
$endgroup$
– avs
Mar 27 at 21:33
$begingroup$
sorry, not understanding your comment.
$endgroup$
– avs
Mar 27 at 21:33
$begingroup$
For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
$endgroup$
– avs
Mar 28 at 16:19
$begingroup$
For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
$endgroup$
– avs
Mar 28 at 16:19
add a comment |
$begingroup$
Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.
So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.
Using the mean-value theorem I think you can show that the same M works for |g(x)| too.
answered Mar 27 at 20:47
bluebirdbluebird
214
$endgroup$
$begingroup$
Yes , our professor mentioned about exactly but I did not understand anything
$endgroup$
– BernhardRiemann
Mar 27 at 21:23
add a comment |
$begingroup$
Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.
So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.
Using the mean-value theorem I think you can show that the same M works for |g(x)| too.
answered Mar 27 at 20:47
bluebirdbluebird
214
$endgroup$
$begingroup$
Yes , our professor mentioned about exactly but I did not understand anything
$endgroup$
– BernhardRiemann
Mar 27 at 21:23
add a comment |
$begingroup$
Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.
So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.
Using the mean-value theorem I think you can show that the same M works for |g(x)| too.
answered Mar 27 at 20:47
bluebirdbluebird
214
$endgroup$
Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.
So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.
Using the mean-value theorem I think you can show that the same M works for |g(x)| too.
answered Mar 27 at 20:47
bluebirdbluebird
214
answered Mar 27 at 20:47
bluebirdbluebird
214
answered Mar 27 at 20:47
bluebirdbluebird
214
answered Mar 27 at 20:47
bluebirdbluebird
214
214
$begingroup$
Yes , our professor mentioned about exactly but I did not understand anything
$endgroup$
– BernhardRiemann
Mar 27 at 21:23
add a comment |
$begingroup$
Yes , our professor mentioned about exactly but I did not understand anything
$endgroup$
– BernhardRiemann
Mar 27 at 21:23
$begingroup$
Yes , our professor mentioned about exactly but I did not understand anything
$endgroup$
– BernhardRiemann
Mar 27 at 21:23
$begingroup$
Yes , our professor mentioned about exactly but I did not understand anything
$endgroup$
– BernhardRiemann
Mar 27 at 21:23
add a comment |
$begingroup$
By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.
If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
$endgroup$
add a comment |
$begingroup$
By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.
If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
$endgroup$
add a comment |
$begingroup$
By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.
If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
$endgroup$
By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.
If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
answered Mar 27 at 21:37
folouer of kaklasfolouer of kaklas
338110
338110
add a comment |
add a comment |
$begingroup$
MVT:
$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.
We have $lim_x rightarrow infty f'(x)=0:$
Let $epsilon >0$ be given.
There is a $M >0$, real, s.t. for $x >M$:
$|f'(x)| < epsilon.$
For $x >M$:
Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have
$|g(x)| =|f'(t_x)| < epsilon,$ i .e
$lim_x rightarrow inftyg(x)=0.$
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
MVT:
$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.
We have $lim_x rightarrow infty f'(x)=0:$
Let $epsilon >0$ be given.
There is a $M >0$, real, s.t. for $x >M$:
$|f'(x)| < epsilon.$
For $x >M$:
Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have
$|g(x)| =|f'(t_x)| < epsilon,$ i .e
$lim_x rightarrow inftyg(x)=0.$
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
MVT:
$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.
We have $lim_x rightarrow infty f'(x)=0:$
Let $epsilon >0$ be given.
There is a $M >0$, real, s.t. for $x >M$:
$|f'(x)| < epsilon.$
For $x >M$:
Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have
$|g(x)| =|f'(t_x)| < epsilon,$ i .e
$lim_x rightarrow inftyg(x)=0.$
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
$endgroup$
MVT:
$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.
We have $lim_x rightarrow infty f'(x)=0:$
Let $epsilon >0$ be given.
There is a $M >0$, real, s.t. for $x >M$:
$|f'(x)| < epsilon.$
For $x >M$:
Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have
$|g(x)| =|f'(t_x)| < epsilon,$ i .e
$lim_x rightarrow inftyg(x)=0.$
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
edited Mar 28 at 4:29
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
answered Mar 27 at 22:27
Peter SzilasPeter Szilas
12k2822
12k2822
add a comment |
add a comment |
$begingroup$
The proof is simple:
$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$
For large $x$, $x+1 approx x$, so, by the previous result,
$lim_x rightarrow infty g(x)=0$
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
$endgroup$
$begingroup$
One has to also prove that $lim_xrightarrow infty f(x)$ exists!
$endgroup$
– folouer of kaklas
Mar 27 at 21:30
add a comment |
$begingroup$
The proof is simple:
$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$
For large $x$, $x+1 approx x$, so, by the previous result,
$lim_x rightarrow infty g(x)=0$
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
$endgroup$
$begingroup$
One has to also prove that $lim_xrightarrow infty f(x)$ exists!
$endgroup$
– folouer of kaklas
Mar 27 at 21:30
add a comment |
$begingroup$
The proof is simple:
$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$
For large $x$, $x+1 approx x$, so, by the previous result,
$lim_x rightarrow infty g(x)=0$
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
$endgroup$
The proof is simple:
$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$
For large $x$, $x+1 approx x$, so, by the previous result,
$lim_x rightarrow infty g(x)=0$
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
answered Mar 27 at 21:24
officialnoriaofficialnoria
112
112
$begingroup$
One has to also prove that $lim_xrightarrow infty f(x)$ exists!
$endgroup$
– folouer of kaklas
Mar 27 at 21:30
add a comment |
$begingroup$
One has to also prove that $lim_xrightarrow infty f(x)$ exists!
$endgroup$
– folouer of kaklas
Mar 27 at 21:30
$begingroup$
One has to also prove that $lim_xrightarrow infty f(x)$ exists!
$endgroup$
– folouer of kaklas
Mar 27 at 21:30
$begingroup$
One has to also prove that $lim_xrightarrow infty f(x)$ exists!
$endgroup$
– folouer of kaklas
Mar 27 at 21:30
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3165068%2fsuppose-f-is-defined-and-differentiable-proof%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
$endgroup$
– avs
Mar 27 at 20:15
$begingroup$
You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
$endgroup$
– avs
Mar 28 at 5:54