$lim_{xto 0}frac{f(ax)}{x},$ knowing the limit of $frac{f(x)}{x}$












0












$begingroup$


If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$



We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.






calculus limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
    $endgroup$
    – user3482749
    Jan 2 at 19:58
















0












$begingroup$


If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$



We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.






calculus limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
    $endgroup$
    – user3482749
    Jan 2 at 19:58














0












0








0





$begingroup$


If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$



We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.






calculus limits







share|cite|improve this question











$endgroup$




If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$



We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 20:15









Namaste

1




1










asked Jan 2 at 19:55









JulesJules

1087




1087








  • 3




    $begingroup$
    Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
    $endgroup$
    – user3482749
    Jan 2 at 19:58














  • 3




    $begingroup$
    Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
    $endgroup$
    – user3482749
    Jan 2 at 19:58








3




3




$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58




$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58










4 Answers
4






active

oldest

votes


















1












$begingroup$

As this is elementary, this may require an $epsilon-delta$ argument.



We want to make the following quantity small:



$$left|frac{f(ax)}{x}-acright|$$



Observe that



$$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$



and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    Hint:



    Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$






    share|cite|improve this answer









    $endgroup$





















      4












      $begingroup$

      $$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$






      share|cite|improve this answer











      $endgroup$





















        2












        $begingroup$

        Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$






        share|cite|improve this answer









        $endgroup$













          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%2f3059897%2flim-x-to-0-fracfaxx-knowing-the-limit-of-fracfxx%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          As this is elementary, this may require an $epsilon-delta$ argument.



          We want to make the following quantity small:



          $$left|frac{f(ax)}{x}-acright|$$



          Observe that



          $$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$



          and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            As this is elementary, this may require an $epsilon-delta$ argument.



            We want to make the following quantity small:



            $$left|frac{f(ax)}{x}-acright|$$



            Observe that



            $$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$



            and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              As this is elementary, this may require an $epsilon-delta$ argument.



              We want to make the following quantity small:



              $$left|frac{f(ax)}{x}-acright|$$



              Observe that



              $$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$



              and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.






              share|cite|improve this answer









              $endgroup$



              As this is elementary, this may require an $epsilon-delta$ argument.



              We want to make the following quantity small:



              $$left|frac{f(ax)}{x}-acright|$$



              Observe that



              $$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$



              and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 2 at 20:17









              Math_QEDMath_QED

              7,69231453




              7,69231453























                  6












                  $begingroup$

                  Hint:



                  Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$






                  share|cite|improve this answer









                  $endgroup$


















                    6












                    $begingroup$

                    Hint:



                    Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$






                    share|cite|improve this answer









                    $endgroup$
















                      6












                      6








                      6





                      $begingroup$

                      Hint:



                      Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint:



                      Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 2 at 19:58









                      HenryHenry

                      101k481168




                      101k481168























                          4












                          $begingroup$

                          $$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$






                          share|cite|improve this answer











                          $endgroup$


















                            4












                            $begingroup$

                            $$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$






                            share|cite|improve this answer











                            $endgroup$
















                              4












                              4








                              4





                              $begingroup$

                              $$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$






                              share|cite|improve this answer











                              $endgroup$



                              $$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 3 at 8:28









                              Math_QED

                              7,69231453




                              7,69231453










                              answered Jan 2 at 19:58









                              Tsemo AristideTsemo Aristide

                              59.1k11445




                              59.1k11445























                                  2












                                  $begingroup$

                                  Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 2 at 20:02









                                      Mostafa AyazMostafa Ayaz

                                      15.7k3939




                                      15.7k3939






























                                          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%2f3059897%2flim-x-to-0-fracfaxx-knowing-the-limit-of-fracfxx%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

                                          Bressuire

                                          Image
                                          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                          Read more

                                          Cabo Verde

                                          Image
                                            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                                          Read more

                                          Gyllenstierna

                                          Image
                                          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                                          Read more