How to calculate the two limits?












3












$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?










share|cite











$endgroup$












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago
















3












$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?










share|cite











$endgroup$












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago














3












3








3





$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?










share|cite











$endgroup$





I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?







limits






share|cite















share|cite













share|cite




share|cite








edited 1 hour ago







lanse7pty

















asked 2 hours ago









lanse7ptylanse7pty

1,8411823




1,8411823












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago


















  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago
















$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago




$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Without L'Hospital
    $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



    Now, by Taylor for large values of $x$
    $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
    $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
    $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
    $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        You can solve the first one using




        • $arctan x + operatorname{arccot}x = frac{pi}{2}$

        • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

        • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


        begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
        & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
        & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
        end{eqnarray*}



        The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





        • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






        share|cite









        $endgroup$














          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%2f3170200%2fhow-to-calculate-the-two-limits%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









          2












          $begingroup$

          Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






          share|cite|improve this answer











          $endgroup$


















            2












            $begingroup$

            Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






            share|cite|improve this answer











            $endgroup$
















              2












              2








              2





              $begingroup$

              Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






              share|cite|improve this answer











              $endgroup$



              Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 1 hour ago

























              answered 2 hours ago









              Paras KhoslaParas Khosla

              2,736423




              2,736423























                  1












                  $begingroup$

                  Without L'Hospital
                  $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                  Now, by Taylor for large values of $x$
                  $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                  $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                  $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                  $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Without L'Hospital
                    $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                    Now, by Taylor for large values of $x$
                    $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                    $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                    $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                    $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Without L'Hospital
                      $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                      $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                      $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                      $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                      share|cite|improve this answer









                      $endgroup$



                      Without L'Hospital
                      $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                      $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                      $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                      $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      Claude LeiboviciClaude Leibovici

                      125k1158136




                      125k1158136























                          0












                          $begingroup$

                          I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                              share|cite|improve this answer









                              $endgroup$



                              I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 hours ago









                              AdmuthAdmuth

                              285




                              285























                                  0












                                  $begingroup$

                                  You can solve the first one using




                                  • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                  • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                  • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                  begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                  & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                  & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                  end{eqnarray*}



                                  The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                  • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                  share|cite









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You can solve the first one using




                                    • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                    • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                    • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                    begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                    & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                    & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                    end{eqnarray*}



                                    The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                    • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                    share|cite









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You can solve the first one using




                                      • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                      • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                      • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                      begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                      & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                      & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                      end{eqnarray*}



                                      The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                      • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                      share|cite









                                      $endgroup$



                                      You can solve the first one using




                                      • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                      • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                      • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                      begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                      & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                      & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                      end{eqnarray*}



                                      The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                      • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.







                                      share|cite












                                      share|cite



                                      share|cite










                                      answered 9 mins ago









                                      trancelocationtrancelocation

                                      13.4k1827




                                      13.4k1827






























                                          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%2f3170200%2fhow-to-calculate-the-two-limits%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

                                          Knooppunt Holsloot

                                          Altaar (religie)

                                          Gregoriusmis