Can I conclude that $lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$ is infinite or it doesn't...
$begingroup$
$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
asked Jan 3 at 9:45
UserUser
452312
$endgroup$
add a comment |
$begingroup$
$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
asked Jan 3 at 9:45
UserUser
452312
$endgroup$
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
add a comment |
$begingroup$
$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
asked Jan 3 at 9:45
UserUser
452312
$endgroup$
$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$
My intuition is that the denominator goes to 0 faster and everything is non-negative, so the limit is positive infinity.
I cant think of elementary proof, l'Hopital doesn't help here, and I'm not sure if and how to use taylor.
WolframAlpha says it's complex infinity, but I can't understand why it doesn't just real positive infinity.
I've tried to use wolfram language to use the assumption that x is real, but couldn't get any result.
real-analysis limits infinity wolfram-alpha
real-analysis limits infinity wolfram-alpha
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
asked Jan 3 at 9:45
UserUser
452312
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
asked Jan 3 at 9:45
UserUser
452312
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
edited Jan 3 at 13:21
Asaf Karagila♦
307k33440773
307k33440773
asked Jan 3 at 9:45
UserUser
452312
asked Jan 3 at 9:45
UserUser
452312
asked Jan 3 at 9:45
UserUser
452312
452312
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
add a comment |
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
2
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac {e^{y}} {y^{2}} to infty$ as $y to infty$. Replace $y$ by $frac 1 {x^{2}}$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
38.4k94478
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
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%2f3060404%2fcan-i-conclude-that-lim-x-to0-fracx2e-frac1x2-cos-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
$begingroup$
The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
$endgroup$
The limit$$lim_{xto0^+}frac{x^2}{expleft(-frac1{x^2}right)}$$is indeed $+infty$. Since $dfrac1{cos^2left(frac1{x^2}right)}geqslant1$ for each $x$ (when it is defined), you are right: the limit is $+infty$.
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
answered Jan 3 at 9:51
José Carlos SantosJosé Carlos Santos
172k23132240
172k23132240
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
$begingroup$
thanks! so if I want to be as rigorous as possible, the limit doesn't exists just because of the points where it's undefined? or can I just ignore them?
$endgroup$
– User
Jan 3 at 9:53
1
1
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
That depends upon the definition of limit that you are using. If it requires that the function is defined in som interval $(0,varepsilon)$, then it doesn't exist.
$endgroup$
– José Carlos Santos
Jan 3 at 9:54
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
ok, will check. thanks!
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
$begingroup$
@user376343 yes it's
$endgroup$
– User
Jan 3 at 9:55
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac {e^{y}} {y^{2}} to infty$ as $y to infty$. Replace $y$ by $frac 1 {x^{2}}$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
38.4k94478
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac {e^{y}} {y^{2}} to infty$ as $y to infty$. Replace $y$ by $frac 1 {x^{2}}$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
38.4k94478
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
$begingroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac {e^{y}} {y^{2}} to infty$ as $y to infty$. Replace $y$ by $frac 1 {x^{2}}$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
38.4k94478
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
The function is not finite valued. If you allow the value $infty$ then the limit is $infty$. Use L'Hôpital's Rule twice to show that $frac {e^{y}} {y^{2}} to infty$ as $y to infty$. Replace $y$ by $frac 1 {x^{2}}$ and use the fact that $|cos (t)| leq 1$.
edited Jan 3 at 11:46
Surb
38.4k94478
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
edited Jan 3 at 11:46
Surb
38.4k94478
edited Jan 3 at 11:46
Surb
38.4k94478
edited Jan 3 at 11:46
Surb
38.4k94478
38.4k94478
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
answered Jan 3 at 9:50
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
71.8k53170
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
add a comment |
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
Thanks! value of $infty$? I'm not using "one point compactification" or "extended real line" if that was your meaning, division by the number 0 is not allowed. I do use a definition which allow infinite limits or limits at infinity. Using this change of variable, I still get 0 in the denominator infinitely many times as y goes to $infty$...
$endgroup$
– User
Jan 3 at 10:59
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
$begingroup$
@User In that case you have to say that function is not defined in $(0,a)$ for any $a>0$ so the limit is not defined.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 11:48
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%2f3060404%2fcan-i-conclude-that-lim-x-to0-fracx2e-frac1x2-cos-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
2
$begingroup$
someone vote to close as off-topic, I can't understand why... I'm mentioning WA just as an aside, I ask about a limit and describe what have I tried. can't see, why this might be considered off-topic?
$endgroup$
– User
Jan 3 at 11:02