Proof of L'Hopital's rule ($x_0 notin D$)












0












$begingroup$


Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)



We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.



Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.



$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)



If we let $n to infty$ we get



$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$



$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$






calculus limits derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    You probably mean L'Hopital's rule
    $endgroup$
    – roman
    Jan 3 at 8:39










  • $begingroup$
    @roman: L'Hôpital's rule, to be precise :)
    $endgroup$
    – Martin R
    Jan 3 at 8:43










  • $begingroup$
    How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
    $endgroup$
    – 5xum
    Jan 3 at 8:47










  • $begingroup$
    @5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:50


















0












$begingroup$


Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)



We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.



Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.



$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)



If we let $n to infty$ we get



$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$



$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$






calculus limits derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    You probably mean L'Hopital's rule
    $endgroup$
    – roman
    Jan 3 at 8:39










  • $begingroup$
    @roman: L'Hôpital's rule, to be precise :)
    $endgroup$
    – Martin R
    Jan 3 at 8:43










  • $begingroup$
    How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
    $endgroup$
    – 5xum
    Jan 3 at 8:47










  • $begingroup$
    @5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:50
















0












0








0





$begingroup$


Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)



We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.



Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.



$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)



If we let $n to infty$ we get



$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$



$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$






calculus limits derivatives







share|cite|improve this question











$endgroup$




Is this a rigorous proof for Hospital rule? (case $lim_{x to x_0} f(x) = g(x) = 0$)



We suppose that $x_0$ doesn't belong to the domain of $f(x),g(x)$, but it is a limit point.



Since $x_0$ is a limit point for $f(x),f(x)$, there is a sequence $x_n: lim_{n to infty} x_n = x_0$.



$frac{f(x) - f(x_n)}{g(x) - g(x_n)} = frac{f'(a_n)}{g'(a_n)}$ $quad$ Cauchy's mean value theorem (with $a_n in (x,x_n) ,,forall , n$)



If we let $n to infty$ we get



$frac{f(x) - 0}{g(x) - 0)} = frac{f(x)}{g(x)} = frac{f'(a_n)}{g'(a_n)}$ $quad a_n in (x,x_0)$



$lim_{x to x_0} frac{f(x)}{g(x)} = frac{f'(x_0)}{g'(x_0)}$ $quad$ because $a_n$ is "squeezed" between $x$ and $x_0$, and $x$ is going to $x_0$






calculus limits derivatives




calculus limits derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 8:42







Francesco Andreuzzi

















asked Jan 3 at 8:36









Francesco AndreuzziFrancesco Andreuzzi

63




63












  • $begingroup$
    You probably mean L'Hopital's rule
    $endgroup$
    – roman
    Jan 3 at 8:39










  • $begingroup$
    @roman: L'Hôpital's rule, to be precise :)
    $endgroup$
    – Martin R
    Jan 3 at 8:43










  • $begingroup$
    How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
    $endgroup$
    – 5xum
    Jan 3 at 8:47










  • $begingroup$
    @5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:50




















  • $begingroup$
    You probably mean L'Hopital's rule
    $endgroup$
    – roman
    Jan 3 at 8:39










  • $begingroup$
    @roman: L'Hôpital's rule, to be precise :)
    $endgroup$
    – Martin R
    Jan 3 at 8:43










  • $begingroup$
    How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
    $endgroup$
    – 5xum
    Jan 3 at 8:47










  • $begingroup$
    @5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:50


















$begingroup$
You probably mean L'Hopital's rule
$endgroup$
– roman
Jan 3 at 8:39




$begingroup$
You probably mean L'Hopital's rule
$endgroup$
– roman
Jan 3 at 8:39












$begingroup$
@roman: L'Hôpital's rule, to be precise :)
$endgroup$
– Martin R
Jan 3 at 8:43




$begingroup$
@roman: L'Hôpital's rule, to be precise :)
$endgroup$
– Martin R
Jan 3 at 8:43












$begingroup$
How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
$endgroup$
– 5xum
Jan 3 at 8:47




$begingroup$
How do you know that $x_n>x$? If you don't know that, then how do you know $(x,x_n)$ is not an empty set?
$endgroup$
– 5xum
Jan 3 at 8:47












$begingroup$
@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:50






$begingroup$
@5xum isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:50












1 Answer
1






active

oldest

votes


















1












$begingroup$

The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:




  1. You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.

  2. You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:53










  • $begingroup$
    @FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
    $endgroup$
    – 5xum
    Jan 3 at 8:54










  • $begingroup$
    2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:55










  • $begingroup$
    @FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
    $endgroup$
    – 5xum
    Jan 3 at 8:56






  • 1




    $begingroup$
    @FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
    $endgroup$
    – 5xum
    Jan 3 at 9:03












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%2f3060366%2fproof-of-lhopitals-rule-x-0-not-in-d%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









1












$begingroup$

The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:




  1. You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.

  2. You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:53










  • $begingroup$
    @FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
    $endgroup$
    – 5xum
    Jan 3 at 8:54










  • $begingroup$
    2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:55










  • $begingroup$
    @FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
    $endgroup$
    – 5xum
    Jan 3 at 8:56






  • 1




    $begingroup$
    @FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
    $endgroup$
    – 5xum
    Jan 3 at 9:03
















1












$begingroup$

The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:




  1. You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.

  2. You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:53










  • $begingroup$
    @FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
    $endgroup$
    – 5xum
    Jan 3 at 8:54










  • $begingroup$
    2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:55










  • $begingroup$
    @FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
    $endgroup$
    – 5xum
    Jan 3 at 8:56






  • 1




    $begingroup$
    @FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
    $endgroup$
    – 5xum
    Jan 3 at 9:03














1












1








1





$begingroup$

The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:




  1. You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.

  2. You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?






share|cite|improve this answer









$endgroup$



The proof is a fine idea, but it is not a rigorous proof as it has some hand-waving and sloppiness in it. Two issues are:




  1. You say that $a_n=(x,x_n)$ for all $n$, however there is no reason for a reader at that point to assume that $x_n>x$, and therefore, the interval $(x,x_n)$ might be empty.

  2. You say that you "let $ntoinfty$", and get $frac{f(x)}{g(x)}=frac{f'(a_n)}{g'(a_n)}$. There are two issues here. One is that saying "let $ntoinfty$" is not particularly rigorous in itself. Two, if you sent $n$ to $infty$, then how can you get an equality which still depends on $n$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 8:51









5xum5xum

91.8k394161




91.8k394161












  • $begingroup$
    1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:53










  • $begingroup$
    @FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
    $endgroup$
    – 5xum
    Jan 3 at 8:54










  • $begingroup$
    2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:55










  • $begingroup$
    @FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
    $endgroup$
    – 5xum
    Jan 3 at 8:56






  • 1




    $begingroup$
    @FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
    $endgroup$
    – 5xum
    Jan 3 at 9:03


















  • $begingroup$
    1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:53










  • $begingroup$
    @FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
    $endgroup$
    – 5xum
    Jan 3 at 8:54










  • $begingroup$
    2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
    $endgroup$
    – Francesco Andreuzzi
    Jan 3 at 8:55










  • $begingroup$
    @FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
    $endgroup$
    – 5xum
    Jan 3 at 8:56






  • 1




    $begingroup$
    @FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
    $endgroup$
    – 5xum
    Jan 3 at 9:03
















$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53




$begingroup$
1. isn't $x$ an arbitrary value? Since $x_n$ converges, it's also bounded, and so I could choose $x leq inf x_n$
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:53












$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54




$begingroup$
@FrancescoAndreuzzi Well I don't know what $x$ is, to be honest. A more rigorous write up might help you avoid such unclarities.
$endgroup$
– 5xum
Jan 3 at 8:54












$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55




$begingroup$
2. I don't know if $a_n$ converges, and if yes, what's the limit, so I left $a_n$ in the equality as it was. I didn't know that this could be a problem
$endgroup$
– Francesco Andreuzzi
Jan 3 at 8:55












$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56




$begingroup$
@FrancescoAndreuzzi I mean, if we can just change $x$, then you didn't even define $a_n$ rigorously enough. LIke I am saying, your proof has a good idea, but it is not rigorous. I would expect a rigorous proof to include some $epsilon-delta$ definitions, or references to other theorems.
$endgroup$
– 5xum
Jan 3 at 8:56




1




1




$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03




$begingroup$
@FrancescoAndreuzzi LIke I said, it's not that you made a mistake, it's more that your proof is not finished yet. It's an idea of a proof. It's not a rigorous proof yet.
$endgroup$
– 5xum
Jan 3 at 9:03


















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%2f3060366%2fproof-of-lhopitals-rule-x-0-not-in-d%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