$lim_{xto 0}frac{f(ax)}{x},$ 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









        $endgroup$













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          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






























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