Can we say a function is “unbounded” when we mean it''s tending to infinity?












1














I'm watching the Limits series on Khan Academy. In many videos Sal repeatedly says that although some people say that functions that tend to infinity have a limit infinity. (For example, in this video, the says that the function $ y = frac {2}{x-1} $ (here's a link to the graph) is unbounded as x approaches 1 from the left side, although "some" people would say that the function is tending towards infinity (i.e., the $ lim_{x to c} f(x) = infty $.



Over a series of videos, I caught hold of that argument and always said that function don't tend towards infinity; they're just unbounded. But, in a different video (here), Sal essentially says:



$$ lim_{x to 0} frac 1{x^2} = infty $$
$$ lim_{x to 0^+} frac 1{x} = infty $$
$$ lim_{x to 0^-} frac 1{x} = -infty $$



There is an answer to this question that says that unbounded sequences don't always tend to infinity. So it seems that we can't say that the limit of unbounded functions as x tends to some number is infinity -- what I'm confused about, and my question is: When do we ever say that the limit of some function is infinity? Specifically, are the limits of the above functions $ frac 1x $ and $$ frac 1{x^2} $$ correct? Or are they unbounded? If we can say that these limits are correct, why can't we say this: $$ lim_{x to 1^-} frac 2{x-1} = - infty $$










share|cite|improve this question




















  • 4




    Things that tend to infinity are unbounded, but unbounded things do not always tend to infinity. Tending to infinity means more: it means it is both unbounded and at the same time doesn't "go back down".
    – Arthur
    Nov 30 '18 at 11:38










  • One of the limits is wrong: $frac 1 {x^{2}}$ is a positive function so it cannot tend to $-infty$. All other limits are correct.
    – Kavi Rama Murthy
    Nov 30 '18 at 11:46










  • @KaviRamaMurthy -- Thank you, edited.
    – WorldGov
    Nov 30 '18 at 12:22










  • @Arthur -- Thanks, that explains it.
    – WorldGov
    Nov 30 '18 at 12:23
















1














I'm watching the Limits series on Khan Academy. In many videos Sal repeatedly says that although some people say that functions that tend to infinity have a limit infinity. (For example, in this video, the says that the function $ y = frac {2}{x-1} $ (here's a link to the graph) is unbounded as x approaches 1 from the left side, although "some" people would say that the function is tending towards infinity (i.e., the $ lim_{x to c} f(x) = infty $.



Over a series of videos, I caught hold of that argument and always said that function don't tend towards infinity; they're just unbounded. But, in a different video (here), Sal essentially says:



$$ lim_{x to 0} frac 1{x^2} = infty $$
$$ lim_{x to 0^+} frac 1{x} = infty $$
$$ lim_{x to 0^-} frac 1{x} = -infty $$



There is an answer to this question that says that unbounded sequences don't always tend to infinity. So it seems that we can't say that the limit of unbounded functions as x tends to some number is infinity -- what I'm confused about, and my question is: When do we ever say that the limit of some function is infinity? Specifically, are the limits of the above functions $ frac 1x $ and $$ frac 1{x^2} $$ correct? Or are they unbounded? If we can say that these limits are correct, why can't we say this: $$ lim_{x to 1^-} frac 2{x-1} = - infty $$










share|cite|improve this question




















  • 4




    Things that tend to infinity are unbounded, but unbounded things do not always tend to infinity. Tending to infinity means more: it means it is both unbounded and at the same time doesn't "go back down".
    – Arthur
    Nov 30 '18 at 11:38










  • One of the limits is wrong: $frac 1 {x^{2}}$ is a positive function so it cannot tend to $-infty$. All other limits are correct.
    – Kavi Rama Murthy
    Nov 30 '18 at 11:46










  • @KaviRamaMurthy -- Thank you, edited.
    – WorldGov
    Nov 30 '18 at 12:22










  • @Arthur -- Thanks, that explains it.
    – WorldGov
    Nov 30 '18 at 12:23














1












1








1







I'm watching the Limits series on Khan Academy. In many videos Sal repeatedly says that although some people say that functions that tend to infinity have a limit infinity. (For example, in this video, the says that the function $ y = frac {2}{x-1} $ (here's a link to the graph) is unbounded as x approaches 1 from the left side, although "some" people would say that the function is tending towards infinity (i.e., the $ lim_{x to c} f(x) = infty $.



Over a series of videos, I caught hold of that argument and always said that function don't tend towards infinity; they're just unbounded. But, in a different video (here), Sal essentially says:



$$ lim_{x to 0} frac 1{x^2} = infty $$
$$ lim_{x to 0^+} frac 1{x} = infty $$
$$ lim_{x to 0^-} frac 1{x} = -infty $$



There is an answer to this question that says that unbounded sequences don't always tend to infinity. So it seems that we can't say that the limit of unbounded functions as x tends to some number is infinity -- what I'm confused about, and my question is: When do we ever say that the limit of some function is infinity? Specifically, are the limits of the above functions $ frac 1x $ and $$ frac 1{x^2} $$ correct? Or are they unbounded? If we can say that these limits are correct, why can't we say this: $$ lim_{x to 1^-} frac 2{x-1} = - infty $$










share|cite|improve this question















I'm watching the Limits series on Khan Academy. In many videos Sal repeatedly says that although some people say that functions that tend to infinity have a limit infinity. (For example, in this video, the says that the function $ y = frac {2}{x-1} $ (here's a link to the graph) is unbounded as x approaches 1 from the left side, although "some" people would say that the function is tending towards infinity (i.e., the $ lim_{x to c} f(x) = infty $.



Over a series of videos, I caught hold of that argument and always said that function don't tend towards infinity; they're just unbounded. But, in a different video (here), Sal essentially says:



$$ lim_{x to 0} frac 1{x^2} = infty $$
$$ lim_{x to 0^+} frac 1{x} = infty $$
$$ lim_{x to 0^-} frac 1{x} = -infty $$



There is an answer to this question that says that unbounded sequences don't always tend to infinity. So it seems that we can't say that the limit of unbounded functions as x tends to some number is infinity -- what I'm confused about, and my question is: When do we ever say that the limit of some function is infinity? Specifically, are the limits of the above functions $ frac 1x $ and $$ frac 1{x^2} $$ correct? Or are they unbounded? If we can say that these limits are correct, why can't we say this: $$ lim_{x to 1^-} frac 2{x-1} = - infty $$







calculus limits functions upper-lower-bounds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 12:21







WorldGov

















asked Nov 30 '18 at 11:35









WorldGovWorldGov

2629




2629








  • 4




    Things that tend to infinity are unbounded, but unbounded things do not always tend to infinity. Tending to infinity means more: it means it is both unbounded and at the same time doesn't "go back down".
    – Arthur
    Nov 30 '18 at 11:38










  • One of the limits is wrong: $frac 1 {x^{2}}$ is a positive function so it cannot tend to $-infty$. All other limits are correct.
    – Kavi Rama Murthy
    Nov 30 '18 at 11:46










  • @KaviRamaMurthy -- Thank you, edited.
    – WorldGov
    Nov 30 '18 at 12:22










  • @Arthur -- Thanks, that explains it.
    – WorldGov
    Nov 30 '18 at 12:23














  • 4




    Things that tend to infinity are unbounded, but unbounded things do not always tend to infinity. Tending to infinity means more: it means it is both unbounded and at the same time doesn't "go back down".
    – Arthur
    Nov 30 '18 at 11:38










  • One of the limits is wrong: $frac 1 {x^{2}}$ is a positive function so it cannot tend to $-infty$. All other limits are correct.
    – Kavi Rama Murthy
    Nov 30 '18 at 11:46










  • @KaviRamaMurthy -- Thank you, edited.
    – WorldGov
    Nov 30 '18 at 12:22










  • @Arthur -- Thanks, that explains it.
    – WorldGov
    Nov 30 '18 at 12:23








4




4




Things that tend to infinity are unbounded, but unbounded things do not always tend to infinity. Tending to infinity means more: it means it is both unbounded and at the same time doesn't "go back down".
– Arthur
Nov 30 '18 at 11:38




Things that tend to infinity are unbounded, but unbounded things do not always tend to infinity. Tending to infinity means more: it means it is both unbounded and at the same time doesn't "go back down".
– Arthur
Nov 30 '18 at 11:38












One of the limits is wrong: $frac 1 {x^{2}}$ is a positive function so it cannot tend to $-infty$. All other limits are correct.
– Kavi Rama Murthy
Nov 30 '18 at 11:46




One of the limits is wrong: $frac 1 {x^{2}}$ is a positive function so it cannot tend to $-infty$. All other limits are correct.
– Kavi Rama Murthy
Nov 30 '18 at 11:46












@KaviRamaMurthy -- Thank you, edited.
– WorldGov
Nov 30 '18 at 12:22




@KaviRamaMurthy -- Thank you, edited.
– WorldGov
Nov 30 '18 at 12:22












@Arthur -- Thanks, that explains it.
– WorldGov
Nov 30 '18 at 12:23




@Arthur -- Thanks, that explains it.
– WorldGov
Nov 30 '18 at 12:23










2 Answers
2






active

oldest

votes


















1














Like Arthur said in the comments, functions being "bounded" and "tending to infinity" mean different things. If $lim_{xto c}f(x)=infty$, then it tends to infinity and gets arbitrarily large in the process. Hence it is not bounded, so there does not exist real numbers $a,b$ so that $f(x)in[a,b]$ for all $x$ in the domain. However, if a function is unbounded it does not necessarily tend to infinity. In fact, the limit need not even exist. Just consider $f(x)=sin(1/x)/x$.






share|cite|improve this answer





























    1














    A necessary condition for a function to have infinite limit is that it is unbounded.



    More precisely, if $lim_{xto c}f(x)=infty$, then $f$ is upper unbounded; if $lim_{xto c}f(x)=-infty$, then $f$ is lower unbounded. (The limit in the previous statements can also be one-sided.)



    One could be more precise and say that $f$ must be (upper/lower) unbounded in every punctured neighborhood of $c$ (or punctured right/left neighborhood of $c$).



    This essentially follows from the definition.



    However, this condition is by no means sufficient. Consider
    $$
    f(x)=frac{1}{x}sinfrac{1}{x}
    $$

    Then this function is unbounded in every punctured neighborhood of $0$, but its limit is neither $infty$ nor $-infty$.



    Restricting to the limit from the right or the left doesn't improve the situation.



    Confusing “unbounded” with “has infinite limit” may be very dangerous.






    share|cite|improve this answer





















      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%2f3019989%2fcan-we-say-a-function-is-unbounded-when-we-mean-its-tending-to-infinity%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









      1














      Like Arthur said in the comments, functions being "bounded" and "tending to infinity" mean different things. If $lim_{xto c}f(x)=infty$, then it tends to infinity and gets arbitrarily large in the process. Hence it is not bounded, so there does not exist real numbers $a,b$ so that $f(x)in[a,b]$ for all $x$ in the domain. However, if a function is unbounded it does not necessarily tend to infinity. In fact, the limit need not even exist. Just consider $f(x)=sin(1/x)/x$.






      share|cite|improve this answer


























        1














        Like Arthur said in the comments, functions being "bounded" and "tending to infinity" mean different things. If $lim_{xto c}f(x)=infty$, then it tends to infinity and gets arbitrarily large in the process. Hence it is not bounded, so there does not exist real numbers $a,b$ so that $f(x)in[a,b]$ for all $x$ in the domain. However, if a function is unbounded it does not necessarily tend to infinity. In fact, the limit need not even exist. Just consider $f(x)=sin(1/x)/x$.






        share|cite|improve this answer
























          1












          1








          1






          Like Arthur said in the comments, functions being "bounded" and "tending to infinity" mean different things. If $lim_{xto c}f(x)=infty$, then it tends to infinity and gets arbitrarily large in the process. Hence it is not bounded, so there does not exist real numbers $a,b$ so that $f(x)in[a,b]$ for all $x$ in the domain. However, if a function is unbounded it does not necessarily tend to infinity. In fact, the limit need not even exist. Just consider $f(x)=sin(1/x)/x$.






          share|cite|improve this answer












          Like Arthur said in the comments, functions being "bounded" and "tending to infinity" mean different things. If $lim_{xto c}f(x)=infty$, then it tends to infinity and gets arbitrarily large in the process. Hence it is not bounded, so there does not exist real numbers $a,b$ so that $f(x)in[a,b]$ for all $x$ in the domain. However, if a function is unbounded it does not necessarily tend to infinity. In fact, the limit need not even exist. Just consider $f(x)=sin(1/x)/x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 12:27









          YiFanYiFan

          2,6841422




          2,6841422























              1














              A necessary condition for a function to have infinite limit is that it is unbounded.



              More precisely, if $lim_{xto c}f(x)=infty$, then $f$ is upper unbounded; if $lim_{xto c}f(x)=-infty$, then $f$ is lower unbounded. (The limit in the previous statements can also be one-sided.)



              One could be more precise and say that $f$ must be (upper/lower) unbounded in every punctured neighborhood of $c$ (or punctured right/left neighborhood of $c$).



              This essentially follows from the definition.



              However, this condition is by no means sufficient. Consider
              $$
              f(x)=frac{1}{x}sinfrac{1}{x}
              $$

              Then this function is unbounded in every punctured neighborhood of $0$, but its limit is neither $infty$ nor $-infty$.



              Restricting to the limit from the right or the left doesn't improve the situation.



              Confusing “unbounded” with “has infinite limit” may be very dangerous.






              share|cite|improve this answer


























                1














                A necessary condition for a function to have infinite limit is that it is unbounded.



                More precisely, if $lim_{xto c}f(x)=infty$, then $f$ is upper unbounded; if $lim_{xto c}f(x)=-infty$, then $f$ is lower unbounded. (The limit in the previous statements can also be one-sided.)



                One could be more precise and say that $f$ must be (upper/lower) unbounded in every punctured neighborhood of $c$ (or punctured right/left neighborhood of $c$).



                This essentially follows from the definition.



                However, this condition is by no means sufficient. Consider
                $$
                f(x)=frac{1}{x}sinfrac{1}{x}
                $$

                Then this function is unbounded in every punctured neighborhood of $0$, but its limit is neither $infty$ nor $-infty$.



                Restricting to the limit from the right or the left doesn't improve the situation.



                Confusing “unbounded” with “has infinite limit” may be very dangerous.






                share|cite|improve this answer
























                  1












                  1








                  1






                  A necessary condition for a function to have infinite limit is that it is unbounded.



                  More precisely, if $lim_{xto c}f(x)=infty$, then $f$ is upper unbounded; if $lim_{xto c}f(x)=-infty$, then $f$ is lower unbounded. (The limit in the previous statements can also be one-sided.)



                  One could be more precise and say that $f$ must be (upper/lower) unbounded in every punctured neighborhood of $c$ (or punctured right/left neighborhood of $c$).



                  This essentially follows from the definition.



                  However, this condition is by no means sufficient. Consider
                  $$
                  f(x)=frac{1}{x}sinfrac{1}{x}
                  $$

                  Then this function is unbounded in every punctured neighborhood of $0$, but its limit is neither $infty$ nor $-infty$.



                  Restricting to the limit from the right or the left doesn't improve the situation.



                  Confusing “unbounded” with “has infinite limit” may be very dangerous.






                  share|cite|improve this answer












                  A necessary condition for a function to have infinite limit is that it is unbounded.



                  More precisely, if $lim_{xto c}f(x)=infty$, then $f$ is upper unbounded; if $lim_{xto c}f(x)=-infty$, then $f$ is lower unbounded. (The limit in the previous statements can also be one-sided.)



                  One could be more precise and say that $f$ must be (upper/lower) unbounded in every punctured neighborhood of $c$ (or punctured right/left neighborhood of $c$).



                  This essentially follows from the definition.



                  However, this condition is by no means sufficient. Consider
                  $$
                  f(x)=frac{1}{x}sinfrac{1}{x}
                  $$

                  Then this function is unbounded in every punctured neighborhood of $0$, but its limit is neither $infty$ nor $-infty$.



                  Restricting to the limit from the right or the left doesn't improve the situation.



                  Confusing “unbounded” with “has infinite limit” may be very dangerous.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 12:31









                  egregegreg

                  179k1485202




                  179k1485202






























                      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%2f3019989%2fcan-we-say-a-function-is-unbounded-when-we-mean-its-tending-to-infinity%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

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Aardman Animations

                      Are they similar matrix