Natural logs with L'Hopital's rule












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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







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$endgroup$












  • $begingroup$
    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






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share|cite|improve this question













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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









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        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






























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