Natural logs with L'Hopital's rule












1












$begingroup$


Given $$lim_{xto 0} (e^x-2x)^frac{1}{x} $$ I know that you take the natural log $$lim_{xto 0} frac{1}{x}ln(e^x-2x) $$ which is $$lim_{xto 0} frac{ln(e^x-2x)}{x} $$ but what is after this?






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator...
    $endgroup$
    – the_candyman
    Dec 9 '18 at 18:52
















1












$begingroup$


Given $$lim_{xto 0} (e^x-2x)^frac{1}{x} $$ I know that you take the natural log $$lim_{xto 0} frac{1}{x}ln(e^x-2x) $$ which is $$lim_{xto 0} frac{ln(e^x-2x)}{x} $$ but what is after this?






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator...
    $endgroup$
    – the_candyman
    Dec 9 '18 at 18:52














1












1








1





$begingroup$


Given $$lim_{xto 0} (e^x-2x)^frac{1}{x} $$ I know that you take the natural log $$lim_{xto 0} frac{1}{x}ln(e^x-2x) $$ which is $$lim_{xto 0} frac{ln(e^x-2x)}{x} $$ but what is after this?






calculus limits







share|cite|improve this question











$endgroup$




Given $$lim_{xto 0} (e^x-2x)^frac{1}{x} $$ I know that you take the natural log $$lim_{xto 0} frac{1}{x}ln(e^x-2x) $$ which is $$lim_{xto 0} frac{ln(e^x-2x)}{x} $$ but what is after this?






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 19:01









gimusi

92.8k84494




92.8k84494










asked Dec 9 '18 at 18:47









ovil101ovil101

373




373












  • $begingroup$
    You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator...
    $endgroup$
    – the_candyman
    Dec 9 '18 at 18:52


















  • $begingroup$
    You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator...
    $endgroup$
    – the_candyman
    Dec 9 '18 at 18:52
















$begingroup$
You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator...
$endgroup$
– the_candyman
Dec 9 '18 at 18:52




$begingroup$
You have just taken the natural log. Now, Hopital rule requires that you differentiate numerator and denominator...
$endgroup$
– the_candyman
Dec 9 '18 at 18:52










4 Answers
4






active

oldest

votes


















1












$begingroup$

Now since the limit is of $0/0$ form we can apply L'Hopital's rule. So differentiate numerator and denominator, we get



$lim_{x to 0}(e^x-2)/(e^x-2x)=-1$.



Now the real limit comes out to be $e^{-1}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I need to take a break my God
    $endgroup$
    – ovil101
    Dec 9 '18 at 19:00










  • $begingroup$
    Don't worry it happens. :)
    $endgroup$
    – Jimmy
    Dec 9 '18 at 19:01



















0












$begingroup$

$$lim_{xto 0} (e^x-2x)^{1/x}neq lim_{xto 0} frac{1}{x}ln(e^x-2x)$$



$$lim_{xto 0} (e^x-2x)^{1/x}=lim_{xto 0} e^{ln{(e^x-2x)}/x}=$$



$$=expleft(lim_{xto 0}frac{ln (e^x-2x)}xright)$$
Now we can apply L'Hospital Rule, which means differentiating both numerator and denominator
$$expleft(lim_{xto 0}frac{e^x-2}{e^x-2x}right)=e^{-1}=frac1e$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For the form of limit 1^(infinity),
    lim f(x)^g(x) = e^ lim [{f(x) - 1}•g(x)]
    Using Maclaurin expansion, you will get e^(-x/x) which is e^-1



    e^x = 1 + x/1! + (x^2)/2! + (x^3)/3! ...






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      HINT



      We have that only by standard limits



      $$lim_{xto 0} (e^x-2x)^frac{1}{x}=lim_{xto 0} eleft(1-frac{2x}{e^x}right)^frac{1}{x}$$



      and by $y=frac1x to infty$



      $$left(1-frac{2x}{e^x}right)^frac{1}{x}=left(1-frac{2}{ye^{1/y}}right)^y=left[left(1-frac{2}{ye^{1/y}}right)^{frac{ye^{1/y}}2}right]^{frac{2}{e^{1/y}}}$$






      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%2f3032781%2fnatural-logs-with-lhopitals-rule%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












        $begingroup$

        Now since the limit is of $0/0$ form we can apply L'Hopital's rule. So differentiate numerator and denominator, we get



        $lim_{x to 0}(e^x-2)/(e^x-2x)=-1$.



        Now the real limit comes out to be $e^{-1}$.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          I need to take a break my God
          $endgroup$
          – ovil101
          Dec 9 '18 at 19:00










        • $begingroup$
          Don't worry it happens. :)
          $endgroup$
          – Jimmy
          Dec 9 '18 at 19:01
















        1












        $begingroup$

        Now since the limit is of $0/0$ form we can apply L'Hopital's rule. So differentiate numerator and denominator, we get



        $lim_{x to 0}(e^x-2)/(e^x-2x)=-1$.



        Now the real limit comes out to be $e^{-1}$.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          I need to take a break my God
          $endgroup$
          – ovil101
          Dec 9 '18 at 19:00










        • $begingroup$
          Don't worry it happens. :)
          $endgroup$
          – Jimmy
          Dec 9 '18 at 19:01














        1












        1








        1





        $begingroup$

        Now since the limit is of $0/0$ form we can apply L'Hopital's rule. So differentiate numerator and denominator, we get



        $lim_{x to 0}(e^x-2)/(e^x-2x)=-1$.



        Now the real limit comes out to be $e^{-1}$.






        share|cite|improve this answer









        $endgroup$



        Now since the limit is of $0/0$ form we can apply L'Hopital's rule. So differentiate numerator and denominator, we get



        $lim_{x to 0}(e^x-2)/(e^x-2x)=-1$.



        Now the real limit comes out to be $e^{-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 18:51









        JimmyJimmy

        30913




        30913












        • $begingroup$
          I need to take a break my God
          $endgroup$
          – ovil101
          Dec 9 '18 at 19:00










        • $begingroup$
          Don't worry it happens. :)
          $endgroup$
          – Jimmy
          Dec 9 '18 at 19:01


















        • $begingroup$
          I need to take a break my God
          $endgroup$
          – ovil101
          Dec 9 '18 at 19:00










        • $begingroup$
          Don't worry it happens. :)
          $endgroup$
          – Jimmy
          Dec 9 '18 at 19:01
















        $begingroup$
        I need to take a break my God
        $endgroup$
        – ovil101
        Dec 9 '18 at 19:00




        $begingroup$
        I need to take a break my God
        $endgroup$
        – ovil101
        Dec 9 '18 at 19:00












        $begingroup$
        Don't worry it happens. :)
        $endgroup$
        – Jimmy
        Dec 9 '18 at 19:01




        $begingroup$
        Don't worry it happens. :)
        $endgroup$
        – Jimmy
        Dec 9 '18 at 19:01











        0












        $begingroup$

        $$lim_{xto 0} (e^x-2x)^{1/x}neq lim_{xto 0} frac{1}{x}ln(e^x-2x)$$



        $$lim_{xto 0} (e^x-2x)^{1/x}=lim_{xto 0} e^{ln{(e^x-2x)}/x}=$$



        $$=expleft(lim_{xto 0}frac{ln (e^x-2x)}xright)$$
        Now we can apply L'Hospital Rule, which means differentiating both numerator and denominator
        $$expleft(lim_{xto 0}frac{e^x-2}{e^x-2x}right)=e^{-1}=frac1e$$






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          $$lim_{xto 0} (e^x-2x)^{1/x}neq lim_{xto 0} frac{1}{x}ln(e^x-2x)$$



          $$lim_{xto 0} (e^x-2x)^{1/x}=lim_{xto 0} e^{ln{(e^x-2x)}/x}=$$



          $$=expleft(lim_{xto 0}frac{ln (e^x-2x)}xright)$$
          Now we can apply L'Hospital Rule, which means differentiating both numerator and denominator
          $$expleft(lim_{xto 0}frac{e^x-2}{e^x-2x}right)=e^{-1}=frac1e$$






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            $$lim_{xto 0} (e^x-2x)^{1/x}neq lim_{xto 0} frac{1}{x}ln(e^x-2x)$$



            $$lim_{xto 0} (e^x-2x)^{1/x}=lim_{xto 0} e^{ln{(e^x-2x)}/x}=$$



            $$=expleft(lim_{xto 0}frac{ln (e^x-2x)}xright)$$
            Now we can apply L'Hospital Rule, which means differentiating both numerator and denominator
            $$expleft(lim_{xto 0}frac{e^x-2}{e^x-2x}right)=e^{-1}=frac1e$$






            share|cite|improve this answer









            $endgroup$



            $$lim_{xto 0} (e^x-2x)^{1/x}neq lim_{xto 0} frac{1}{x}ln(e^x-2x)$$



            $$lim_{xto 0} (e^x-2x)^{1/x}=lim_{xto 0} e^{ln{(e^x-2x)}/x}=$$



            $$=expleft(lim_{xto 0}frac{ln (e^x-2x)}xright)$$
            Now we can apply L'Hospital Rule, which means differentiating both numerator and denominator
            $$expleft(lim_{xto 0}frac{e^x-2}{e^x-2x}right)=e^{-1}=frac1e$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 19:01









            Lorenzo B.Lorenzo B.

            1,8402520




            1,8402520























                0












                $begingroup$

                For the form of limit 1^(infinity),
                lim f(x)^g(x) = e^ lim [{f(x) - 1}•g(x)]
                Using Maclaurin expansion, you will get e^(-x/x) which is e^-1



                e^x = 1 + x/1! + (x^2)/2! + (x^3)/3! ...






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  For the form of limit 1^(infinity),
                  lim f(x)^g(x) = e^ lim [{f(x) - 1}•g(x)]
                  Using Maclaurin expansion, you will get e^(-x/x) which is e^-1



                  e^x = 1 + x/1! + (x^2)/2! + (x^3)/3! ...






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    For the form of limit 1^(infinity),
                    lim f(x)^g(x) = e^ lim [{f(x) - 1}•g(x)]
                    Using Maclaurin expansion, you will get e^(-x/x) which is e^-1



                    e^x = 1 + x/1! + (x^2)/2! + (x^3)/3! ...






                    share|cite|improve this answer









                    $endgroup$



                    For the form of limit 1^(infinity),
                    lim f(x)^g(x) = e^ lim [{f(x) - 1}•g(x)]
                    Using Maclaurin expansion, you will get e^(-x/x) which is e^-1



                    e^x = 1 + x/1! + (x^2)/2! + (x^3)/3! ...







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 9 '18 at 19:16









                    Debjit KarDebjit Kar

                    11




                    11























                        -1












                        $begingroup$

                        HINT



                        We have that only by standard limits



                        $$lim_{xto 0} (e^x-2x)^frac{1}{x}=lim_{xto 0} eleft(1-frac{2x}{e^x}right)^frac{1}{x}$$



                        and by $y=frac1x to infty$



                        $$left(1-frac{2x}{e^x}right)^frac{1}{x}=left(1-frac{2}{ye^{1/y}}right)^y=left[left(1-frac{2}{ye^{1/y}}right)^{frac{ye^{1/y}}2}right]^{frac{2}{e^{1/y}}}$$






                        share|cite|improve this answer









                        $endgroup$


















                          -1












                          $begingroup$

                          HINT



                          We have that only by standard limits



                          $$lim_{xto 0} (e^x-2x)^frac{1}{x}=lim_{xto 0} eleft(1-frac{2x}{e^x}right)^frac{1}{x}$$



                          and by $y=frac1x to infty$



                          $$left(1-frac{2x}{e^x}right)^frac{1}{x}=left(1-frac{2}{ye^{1/y}}right)^y=left[left(1-frac{2}{ye^{1/y}}right)^{frac{ye^{1/y}}2}right]^{frac{2}{e^{1/y}}}$$






                          share|cite|improve this answer









                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            HINT



                            We have that only by standard limits



                            $$lim_{xto 0} (e^x-2x)^frac{1}{x}=lim_{xto 0} eleft(1-frac{2x}{e^x}right)^frac{1}{x}$$



                            and by $y=frac1x to infty$



                            $$left(1-frac{2x}{e^x}right)^frac{1}{x}=left(1-frac{2}{ye^{1/y}}right)^y=left[left(1-frac{2}{ye^{1/y}}right)^{frac{ye^{1/y}}2}right]^{frac{2}{e^{1/y}}}$$






                            share|cite|improve this answer









                            $endgroup$



                            HINT



                            We have that only by standard limits



                            $$lim_{xto 0} (e^x-2x)^frac{1}{x}=lim_{xto 0} eleft(1-frac{2x}{e^x}right)^frac{1}{x}$$



                            and by $y=frac1x to infty$



                            $$left(1-frac{2x}{e^x}right)^frac{1}{x}=left(1-frac{2}{ye^{1/y}}right)^y=left[left(1-frac{2}{ye^{1/y}}right)^{frac{ye^{1/y}}2}right]^{frac{2}{e^{1/y}}}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 9 '18 at 19:01









                            gimusigimusi

                            92.8k84494




                            92.8k84494






























                                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%2f3032781%2fnatural-logs-with-lhopitals-rule%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

                                Bundesstraße 106

                                Image
                                Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                                Read more

                                Verónica Boquete

                                Image
                                Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                                Read more

                                Ida-Boy-Ed-Garten

                                Image
                                Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                                Read more