Evaluate $ lim_{xto 0}|frac{5^x - 5^{-x}}{5^x-1}| $ without using L'Hospital's rule.












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$begingroup$


$$
lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
$$



I know the limit is equal to 2. But I am not allowed to use L'Hospital.
How can I evaluate the limit without L'Hospital?






calculus limits limits-without-lhopital







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$endgroup$

















    0












    $begingroup$


    $$
    lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
    $$



    I know the limit is equal to 2. But I am not allowed to use L'Hospital.
    How can I evaluate the limit without L'Hospital?






    calculus limits limits-without-lhopital







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $$
      lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
      $$



      I know the limit is equal to 2. But I am not allowed to use L'Hospital.
      How can I evaluate the limit without L'Hospital?






      calculus limits limits-without-lhopital







      share|cite|improve this question











      $endgroup$




      $$
      lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
      $$



      I know the limit is equal to 2. But I am not allowed to use L'Hospital.
      How can I evaluate the limit without L'Hospital?






      calculus limits limits-without-lhopital




      calculus limits limits-without-lhopital






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      edited Nov 30 '18 at 20:57









      Martin Sleziak

      44.8k10118272




      44.8k10118272










      asked Nov 30 '18 at 12:25









      nofnof

      32




      32






















          4 Answers
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          8












          $begingroup$

          Hint:



          $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






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          $endgroup$





















            2












            $begingroup$

            Set $y= 5^x$ and consider $ylongrightarrow 1$
            $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              HINT



              For positive $x$ we have:
              $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
              As $xrightarrow 0$ this approaches to $1+1=2$.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                share|cite|improve this answer









                $endgroup$













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                  4 Answers
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                  4 Answers
                  4






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                  active

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                  active

                  oldest

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                  8












                  $begingroup$

                  Hint:



                  $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






                  share|cite|improve this answer









                  $endgroup$


















                    8












                    $begingroup$

                    Hint:



                    $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






                    share|cite|improve this answer









                    $endgroup$
















                      8












                      8








                      8





                      $begingroup$

                      Hint:



                      $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint:



                      $$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$







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                      answered Nov 30 '18 at 12:33









                      KM101KM101

                      5,9251524




                      5,9251524























                          2












                          $begingroup$

                          Set $y= 5^x$ and consider $ylongrightarrow 1$
                          $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






                          share|cite|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Set $y= 5^x$ and consider $ylongrightarrow 1$
                            $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






                            share|cite|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Set $y= 5^x$ and consider $ylongrightarrow 1$
                              $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$






                              share|cite|improve this answer









                              $endgroup$



                              Set $y= 5^x$ and consider $ylongrightarrow 1$
                              $$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 30 '18 at 12:32









                              trancelocationtrancelocation

                              11.2k1724




                              11.2k1724























                                  2












                                  $begingroup$

                                  HINT



                                  For positive $x$ we have:
                                  $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                  As $xrightarrow 0$ this approaches to $1+1=2$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    2












                                    $begingroup$

                                    HINT



                                    For positive $x$ we have:
                                    $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                    As $xrightarrow 0$ this approaches to $1+1=2$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      2












                                      2








                                      2





                                      $begingroup$

                                      HINT



                                      For positive $x$ we have:
                                      $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                      As $xrightarrow 0$ this approaches to $1+1=2$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      HINT



                                      For positive $x$ we have:
                                      $frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
                                      As $xrightarrow 0$ this approaches to $1+1=2$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 30 '18 at 12:52









                                      A. PongráczA. Pongrácz

                                      5,9531929




                                      5,9531929























                                          0












                                          $begingroup$

                                          Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                          $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                            $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                              $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $






                                              share|cite|improve this answer









                                              $endgroup$



                                              Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$



                                              $lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Nov 30 '18 at 12:36









                                              Shubham JohriShubham Johri

                                              5,172717




                                              5,172717






























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