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...












3














$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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







share|cite|improve this question




















  • 2




    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?
    – User
    yesterday
















3














$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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







share|cite|improve this question




















  • 2




    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?
    – User
    yesterday














3












3








3







$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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







share|cite|improve this question















$$lim_{xto0^+}frac{x^2}{e^{-frac{1}{x^2}}cos(frac{1}{x^2})^2}$$



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Asaf Karagila♦

302k32426756




302k32426756










asked yesterday









User

425311




425311








  • 2




    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?
    – User
    yesterday














  • 2




    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?
    – User
    yesterday








2




2




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?
– User
yesterday




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?
– User
yesterday










2 Answers
2






active

oldest

votes


















6














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 $+infty$.






share|cite|improve this answer





















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday



















4














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 $|cos (t)| leq 1$.






share|cite|improve this answer























  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday













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%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









6














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 $+infty$.






share|cite|improve this answer





















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday
















6














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 $+infty$.






share|cite|improve this answer





















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday














6












6








6






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 $+infty$.






share|cite|improve this answer












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 $+infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









José Carlos Santos

151k22123224




151k22123224












  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday


















  • 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?
    – User
    yesterday








  • 1




    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.
    – José Carlos Santos
    yesterday










  • ok, will check. thanks!
    – User
    yesterday










  • @user376343 yes it's
    – User
    yesterday
















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?
– User
yesterday






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?
– User
yesterday






1




1




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.
– José Carlos Santos
yesterday




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.
– José Carlos Santos
yesterday












ok, will check. thanks!
– User
yesterday




ok, will check. thanks!
– User
yesterday












@user376343 yes it's
– User
yesterday




@user376343 yes it's
– User
yesterday











4














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 $|cos (t)| leq 1$.






share|cite|improve this answer























  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday


















4














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 $|cos (t)| leq 1$.






share|cite|improve this answer























  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday
















4












4








4






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 $|cos (t)| leq 1$.






share|cite|improve this answer














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 $|cos (t)| leq 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday









Surb

37.4k94375




37.4k94375










answered yesterday









Kavi Rama Murthy

51k31854




51k31854












  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday




















  • 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 $infty$...
    – User
    yesterday










  • @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.
    – Kavi Rama Murthy
    yesterday


















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 $infty$...
– User
yesterday




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 $infty$...
– User
yesterday












@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.
– Kavi Rama Murthy
yesterday






@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.
– Kavi Rama Murthy
yesterday




















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3060404%2fcan-i-conclude-that-lim-x-to0-fracx2e-frac1x2-cos-frac1x%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

An IMO inspired problem

Image
14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
Read more

Management

Image
Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
Read more

Investment

Image
Investment From Wikipedia, the free encyclopedia Jump to navigation Jump to search This article is about investment in finance. For investment in macroeconomics, see Investment (macroeconomics). For other uses, see Investment (disambiguation). "Invest" redirects here. For the term in meteorology, see Invest (meteorology). This article needs additional citations for verification . Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. ( February 2017 ) (Learn how and when to remove this template message) In general, to invest is to distribute money in the expectation of some benefit in the future – for example, investment in durable goods, in real estate by the service industry, in factories for manufacturing, in product development, and in research and development. However, this article focuses specifically on investment in financial
Read more