Evaluate $lim_{xto infty} (frac {x+1}{x-1})^x$












1












$begingroup$


Evaluate $$lim_{xto infty} left(frac {x+1}{x-1}right)^x$$



My method:



$$lim_{xto infty} left(frac {x+1}{x-1}right)^x=lim_{xto infty} left(frac {1+1/x}{1-1/x}right)^x=1$$



Is that right?






limits exponential-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is $lim_{xtoinfty}(1+1/x)^x = 1$?
    $endgroup$
    – Ennar
    Jan 9 at 13:32


















1












$begingroup$


Evaluate $$lim_{xto infty} left(frac {x+1}{x-1}right)^x$$



My method:



$$lim_{xto infty} left(frac {x+1}{x-1}right)^x=lim_{xto infty} left(frac {1+1/x}{1-1/x}right)^x=1$$



Is that right?






limits exponential-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is $lim_{xtoinfty}(1+1/x)^x = 1$?
    $endgroup$
    – Ennar
    Jan 9 at 13:32
















1












1








1





$begingroup$


Evaluate $$lim_{xto infty} left(frac {x+1}{x-1}right)^x$$



My method:



$$lim_{xto infty} left(frac {x+1}{x-1}right)^x=lim_{xto infty} left(frac {1+1/x}{1-1/x}right)^x=1$$



Is that right?






limits exponential-function







share|cite|improve this question











$endgroup$




Evaluate $$lim_{xto infty} left(frac {x+1}{x-1}right)^x$$



My method:



$$lim_{xto infty} left(frac {x+1}{x-1}right)^x=lim_{xto infty} left(frac {1+1/x}{1-1/x}right)^x=1$$



Is that right?






limits exponential-function




limits exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 14:10









Martin Sleziak

44.7k9117272




44.7k9117272










asked Jan 9 at 13:28









MaggieMaggie

888




888








  • 1




    $begingroup$
    Is $lim_{xtoinfty}(1+1/x)^x = 1$?
    $endgroup$
    – Ennar
    Jan 9 at 13:32
















  • 1




    $begingroup$
    Is $lim_{xtoinfty}(1+1/x)^x = 1$?
    $endgroup$
    – Ennar
    Jan 9 at 13:32










1




1




$begingroup$
Is $lim_{xtoinfty}(1+1/x)^x = 1$?
$endgroup$
– Ennar
Jan 9 at 13:32






$begingroup$
Is $lim_{xtoinfty}(1+1/x)^x = 1$?
$endgroup$
– Ennar
Jan 9 at 13:32












3 Answers
3






active

oldest

votes


















3












$begingroup$

Not really, notice that



$$lim_{x to infty}left( 1+frac{y}{x}right)^x=exp(y)$$



Hence



$$lim_{x to infty} left(frac{1+frac1x}{1-frac1x} right)^x=frac{lim_{x to infty}left(1+frac1x right)^x}{lim_{x to infty}left(1-frac1x right)^x}=frac{e}{e^{-1}}=e^2$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Hint: No, you can’t treat the limits separately as you did in your final step:



    $$lim_{x to infty}left(frac{1+frac{1}{x}}{1-frac{1}{x}}right)^x color{red}{neq left(frac{1+0}{1-0}right)^x = 1}$$



    Instead, note that



    $$lim_{x to infty}left(frac{x+1}{x-1}right)^x = lim_{x to infty}left(1+frac{2}{x-1}right)^x = lim_{x to infty}left[left(1+frac{2}{x-1}right)^{x-1}right]^{frac{x}{x-1}}$$



    and make use of the standard limit of $e^x$.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      Note that




      • $lim_{xto +infty}left(1 + frac{a}{x} right)^x = e^a$


      So you have
      begin{eqnarray*}left(frac{x+1}{x-1}right)^x
      & = & left(frac{xleft(1+frac{1}{x} right)}{xleft(1-frac{1}{x} right)}right)^x \
      & = & frac{left(1+frac{1}{x}right)^x}{left(1-frac{1}{x}right)^x} \
      & stackrel{x to +infty}{longrightarrow} & frac{e}{e^{-1}} = e^2
      end{eqnarray*}






      share|cite|improve this answer











      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067442%2fevaluate-lim-x-to-infty-frac-x1x-1x%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Not really, notice that



        $$lim_{x to infty}left( 1+frac{y}{x}right)^x=exp(y)$$



        Hence



        $$lim_{x to infty} left(frac{1+frac1x}{1-frac1x} right)^x=frac{lim_{x to infty}left(1+frac1x right)^x}{lim_{x to infty}left(1-frac1x right)^x}=frac{e}{e^{-1}}=e^2$$






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Not really, notice that



          $$lim_{x to infty}left( 1+frac{y}{x}right)^x=exp(y)$$



          Hence



          $$lim_{x to infty} left(frac{1+frac1x}{1-frac1x} right)^x=frac{lim_{x to infty}left(1+frac1x right)^x}{lim_{x to infty}left(1-frac1x right)^x}=frac{e}{e^{-1}}=e^2$$






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Not really, notice that



            $$lim_{x to infty}left( 1+frac{y}{x}right)^x=exp(y)$$



            Hence



            $$lim_{x to infty} left(frac{1+frac1x}{1-frac1x} right)^x=frac{lim_{x to infty}left(1+frac1x right)^x}{lim_{x to infty}left(1-frac1x right)^x}=frac{e}{e^{-1}}=e^2$$






            share|cite|improve this answer









            $endgroup$



            Not really, notice that



            $$lim_{x to infty}left( 1+frac{y}{x}right)^x=exp(y)$$



            Hence



            $$lim_{x to infty} left(frac{1+frac1x}{1-frac1x} right)^x=frac{lim_{x to infty}left(1+frac1x right)^x}{lim_{x to infty}left(1-frac1x right)^x}=frac{e}{e^{-1}}=e^2$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 13:34









            Siong Thye GohSiong Thye Goh

            100k1465117




            100k1465117























                4












                $begingroup$

                Hint: No, you can’t treat the limits separately as you did in your final step:



                $$lim_{x to infty}left(frac{1+frac{1}{x}}{1-frac{1}{x}}right)^x color{red}{neq left(frac{1+0}{1-0}right)^x = 1}$$



                Instead, note that



                $$lim_{x to infty}left(frac{x+1}{x-1}right)^x = lim_{x to infty}left(1+frac{2}{x-1}right)^x = lim_{x to infty}left[left(1+frac{2}{x-1}right)^{x-1}right]^{frac{x}{x-1}}$$



                and make use of the standard limit of $e^x$.






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  Hint: No, you can’t treat the limits separately as you did in your final step:



                  $$lim_{x to infty}left(frac{1+frac{1}{x}}{1-frac{1}{x}}right)^x color{red}{neq left(frac{1+0}{1-0}right)^x = 1}$$



                  Instead, note that



                  $$lim_{x to infty}left(frac{x+1}{x-1}right)^x = lim_{x to infty}left(1+frac{2}{x-1}right)^x = lim_{x to infty}left[left(1+frac{2}{x-1}right)^{x-1}right]^{frac{x}{x-1}}$$



                  and make use of the standard limit of $e^x$.






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Hint: No, you can’t treat the limits separately as you did in your final step:



                    $$lim_{x to infty}left(frac{1+frac{1}{x}}{1-frac{1}{x}}right)^x color{red}{neq left(frac{1+0}{1-0}right)^x = 1}$$



                    Instead, note that



                    $$lim_{x to infty}left(frac{x+1}{x-1}right)^x = lim_{x to infty}left(1+frac{2}{x-1}right)^x = lim_{x to infty}left[left(1+frac{2}{x-1}right)^{x-1}right]^{frac{x}{x-1}}$$



                    and make use of the standard limit of $e^x$.






                    share|cite|improve this answer











                    $endgroup$



                    Hint: No, you can’t treat the limits separately as you did in your final step:



                    $$lim_{x to infty}left(frac{1+frac{1}{x}}{1-frac{1}{x}}right)^x color{red}{neq left(frac{1+0}{1-0}right)^x = 1}$$



                    Instead, note that



                    $$lim_{x to infty}left(frac{x+1}{x-1}right)^x = lim_{x to infty}left(1+frac{2}{x-1}right)^x = lim_{x to infty}left[left(1+frac{2}{x-1}right)^{x-1}right]^{frac{x}{x-1}}$$



                    and make use of the standard limit of $e^x$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 9 at 13:59

























                    answered Jan 9 at 13:38









                    KM101KM101

                    5,9431523




                    5,9431523























                        1












                        $begingroup$

                        Note that




                        • $lim_{xto +infty}left(1 + frac{a}{x} right)^x = e^a$


                        So you have
                        begin{eqnarray*}left(frac{x+1}{x-1}right)^x
                        & = & left(frac{xleft(1+frac{1}{x} right)}{xleft(1-frac{1}{x} right)}right)^x \
                        & = & frac{left(1+frac{1}{x}right)^x}{left(1-frac{1}{x}right)^x} \
                        & stackrel{x to +infty}{longrightarrow} & frac{e}{e^{-1}} = e^2
                        end{eqnarray*}






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Note that




                          • $lim_{xto +infty}left(1 + frac{a}{x} right)^x = e^a$


                          So you have
                          begin{eqnarray*}left(frac{x+1}{x-1}right)^x
                          & = & left(frac{xleft(1+frac{1}{x} right)}{xleft(1-frac{1}{x} right)}right)^x \
                          & = & frac{left(1+frac{1}{x}right)^x}{left(1-frac{1}{x}right)^x} \
                          & stackrel{x to +infty}{longrightarrow} & frac{e}{e^{-1}} = e^2
                          end{eqnarray*}






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Note that




                            • $lim_{xto +infty}left(1 + frac{a}{x} right)^x = e^a$


                            So you have
                            begin{eqnarray*}left(frac{x+1}{x-1}right)^x
                            & = & left(frac{xleft(1+frac{1}{x} right)}{xleft(1-frac{1}{x} right)}right)^x \
                            & = & frac{left(1+frac{1}{x}right)^x}{left(1-frac{1}{x}right)^x} \
                            & stackrel{x to +infty}{longrightarrow} & frac{e}{e^{-1}} = e^2
                            end{eqnarray*}






                            share|cite|improve this answer











                            $endgroup$



                            Note that




                            • $lim_{xto +infty}left(1 + frac{a}{x} right)^x = e^a$


                            So you have
                            begin{eqnarray*}left(frac{x+1}{x-1}right)^x
                            & = & left(frac{xleft(1+frac{1}{x} right)}{xleft(1-frac{1}{x} right)}right)^x \
                            & = & frac{left(1+frac{1}{x}right)^x}{left(1-frac{1}{x}right)^x} \
                            & stackrel{x to +infty}{longrightarrow} & frac{e}{e^{-1}} = e^2
                            end{eqnarray*}







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 9 at 16:47

























                            answered Jan 9 at 13:40









                            trancelocationtrancelocation

                            9,8101722




                            9,8101722






























                                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%2f3067442%2fevaluate-lim-x-to-infty-frac-x1x-1x%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

                                Mario Kart Wii

                                Image
                                Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                                Read more

                                Understanding the size os this class of aleatory events

                                Image
                                0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
                                Read more

                                Partial Derivative Guidance.

                                Image
                                0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
                                Read more