Limit using definition of $e$












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$begingroup$


I'm trying to calculate the following limit:
$$
lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
$$
My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.



Does anyone have any advice?






real-analysis







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    1












    $begingroup$


    I'm trying to calculate the following limit:
    $$
    lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
    $$
    My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.



    Does anyone have any advice?






    real-analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm trying to calculate the following limit:
      $$
      lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
      $$
      My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.



      Does anyone have any advice?






      real-analysis







      share|cite|improve this question











      $endgroup$




      I'm trying to calculate the following limit:
      $$
      lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
      $$
      My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.



      Does anyone have any advice?






      real-analysis




      real-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 21:19









      egreg

      185k1486206




      185k1486206










      asked Jan 7 at 21:14









      Bob Garth Bob Garth

      61




      61






















          5 Answers
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          1












          $begingroup$

          You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @egreg Thanks; fixed.
            $endgroup$
            – J.G.
            Jan 7 at 21:35



















          0












          $begingroup$

          Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
          $$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
          By using L'Hôpital's Rule, this is equal to:
          $$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
          $$=-4 cdot b$$
          $$therefore L=e^{-4 cdot b}$$






          share|cite|improve this answer









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          • $begingroup$
            use limlimits_{xto 0} for $limlimits_{xto 0}$
            $endgroup$
            – zwim
            Jan 7 at 21:27












          • $begingroup$
            A few judicious parentheses might help here.
            $endgroup$
            – TonyK
            Jan 7 at 21:30



















          0












          $begingroup$

          It's basically right:
          $$
          -4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
          $$
          Now you can substitute $-4x=y$, so the limit becomes
          $$
          lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
          $$
          and $lim_{yto0}(1+y)^{1/y}=e$ is known.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
            $$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
            $$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
            $$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
            $$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
            $$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:





              • $y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
                begin{eqnarray*}
                bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
                & = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
                & stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
                end{eqnarray*}






              share|cite|improve this answer









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                5 Answers
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                5 Answers
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                active

                oldest

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                active

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                1












                $begingroup$

                You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  @egreg Thanks; fixed.
                  $endgroup$
                  – J.G.
                  Jan 7 at 21:35
















                1












                $begingroup$

                You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  @egreg Thanks; fixed.
                  $endgroup$
                  – J.G.
                  Jan 7 at 21:35














                1












                1








                1





                $begingroup$

                You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$






                share|cite|improve this answer











                $endgroup$



                You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 7 at 21:35

























                answered Jan 7 at 21:25









                J.G.J.G.

                32.3k23250




                32.3k23250












                • $begingroup$
                  @egreg Thanks; fixed.
                  $endgroup$
                  – J.G.
                  Jan 7 at 21:35


















                • $begingroup$
                  @egreg Thanks; fixed.
                  $endgroup$
                  – J.G.
                  Jan 7 at 21:35
















                $begingroup$
                @egreg Thanks; fixed.
                $endgroup$
                – J.G.
                Jan 7 at 21:35




                $begingroup$
                @egreg Thanks; fixed.
                $endgroup$
                – J.G.
                Jan 7 at 21:35











                0












                $begingroup$

                Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
                $$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
                By using L'Hôpital's Rule, this is equal to:
                $$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
                $$=-4 cdot b$$
                $$therefore L=e^{-4 cdot b}$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  use limlimits_{xto 0} for $limlimits_{xto 0}$
                  $endgroup$
                  – zwim
                  Jan 7 at 21:27












                • $begingroup$
                  A few judicious parentheses might help here.
                  $endgroup$
                  – TonyK
                  Jan 7 at 21:30
















                0












                $begingroup$

                Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
                $$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
                By using L'Hôpital's Rule, this is equal to:
                $$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
                $$=-4 cdot b$$
                $$therefore L=e^{-4 cdot b}$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  use limlimits_{xto 0} for $limlimits_{xto 0}$
                  $endgroup$
                  – zwim
                  Jan 7 at 21:27












                • $begingroup$
                  A few judicious parentheses might help here.
                  $endgroup$
                  – TonyK
                  Jan 7 at 21:30














                0












                0








                0





                $begingroup$

                Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
                $$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
                By using L'Hôpital's Rule, this is equal to:
                $$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
                $$=-4 cdot b$$
                $$therefore L=e^{-4 cdot b}$$






                share|cite|improve this answer









                $endgroup$



                Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
                $$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
                By using L'Hôpital's Rule, this is equal to:
                $$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
                $$=-4 cdot b$$
                $$therefore L=e^{-4 cdot b}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 21:26









                Peter ForemanPeter Foreman

                4,5401216




                4,5401216












                • $begingroup$
                  use limlimits_{xto 0} for $limlimits_{xto 0}$
                  $endgroup$
                  – zwim
                  Jan 7 at 21:27












                • $begingroup$
                  A few judicious parentheses might help here.
                  $endgroup$
                  – TonyK
                  Jan 7 at 21:30


















                • $begingroup$
                  use limlimits_{xto 0} for $limlimits_{xto 0}$
                  $endgroup$
                  – zwim
                  Jan 7 at 21:27












                • $begingroup$
                  A few judicious parentheses might help here.
                  $endgroup$
                  – TonyK
                  Jan 7 at 21:30
















                $begingroup$
                use limlimits_{xto 0} for $limlimits_{xto 0}$
                $endgroup$
                – zwim
                Jan 7 at 21:27






                $begingroup$
                use limlimits_{xto 0} for $limlimits_{xto 0}$
                $endgroup$
                – zwim
                Jan 7 at 21:27














                $begingroup$
                A few judicious parentheses might help here.
                $endgroup$
                – TonyK
                Jan 7 at 21:30




                $begingroup$
                A few judicious parentheses might help here.
                $endgroup$
                – TonyK
                Jan 7 at 21:30











                0












                $begingroup$

                It's basically right:
                $$
                -4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
                $$
                Now you can substitute $-4x=y$, so the limit becomes
                $$
                lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
                $$
                and $lim_{yto0}(1+y)^{1/y}=e$ is known.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  It's basically right:
                  $$
                  -4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
                  $$
                  Now you can substitute $-4x=y$, so the limit becomes
                  $$
                  lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
                  $$
                  and $lim_{yto0}(1+y)^{1/y}=e$ is known.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    It's basically right:
                    $$
                    -4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
                    $$
                    Now you can substitute $-4x=y$, so the limit becomes
                    $$
                    lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
                    $$
                    and $lim_{yto0}(1+y)^{1/y}=e$ is known.






                    share|cite|improve this answer









                    $endgroup$



                    It's basically right:
                    $$
                    -4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
                    $$
                    Now you can substitute $-4x=y$, so the limit becomes
                    $$
                    lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
                    $$
                    and $lim_{yto0}(1+y)^{1/y}=e$ is known.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 7 at 21:39









                    egregegreg

                    185k1486206




                    185k1486206























                        0












                        $begingroup$

                        Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
                        $$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
                        $$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
                        $$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
                        $$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
                        $$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
                          $$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
                          $$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
                          $$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
                          $$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
                          $$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
                            $$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
                            $$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
                            $$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
                            $$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
                            $$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.






                            share|cite|improve this answer









                            $endgroup$



                            Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
                            $$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
                            $$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
                            $$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
                            $$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
                            $$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 8 at 6:42









                            Claude LeiboviciClaude Leibovici

                            125k1158135




                            125k1158135























                                0












                                $begingroup$

                                Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:





                                • $y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
                                  begin{eqnarray*}
                                  bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
                                  & = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
                                  & stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
                                  end{eqnarray*}






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:





                                  • $y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
                                    begin{eqnarray*}
                                    bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
                                    & = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
                                    & stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
                                    end{eqnarray*}






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:





                                    • $y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
                                      begin{eqnarray*}
                                      bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
                                      & = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
                                      & stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
                                      end{eqnarray*}






                                    share|cite|improve this answer









                                    $endgroup$



                                    Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:





                                    • $y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
                                      begin{eqnarray*}
                                      bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
                                      & = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
                                      & stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
                                      end{eqnarray*}







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                                    answered Jan 8 at 12:11









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