Advance Calculus Limit question












3












$begingroup$


I'm trying to compute this limit without the use of L'Hopital's rule:



$$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






calculus limits limits-without-lhopital







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    3












    $begingroup$


    I'm trying to compute this limit without the use of L'Hopital's rule:



    $$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



    I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






    calculus limits limits-without-lhopital







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I'm trying to compute this limit without the use of L'Hopital's rule:



      $$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



      I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






      calculus limits limits-without-lhopital







      share|cite|improve this question











      $endgroup$




      I'm trying to compute this limit without the use of L'Hopital's rule:



      $$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



      I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






      calculus limits limits-without-lhopital




      calculus limits limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 6 hours ago









      Foobaz John

      22.9k41552




      22.9k41552










      asked 6 hours ago









      Kevin CalderonKevin Calderon

      563




      563






















          3 Answers
          3






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          6












          $begingroup$

          Write the limit as
          $$
          lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
          $$
          and use the fact that
          $$
          lim_{xto 0+}frac{-2}{x}=-infty.
          $$
          to find that the limit equals $-1$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            A substitution can be helpful, as it transforms the expression into a rational function:




            • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


            begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
            & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
            & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
            & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
            end{eqnarray*}






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



              Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






              share|cite|improve this answer









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

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                active

                oldest

                votes






                active

                oldest

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                6












                $begingroup$

                Write the limit as
                $$
                lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                $$
                and use the fact that
                $$
                lim_{xto 0+}frac{-2}{x}=-infty.
                $$
                to find that the limit equals $-1$.






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  Write the limit as
                  $$
                  lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                  $$
                  and use the fact that
                  $$
                  lim_{xto 0+}frac{-2}{x}=-infty.
                  $$
                  to find that the limit equals $-1$.






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    Write the limit as
                    $$
                    lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                    $$
                    and use the fact that
                    $$
                    lim_{xto 0+}frac{-2}{x}=-infty.
                    $$
                    to find that the limit equals $-1$.






                    share|cite|improve this answer









                    $endgroup$



                    Write the limit as
                    $$
                    lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                    $$
                    and use the fact that
                    $$
                    lim_{xto 0+}frac{-2}{x}=-infty.
                    $$
                    to find that the limit equals $-1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 6 hours ago









                    Foobaz JohnFoobaz John

                    22.9k41552




                    22.9k41552























                        3












                        $begingroup$

                        A substitution can be helpful, as it transforms the expression into a rational function:




                        • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                        begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                        & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                        & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                        & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                        end{eqnarray*}






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          A substitution can be helpful, as it transforms the expression into a rational function:




                          • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                          begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                          & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                          & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                          & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                          end{eqnarray*}






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            A substitution can be helpful, as it transforms the expression into a rational function:




                            • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                            begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                            & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                            & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                            & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                            end{eqnarray*}






                            share|cite|improve this answer









                            $endgroup$



                            A substitution can be helpful, as it transforms the expression into a rational function:




                            • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                            begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                            & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                            & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                            & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                            end{eqnarray*}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            trancelocationtrancelocation

                            13.5k1827




                            13.5k1827























                                0












                                $begingroup$

                                $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                  Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                    Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                    Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 40 mins ago









                                    Paras KhoslaParas Khosla

                                    2,758423




                                    2,758423






























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