How to calculate the two limits?












2












$begingroup$



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 ?






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$



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 ?






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$



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 ?






limits







share|cite|improve this question











$endgroup$





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




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







lanse7pty

















asked 2 hours ago









lanse7ptylanse7pty

1,8361823




1,8361823












  • $begingroup$
    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


















1












$begingroup$

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









          1












          $begingroup$

          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












            $begingroup$

            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












              1








              1





              $begingroup$

              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

              2,726423




              2,726423























                  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

                              185




                              185























                                  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 2 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

                                          flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

                                          Image
                                          0 I recently upgraded an Ubuntu 16.04 virtual machine to 18.04. I regularly use this VM to run apt-mirror at school on their high-speed internet, then rsync these files to my rack-mount server at home for updating multiple computers at home without wasting the tiny amount of bandwidth I have there. Since updating I can't get apt-mirror to run on my VM. It keeps giving the error: flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror line 214 immediately followed by: apt-mirror is already running, exiting at /usr/bin/apt-mirror line 217 apt-mirror is NOT running, and does not have a cron job. I've tried rebooting, and even looked for the old-style lockfile to delete, but I guess that method is no longer used. Is there some way to force unlock this so I can run apt-mirror? UPDATE: So... I f
                                          Read more

                                          Mangá

                                          Image
                                            Nota:  Não confundir com Manga. Mangás e Animes Mangá Mangaká • Tankōbon Film comic La nouvelle manga Mangá original em inglês Anime História Animação influenciada por animes • Filler • OVA • Seiyū Públicos Kodomo • Shōjo • Shōnen Josei • Seinen Gêneros Mahō shōjo • Mecha • Harém Ecchi • Hentai ( Yaoi • Yuri Futanari • Shotacon • Lolicon Toddlercon ) Termos Bishōjo • Chibi • Fan service Gaiden • Kawaii • Kaitō • Light novel Mahō shōnen • Moe (antropomorfismo • Kemonomimi Nekomimi ) • Super deformed Omake • Tsundere • Yonkoma Listas Mangás • Animes • Ecchi • Mangaka • Estúdios • Revistas Fãs ( otaku ) Anime Music Video Convenções • Cosplay Dōjinshi • Fansub • Fandub • Fujoshi OS-tan • Scanlation Portal de Anime e mangá v d e O mangá (português brasileiro) ou manga (português europeu) [ 1 ] (em japonês: 漫画 , manga ? , lit. “história em quadrinhos”) , é a
                                          Read more

                                           ⁒  ․,‪⁊‑⁙ ⁖, ⁇‒※‌, †,⁖‗‌⁝    ‾‸⁘,‖⁔⁣,⁂‾
”‑,‥–,‬ ,⁀‹⁋‴⁑ ‒ ,‴⁋”‼ ⁨,‷⁔„ ‰′,‐‚ ‥‡‎“‷⁃⁨⁅⁣,⁔
⁇‘⁔⁡⁏⁌⁡‿‶‏⁨ ⁣⁕⁖⁨⁩⁥‽⁀  ‴‬⁜‟ ⁃‣‧⁕‮ …‍⁨‴ ⁩,⁚⁖‫ ,‵ ⁀,‮⁝‣‣ ⁑  ⁂– ․, ‾‽ ‏⁁“⁗‸ ‾… ‹‡⁌⁎‸‘ ‡⁏⁌‪ ‵⁛ ‎⁨ ―⁦⁤⁄⁕

                                          ⁉…‧ ⁩,“⁡‹⁃‽⁞•,⁠‚  ⁓⁎  ⁜ ‴ ⁥⁡⁎‭,‍‑⁁⁌⁦ ⁁‟⁙‥‒“⁈ ,‪⁧ ⁙‿⁇⁦,⁝‘ ⁎“⁘‫‪,‗‹‛‹⁒  ⁇⁄‗‒•“⁇   ‐ ⁂
‘‛, ‶⁦ ⁇ ​⁞‧ 
,⁏ …‹–⁗⁡⁘⁄ ‏‿‬‟″ ‏⁢⁦⁗⁚⁝‬⁢⁏,⁧″⁍′‖⁂※⁄‘– ‵⁡ ‭⁢⁚,⁅‧,‼‛⁗,⁡⁗“⁔⁃⁍‷⁏‗‱⁌‚  
•⁝ ‚‒ ‥‰  ′⁌, ⁄, ⁛—‭‑”⁂,⁑ ‡⁨‴⁖⁙…,‑, ,⁎,․‣‥  ‴‎⁎,″‛⁎‭‿ ⁤ ‷‾‼‖,⁠⁊‾,⁦“‗‑‌†,⁙,†⁝′ 
,―‚․‘, ⁢  ‿‌,⁀„‣,⁌,‭‖ ⁆“‪,―  ⁀ ⁃,
‷⁥,​,‟‗⁄⁉⁐⁗―
,‮⁢⁩⁆⁣‹‭⁅ ⁝‾‏‭⁖ ⁕⁄‷,†‟‒ ⁣‹,​…​,⁤⁂⁨‮⁎⁙–‟‼ ⁍‱‱,‽,‫‽‣ ⁒,⁨ ⁄ ⁝”„‿‑‎⁈‿
 ⁀‛‼‟ ⁑“,⁩​ ‒
–⁡…⁢ ⁂ ⁈⁎,‵‡,⁛,,‱⁧⁋,․‬⁄ “,⁄‴⁊  ‗―,‵⁙⁔—⁀ —  ⁀‬‬ ‒⁋‼⁀― ,⁌′ ‍‶ ’
                                          Read more