How to calculate the two limits? The Next CEO of Stack OverflowCompute $lim limits_{xtoinfty}...

(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?

TikZ: How to fill area with a special pattern?

From jafe to El-Guest

How to calculate the two limits?

How do I fit a non linear curve?

Point distance program written without a framework

"Eavesdropping" vs "Listen in on"

How can I make proper oatmeal cookies?

What steps are necessary to read a Modern SSD in Medieval Europe?

Why did early computer designers eschew integers?

How to avoid supervisors with prejudiced views?

Is a distribution that is normal, but highly skewed, considered Gaussian?

Why don't programming languages automatically manage the synchronous/asynchronous problem?

Is there such a thing as a proper verb, like a proper noun?

What would be the main consequences for a country leaving the WTO?

Can this note be analyzed as a non-chord tone?

Plausibility of squid whales

Audio Conversion With ADS1243

Is Nisuin Biblical or Rabbinic?

Film where the government was corrupt with aliens, people sent to kill aliens are given rigged visors not showing the right aliens

Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?

Does Germany produce more waste than the US?

Is it ok to trim down a tube patch?

What does "shotgun unity" refer to here in this sentence?



How to calculate the two limits?



The Next CEO of Stack OverflowCompute $lim limits_{xtoinfty} (frac{x-2}{x+2})^x$limits of the sequence $n/(n+1)$How to calculate $lim_{xto1}left(frac{1+cos(pi x)}{tan^2(pi x)}right)^{!x^2}$Calculate the limit of integralHow to evaluate $lim_{xtoinfty}arctan (4/x)/ |arcsin (-3/x)|$?Is there a way to get at this limit problem algebraically?Calculate $lim_{x to infty}(x+1)e^{-2x}$How to calculate $lim_{nto infty } frac{n^n}{n!^2}$?Calculate the limit: $lim limits_{n rightarrow infty } frac {4(n+3)!-n!}{n((n+2)!-(n-1)!)}$How to solve the limit $limlimits_{xto infty} (x arctan x - frac{xpi}{2})$












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











$endgroup$












  • $begingroup$
    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

                                          Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...

                                          Image
                                          Does the Mueller report show a conspiracy between Russia and the Trump Campaign? Relating to the President and obstruction, were Mueller's conclusions preordained? How much damage would a cupful of neutron star matter do to the Earth? Nose gear failure in single prop aircraft: belly landing or nose landing? Should a wizard buy fine inks every time he want to copy spells into his spellbook? Can you force honesty by using the Speak with Dead and Zone of Truth spells together? A trigonometry question from STEP examination Special flights What is the "studentd" process? systemd and copy (/bin/cp): no such file or directory...
                                          Read more

                                          Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

                                          Image
                                          Why didn't Boeing produce its own regional jet? What is the most common color to indicate the input-field is disabled? Theorists sure want true answers to this! What is the opposite of "eschatology"? What do you call someone who asks many questions? Is there a hemisphere-neutral way of specifying a season? Is this draw by repetition? Bug? String.split chopping off empty trailing value Is it inappropriate for a student to attend their mentor's dissertation defense? How dangerous is XSS Why do I get negative height? Is it possible to map ...
                                          Read more

                                          VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...

                                          Image
                                          risk of flooding in petra in november How to get directions in deep space? How could a planet have erratic days? 15% tax on $7.5k earnings. Is that right? Did the UK lift the requirement for registering SIM cards? Is it allowed to activate the ability of multiple planeswalkers in a single turn? Stack Interview Code methods made from class Node and Smart Pointers Non-trope happy ending? What is the highest possible scrabble score for placing a single tile Open a doc from terminal, but not by its name The IT department bottlenecks progress, how should I handle this? ...
                                          Read more