Let $f:(0,infty)tomathbb{R}$ be defined by $ f(x)=frac{sin(x^{3})}{x}$. Then f is not bounded and not...












0












$begingroup$


Let $f:(0,infty)tomathbb{R}$ be defined by $ f(x)=frac{sin(x^{3})}{x}$. Then which of the following is correct:



a)f is not bounded and not uniformly continuous



b)f is bounded and not uniformly continuous



c)f is not bounded and uniformly continuous



d)f is bounded and uniformly continuous



I think option a is correct $because sin{x}$ is bounded between $-1$ and $1$ and $frac{1}{x}$ approches $infty$ in neighborhood of zero.



This question was asked in TIFR 2019.






real-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No, option (a) is not correct; the numerator vanishes at $0$ as well, and does so faster than the denominator.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 18:04






  • 1




    $begingroup$
    yes , you are right then this becomes unbounded in $(0,infty)$
    $endgroup$
    – sejy
    Dec 10 '18 at 18:06






  • 2




    $begingroup$
    @sejy No it doesn't. Think about what T. Bongers said - how could $sin(x^3)/x$ be unbounded in $(0,+infty)$. Can you show me an $x$ such that $sin(x^3)/x=2$?
    $endgroup$
    – Jam
    Dec 10 '18 at 18:12








  • 1




    $begingroup$
    $-frac{1}{x} < frac{sin{x^3}}{x} < frac{1}{x}$ on $[1,infty]$. All that's left is that pesky $(0,1)$ region.
    $endgroup$
    – David Diaz
    Dec 10 '18 at 18:22






  • 1




    $begingroup$
    A lot of confusion here. This function is very well behaved, in fact $int_0^infty frac{sin x^3}{x} , dx = frac{pi}{6}$
    $endgroup$
    – RRL
    Dec 10 '18 at 18:53
















0












$begingroup$


Let $f:(0,infty)tomathbb{R}$ be defined by $ f(x)=frac{sin(x^{3})}{x}$. Then which of the following is correct:



a)f is not bounded and not uniformly continuous



b)f is bounded and not uniformly continuous



c)f is not bounded and uniformly continuous



d)f is bounded and uniformly continuous



I think option a is correct $because sin{x}$ is bounded between $-1$ and $1$ and $frac{1}{x}$ approches $infty$ in neighborhood of zero.



This question was asked in TIFR 2019.






real-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No, option (a) is not correct; the numerator vanishes at $0$ as well, and does so faster than the denominator.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 18:04






  • 1




    $begingroup$
    yes , you are right then this becomes unbounded in $(0,infty)$
    $endgroup$
    – sejy
    Dec 10 '18 at 18:06






  • 2




    $begingroup$
    @sejy No it doesn't. Think about what T. Bongers said - how could $sin(x^3)/x$ be unbounded in $(0,+infty)$. Can you show me an $x$ such that $sin(x^3)/x=2$?
    $endgroup$
    – Jam
    Dec 10 '18 at 18:12








  • 1




    $begingroup$
    $-frac{1}{x} < frac{sin{x^3}}{x} < frac{1}{x}$ on $[1,infty]$. All that's left is that pesky $(0,1)$ region.
    $endgroup$
    – David Diaz
    Dec 10 '18 at 18:22






  • 1




    $begingroup$
    A lot of confusion here. This function is very well behaved, in fact $int_0^infty frac{sin x^3}{x} , dx = frac{pi}{6}$
    $endgroup$
    – RRL
    Dec 10 '18 at 18:53














0












0








0





$begingroup$


Let $f:(0,infty)tomathbb{R}$ be defined by $ f(x)=frac{sin(x^{3})}{x}$. Then which of the following is correct:



a)f is not bounded and not uniformly continuous



b)f is bounded and not uniformly continuous



c)f is not bounded and uniformly continuous



d)f is bounded and uniformly continuous



I think option a is correct $because sin{x}$ is bounded between $-1$ and $1$ and $frac{1}{x}$ approches $infty$ in neighborhood of zero.



This question was asked in TIFR 2019.






real-analysis







share|cite|improve this question











$endgroup$




Let $f:(0,infty)tomathbb{R}$ be defined by $ f(x)=frac{sin(x^{3})}{x}$. Then which of the following is correct:



a)f is not bounded and not uniformly continuous



b)f is bounded and not uniformly continuous



c)f is not bounded and uniformly continuous



d)f is bounded and uniformly continuous



I think option a is correct $because sin{x}$ is bounded between $-1$ and $1$ and $frac{1}{x}$ approches $infty$ in neighborhood of zero.



This question was asked in TIFR 2019.






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 18:10









Andrei

11.8k21026




11.8k21026










asked Dec 10 '18 at 18:02









sejysejy

1469




1469








  • 3




    $begingroup$
    No, option (a) is not correct; the numerator vanishes at $0$ as well, and does so faster than the denominator.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 18:04






  • 1




    $begingroup$
    yes , you are right then this becomes unbounded in $(0,infty)$
    $endgroup$
    – sejy
    Dec 10 '18 at 18:06






  • 2




    $begingroup$
    @sejy No it doesn't. Think about what T. Bongers said - how could $sin(x^3)/x$ be unbounded in $(0,+infty)$. Can you show me an $x$ such that $sin(x^3)/x=2$?
    $endgroup$
    – Jam
    Dec 10 '18 at 18:12








  • 1




    $begingroup$
    $-frac{1}{x} < frac{sin{x^3}}{x} < frac{1}{x}$ on $[1,infty]$. All that's left is that pesky $(0,1)$ region.
    $endgroup$
    – David Diaz
    Dec 10 '18 at 18:22






  • 1




    $begingroup$
    A lot of confusion here. This function is very well behaved, in fact $int_0^infty frac{sin x^3}{x} , dx = frac{pi}{6}$
    $endgroup$
    – RRL
    Dec 10 '18 at 18:53














  • 3




    $begingroup$
    No, option (a) is not correct; the numerator vanishes at $0$ as well, and does so faster than the denominator.
    $endgroup$
    – T. Bongers
    Dec 10 '18 at 18:04






  • 1




    $begingroup$
    yes , you are right then this becomes unbounded in $(0,infty)$
    $endgroup$
    – sejy
    Dec 10 '18 at 18:06






  • 2




    $begingroup$
    @sejy No it doesn't. Think about what T. Bongers said - how could $sin(x^3)/x$ be unbounded in $(0,+infty)$. Can you show me an $x$ such that $sin(x^3)/x=2$?
    $endgroup$
    – Jam
    Dec 10 '18 at 18:12








  • 1




    $begingroup$
    $-frac{1}{x} < frac{sin{x^3}}{x} < frac{1}{x}$ on $[1,infty]$. All that's left is that pesky $(0,1)$ region.
    $endgroup$
    – David Diaz
    Dec 10 '18 at 18:22






  • 1




    $begingroup$
    A lot of confusion here. This function is very well behaved, in fact $int_0^infty frac{sin x^3}{x} , dx = frac{pi}{6}$
    $endgroup$
    – RRL
    Dec 10 '18 at 18:53








3




3




$begingroup$
No, option (a) is not correct; the numerator vanishes at $0$ as well, and does so faster than the denominator.
$endgroup$
– T. Bongers
Dec 10 '18 at 18:04




$begingroup$
No, option (a) is not correct; the numerator vanishes at $0$ as well, and does so faster than the denominator.
$endgroup$
– T. Bongers
Dec 10 '18 at 18:04




1




1




$begingroup$
yes , you are right then this becomes unbounded in $(0,infty)$
$endgroup$
– sejy
Dec 10 '18 at 18:06




$begingroup$
yes , you are right then this becomes unbounded in $(0,infty)$
$endgroup$
– sejy
Dec 10 '18 at 18:06




2




2




$begingroup$
@sejy No it doesn't. Think about what T. Bongers said - how could $sin(x^3)/x$ be unbounded in $(0,+infty)$. Can you show me an $x$ such that $sin(x^3)/x=2$?
$endgroup$
– Jam
Dec 10 '18 at 18:12






$begingroup$
@sejy No it doesn't. Think about what T. Bongers said - how could $sin(x^3)/x$ be unbounded in $(0,+infty)$. Can you show me an $x$ such that $sin(x^3)/x=2$?
$endgroup$
– Jam
Dec 10 '18 at 18:12






1




1




$begingroup$
$-frac{1}{x} < frac{sin{x^3}}{x} < frac{1}{x}$ on $[1,infty]$. All that's left is that pesky $(0,1)$ region.
$endgroup$
– David Diaz
Dec 10 '18 at 18:22




$begingroup$
$-frac{1}{x} < frac{sin{x^3}}{x} < frac{1}{x}$ on $[1,infty]$. All that's left is that pesky $(0,1)$ region.
$endgroup$
– David Diaz
Dec 10 '18 at 18:22




1




1




$begingroup$
A lot of confusion here. This function is very well behaved, in fact $int_0^infty frac{sin x^3}{x} , dx = frac{pi}{6}$
$endgroup$
– RRL
Dec 10 '18 at 18:53




$begingroup$
A lot of confusion here. This function is very well behaved, in fact $int_0^infty frac{sin x^3}{x} , dx = frac{pi}{6}$
$endgroup$
– RRL
Dec 10 '18 at 18:53










1 Answer
1






active

oldest

votes


















4












$begingroup$

The function is bounded and uniformly continuous on $(0,infty)$.



Clearly, $f$ is continuous and, hence, uniformly continuous on any compact interval $[a,b]$ with $a > 0$.



On the interval $(0,a]$ we have $displaystyle f(x) = frac{sin x^3}{x} = x^2frac{sin x^3}{x^3} to 0cdot 1 = 0 $ as $x to 0$ and
$f$ is extendible as a continuous function to the compact interval $[0,a]$, and, hence, uniformly continuous there.



On $[b, infty)$, $f$ is uniformly continuous as well since $displaystyle |f(x)| = frac{|sin x^3|}{x} leqslant frac{1}{x} to 0 $ as $x to infty$.



A continuous function that approaches a finite limit as $x to infty$ must be uniformly continuous -- proved many times on this site -- for example here. This is also an interesting example of a function with an unbounded derivative that is uniformly continuous.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @hamam_Abdallah I mean, it is trivial to see that near $0$ you have $sin(x^3)/xsim x^3/x=x^2$, so it is definitely bounded there
    $endgroup$
    – Federico
    Dec 10 '18 at 19:08








  • 1




    $begingroup$
    @Federico You are right. before answering, i had to think twice.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 19:10






  • 1




    $begingroup$
    @hamam_Abdallah Hey, cheer up! ;)
    $endgroup$
    – Federico
    Dec 10 '18 at 19:12











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034266%2flet-f0-infty-to-mathbbr-be-defined-by-fx-frac-sinx3x-the%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The function is bounded and uniformly continuous on $(0,infty)$.



Clearly, $f$ is continuous and, hence, uniformly continuous on any compact interval $[a,b]$ with $a > 0$.



On the interval $(0,a]$ we have $displaystyle f(x) = frac{sin x^3}{x} = x^2frac{sin x^3}{x^3} to 0cdot 1 = 0 $ as $x to 0$ and
$f$ is extendible as a continuous function to the compact interval $[0,a]$, and, hence, uniformly continuous there.



On $[b, infty)$, $f$ is uniformly continuous as well since $displaystyle |f(x)| = frac{|sin x^3|}{x} leqslant frac{1}{x} to 0 $ as $x to infty$.



A continuous function that approaches a finite limit as $x to infty$ must be uniformly continuous -- proved many times on this site -- for example here. This is also an interesting example of a function with an unbounded derivative that is uniformly continuous.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @hamam_Abdallah I mean, it is trivial to see that near $0$ you have $sin(x^3)/xsim x^3/x=x^2$, so it is definitely bounded there
    $endgroup$
    – Federico
    Dec 10 '18 at 19:08








  • 1




    $begingroup$
    @Federico You are right. before answering, i had to think twice.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 19:10






  • 1




    $begingroup$
    @hamam_Abdallah Hey, cheer up! ;)
    $endgroup$
    – Federico
    Dec 10 '18 at 19:12
















4












$begingroup$

The function is bounded and uniformly continuous on $(0,infty)$.



Clearly, $f$ is continuous and, hence, uniformly continuous on any compact interval $[a,b]$ with $a > 0$.



On the interval $(0,a]$ we have $displaystyle f(x) = frac{sin x^3}{x} = x^2frac{sin x^3}{x^3} to 0cdot 1 = 0 $ as $x to 0$ and
$f$ is extendible as a continuous function to the compact interval $[0,a]$, and, hence, uniformly continuous there.



On $[b, infty)$, $f$ is uniformly continuous as well since $displaystyle |f(x)| = frac{|sin x^3|}{x} leqslant frac{1}{x} to 0 $ as $x to infty$.



A continuous function that approaches a finite limit as $x to infty$ must be uniformly continuous -- proved many times on this site -- for example here. This is also an interesting example of a function with an unbounded derivative that is uniformly continuous.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @hamam_Abdallah I mean, it is trivial to see that near $0$ you have $sin(x^3)/xsim x^3/x=x^2$, so it is definitely bounded there
    $endgroup$
    – Federico
    Dec 10 '18 at 19:08








  • 1




    $begingroup$
    @Federico You are right. before answering, i had to think twice.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 19:10






  • 1




    $begingroup$
    @hamam_Abdallah Hey, cheer up! ;)
    $endgroup$
    – Federico
    Dec 10 '18 at 19:12














4












4








4





$begingroup$

The function is bounded and uniformly continuous on $(0,infty)$.



Clearly, $f$ is continuous and, hence, uniformly continuous on any compact interval $[a,b]$ with $a > 0$.



On the interval $(0,a]$ we have $displaystyle f(x) = frac{sin x^3}{x} = x^2frac{sin x^3}{x^3} to 0cdot 1 = 0 $ as $x to 0$ and
$f$ is extendible as a continuous function to the compact interval $[0,a]$, and, hence, uniformly continuous there.



On $[b, infty)$, $f$ is uniformly continuous as well since $displaystyle |f(x)| = frac{|sin x^3|}{x} leqslant frac{1}{x} to 0 $ as $x to infty$.



A continuous function that approaches a finite limit as $x to infty$ must be uniformly continuous -- proved many times on this site -- for example here. This is also an interesting example of a function with an unbounded derivative that is uniformly continuous.






share|cite|improve this answer











$endgroup$



The function is bounded and uniformly continuous on $(0,infty)$.



Clearly, $f$ is continuous and, hence, uniformly continuous on any compact interval $[a,b]$ with $a > 0$.



On the interval $(0,a]$ we have $displaystyle f(x) = frac{sin x^3}{x} = x^2frac{sin x^3}{x^3} to 0cdot 1 = 0 $ as $x to 0$ and
$f$ is extendible as a continuous function to the compact interval $[0,a]$, and, hence, uniformly continuous there.



On $[b, infty)$, $f$ is uniformly continuous as well since $displaystyle |f(x)| = frac{|sin x^3|}{x} leqslant frac{1}{x} to 0 $ as $x to infty$.



A continuous function that approaches a finite limit as $x to infty$ must be uniformly continuous -- proved many times on this site -- for example here. This is also an interesting example of a function with an unbounded derivative that is uniformly continuous.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 19:00

























answered Dec 10 '18 at 18:50









RRLRRL

50.6k42573




50.6k42573








  • 1




    $begingroup$
    @hamam_Abdallah I mean, it is trivial to see that near $0$ you have $sin(x^3)/xsim x^3/x=x^2$, so it is definitely bounded there
    $endgroup$
    – Federico
    Dec 10 '18 at 19:08








  • 1




    $begingroup$
    @Federico You are right. before answering, i had to think twice.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 19:10






  • 1




    $begingroup$
    @hamam_Abdallah Hey, cheer up! ;)
    $endgroup$
    – Federico
    Dec 10 '18 at 19:12














  • 1




    $begingroup$
    @hamam_Abdallah I mean, it is trivial to see that near $0$ you have $sin(x^3)/xsim x^3/x=x^2$, so it is definitely bounded there
    $endgroup$
    – Federico
    Dec 10 '18 at 19:08








  • 1




    $begingroup$
    @Federico You are right. before answering, i had to think twice.
    $endgroup$
    – hamam_Abdallah
    Dec 10 '18 at 19:10






  • 1




    $begingroup$
    @hamam_Abdallah Hey, cheer up! ;)
    $endgroup$
    – Federico
    Dec 10 '18 at 19:12








1




1




$begingroup$
@hamam_Abdallah I mean, it is trivial to see that near $0$ you have $sin(x^3)/xsim x^3/x=x^2$, so it is definitely bounded there
$endgroup$
– Federico
Dec 10 '18 at 19:08






$begingroup$
@hamam_Abdallah I mean, it is trivial to see that near $0$ you have $sin(x^3)/xsim x^3/x=x^2$, so it is definitely bounded there
$endgroup$
– Federico
Dec 10 '18 at 19:08






1




1




$begingroup$
@Federico You are right. before answering, i had to think twice.
$endgroup$
– hamam_Abdallah
Dec 10 '18 at 19:10




$begingroup$
@Federico You are right. before answering, i had to think twice.
$endgroup$
– hamam_Abdallah
Dec 10 '18 at 19:10




1




1




$begingroup$
@hamam_Abdallah Hey, cheer up! ;)
$endgroup$
– Federico
Dec 10 '18 at 19:12




$begingroup$
@hamam_Abdallah Hey, cheer up! ;)
$endgroup$
– Federico
Dec 10 '18 at 19:12


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034266%2flet-f0-infty-to-mathbbr-be-defined-by-fx-frac-sinx3x-the%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Wiesbaden

Image
Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
Read more

Marschland

Image
Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more