$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$
$begingroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
$endgroup$
add a comment |
$begingroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
$endgroup$
add a comment |
$begingroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
$endgroup$
Suppose
$,f:mathbb{R}rightarrowmathbb{R},,limlimits_{xdownarrow0},f(x) = c in mathbb{R}cup{pminfty}$. I'd like to know if the follwing statement is true.
$$limlimits_{xdownarrow0},f(x) = limlimits_{xrightarrowinfty},f(frac1x)$$
My intuition about this is that for any $x$ in $mathbb{R}^*_+$ there is a $y$ in $mathbb{R}^*_+$ so that $frac1y<x$. So it's like following the reversed graph in the reverse direction.
If the statement is true could you either:
- proof it
- explain your intuition why this is true
... and also would this be the same for $limlimits_{xuparrow0},f(x) = limlimits_{xrightarrow-infty},f(frac1x)$?
If the statement is false could you either:
- hint me a counter-example
- disproof the statement
... and also is there a way to correct the statement? E.g. constraining $,f$ to be conitious.
Cheers,
Pascal
limits
limits
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
asked Dec 18 '18 at 22:13
plauerplauer
355
355
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,206412
$endgroup$
add a comment |
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
});
}
});
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%2f3045776%2flim-limits-x-downarrow0-fx-lim-limits-x-rightarrow-infty-f-frac1x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
$endgroup$
say $c$ is a finite real number then
$limlimits_{xdownarrow0},f(x) = c $ means :
$forall epsilon > 0, exists delta > 0, text{such that } , |x | < delta implies |f(x)|<epsilon $
meaning
$forall epsilon > 0, exists delta > 0, text{such that } , |frac1x | < delta implies |f(frac1x)|<epsilon $
meaning
$forall epsilon > 0, exists delta' = frac{1}{delta} > 0, text{such that } , |x | > delta' implies |f(frac1x)|<epsilon $
meaning
$limlimits_{xrightarrowinfty},f(frac1x) = c$
if $c$ is $pm infty$ it's the same reasoning
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
answered Dec 18 '18 at 22:23
rapidracimrapidracim
1,7191419
1,7191419
add a comment |
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,206412
$endgroup$
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,206412
$endgroup$
add a comment |
$begingroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,206412
$endgroup$
It is true.
First assume that $lim_{x in mathbb{R}^+ rightarrow 0} = K$ for some finite $K$.
Then for all $epsilon >0$ there is an $x' > 0$ such that $|f(x'') - K| leq epsilon$ for all positive $x'' le x'$. This implies that $|f(frac{1}{y}) - K| le epsilon$ for all $y geq frac{1}{x'}$. Which implies $lim_{y rightarrow infty} f(frac{1}{y}) = K$.
Can you handle the case where $lim_{x in mathbb{R}^+ rightarrow 0} = infty$ ?
answered Dec 18 '18 at 22:27
MikeMike
4,206412
answered Dec 18 '18 at 22:27
MikeMike
4,206412
answered Dec 18 '18 at 22:27
MikeMike
4,206412
answered Dec 18 '18 at 22:27
MikeMike
4,206412
4,206412
add a comment |
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%2f3045776%2flim-limits-x-downarrow0-fx-lim-limits-x-rightarrow-infty-f-frac1x%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
