Expressing a divergent function in algebraic terms












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We're asked to find the $lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
})cos(x)$. I've tried to break down the expression a little bit but couldn't get far. The graph clearly shows that the function oscillates and thus the limit doesn't exist but how would one get to the conclusion algebraically? Perhaps we could do so with the Squeeze theorem by stating that $cos(x)leq1$



My attempt:



$$=lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
})cos(x)$$
$$=lim_{xtoinfty}xsqrt{Big(1+frac{1}{x}Big)}-xsqrt{Big(1-frac{1}{x}Big)}cos(x)$$






calculus limits convergence







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    We're asked to find the $lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
    })cos(x)$    . I've tried to break down the expression a little bit but couldn't get far. The graph clearly shows that the function oscillates and thus the limit doesn't exist but how would one get to the conclusion algebraically? Perhaps we could do so with the Squeeze theorem by stating that $cos(x)leq1$



    My attempt:



    $$=lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
    })cos(x)$$
    $$=lim_{xtoinfty}xsqrt{Big(1+frac{1}{x}Big)}-xsqrt{Big(1-frac{1}{x}Big)}cos(x)$$






    calculus limits convergence







    share|cite|improve this question

























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      We're asked to find the $lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
      })cos(x)$      . I've tried to break down the expression a little bit but couldn't get far. The graph clearly shows that the function oscillates and thus the limit doesn't exist but how would one get to the conclusion algebraically? Perhaps we could do so with the Squeeze theorem by stating that $cos(x)leq1$



      My attempt:



      $$=lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
      })cos(x)$$
      $$=lim_{xtoinfty}xsqrt{Big(1+frac{1}{x}Big)}-xsqrt{Big(1-frac{1}{x}Big)}cos(x)$$






      calculus limits convergence







      share|cite|improve this question













      We're asked to find the $lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
      })cos(x)$      . I've tried to break down the expression a little bit but couldn't get far. The graph clearly shows that the function oscillates and thus the limit doesn't exist but how would one get to the conclusion algebraically? Perhaps we could do so with the Squeeze theorem by stating that $cos(x)leq1$



      My attempt:



      $$=lim_{xtoinfty}(sqrt{x^2+x}-sqrt{x^2-x
      })cos(x)$$
      $$=lim_{xtoinfty}xsqrt{Big(1+frac{1}{x}Big)}-xsqrt{Big(1-frac{1}{x}Big)}cos(x)$$






      calculus limits convergence




      calculus limits convergence






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      asked Nov 24 at 22:20









      kareem bokai

      344




      344






















          2 Answers
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          Use the fact thatbegin{align}sqrt{x^2+x}-sqrt{x^2-x}&=frac{left(sqrt{x^2+x}-sqrt{x^2-x}right)left(sqrt{x^2+x}+sqrt{x^2-x}right)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{x^2+x-(x^2-x)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{2x}{sqrt{x^2+x}+sqrt{x^2-x}}end{align}






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            By binomial first order expansion we have that




            • $sqrt{x^2+x}=x(1+1/x)^frac12=x+frac12+o(1)$

            • $sqrt{x^2-x}=x(1-1/x)^frac12=x-frac12+o(1)$


            and then



            $$sqrt{x^2+x}-sqrt{x^2-x
            }=x+frac12-x+frac12+o(1)=1+o(1) to 1$$



            then consider




            • $x_n = 2pi n to infty$


            $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 1=1$$



            but for




            • $x_n = frac{pi}2+2pi n to infty$


            $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 0=0$$



            therefore, since there exist two subsequnces with different limit, the given limit doesn't exist.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              Use the fact thatbegin{align}sqrt{x^2+x}-sqrt{x^2-x}&=frac{left(sqrt{x^2+x}-sqrt{x^2-x}right)left(sqrt{x^2+x}+sqrt{x^2-x}right)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{x^2+x-(x^2-x)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{2x}{sqrt{x^2+x}+sqrt{x^2-x}}end{align}






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                Use the fact thatbegin{align}sqrt{x^2+x}-sqrt{x^2-x}&=frac{left(sqrt{x^2+x}-sqrt{x^2-x}right)left(sqrt{x^2+x}+sqrt{x^2-x}right)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{x^2+x-(x^2-x)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{2x}{sqrt{x^2+x}+sqrt{x^2-x}}end{align}






                share|cite|improve this answer
























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                  Use the fact thatbegin{align}sqrt{x^2+x}-sqrt{x^2-x}&=frac{left(sqrt{x^2+x}-sqrt{x^2-x}right)left(sqrt{x^2+x}+sqrt{x^2-x}right)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{x^2+x-(x^2-x)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{2x}{sqrt{x^2+x}+sqrt{x^2-x}}end{align}






                  share|cite|improve this answer












                  Use the fact thatbegin{align}sqrt{x^2+x}-sqrt{x^2-x}&=frac{left(sqrt{x^2+x}-sqrt{x^2-x}right)left(sqrt{x^2+x}+sqrt{x^2-x}right)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{x^2+x-(x^2-x)}{sqrt{x^2+x}+sqrt{x^2-x}}\&=frac{2x}{sqrt{x^2+x}+sqrt{x^2-x}}end{align}







                  share|cite|improve this answer












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                  answered Nov 24 at 22:24









                  José Carlos Santos

                  148k22117218




                  148k22117218























                      0














                      By binomial first order expansion we have that




                      • $sqrt{x^2+x}=x(1+1/x)^frac12=x+frac12+o(1)$

                      • $sqrt{x^2-x}=x(1-1/x)^frac12=x-frac12+o(1)$


                      and then



                      $$sqrt{x^2+x}-sqrt{x^2-x
                      }=x+frac12-x+frac12+o(1)=1+o(1) to 1$$



                      then consider




                      • $x_n = 2pi n to infty$


                      $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 1=1$$



                      but for




                      • $x_n = frac{pi}2+2pi n to infty$


                      $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 0=0$$



                      therefore, since there exist two subsequnces with different limit, the given limit doesn't exist.






                      share|cite|improve this answer


























                        0














                        By binomial first order expansion we have that




                        • $sqrt{x^2+x}=x(1+1/x)^frac12=x+frac12+o(1)$

                        • $sqrt{x^2-x}=x(1-1/x)^frac12=x-frac12+o(1)$


                        and then



                        $$sqrt{x^2+x}-sqrt{x^2-x
                        }=x+frac12-x+frac12+o(1)=1+o(1) to 1$$



                        then consider




                        • $x_n = 2pi n to infty$


                        $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 1=1$$



                        but for




                        • $x_n = frac{pi}2+2pi n to infty$


                        $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 0=0$$



                        therefore, since there exist two subsequnces with different limit, the given limit doesn't exist.






                        share|cite|improve this answer
























                          0












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                          0






                          By binomial first order expansion we have that




                          • $sqrt{x^2+x}=x(1+1/x)^frac12=x+frac12+o(1)$

                          • $sqrt{x^2-x}=x(1-1/x)^frac12=x-frac12+o(1)$


                          and then



                          $$sqrt{x^2+x}-sqrt{x^2-x
                          }=x+frac12-x+frac12+o(1)=1+o(1) to 1$$



                          then consider




                          • $x_n = 2pi n to infty$


                          $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 1=1$$



                          but for




                          • $x_n = frac{pi}2+2pi n to infty$


                          $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 0=0$$



                          therefore, since there exist two subsequnces with different limit, the given limit doesn't exist.






                          share|cite|improve this answer












                          By binomial first order expansion we have that




                          • $sqrt{x^2+x}=x(1+1/x)^frac12=x+frac12+o(1)$

                          • $sqrt{x^2-x}=x(1-1/x)^frac12=x-frac12+o(1)$


                          and then



                          $$sqrt{x^2+x}-sqrt{x^2-x
                          }=x+frac12-x+frac12+o(1)=1+o(1) to 1$$



                          then consider




                          • $x_n = 2pi n to infty$


                          $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 1=1$$



                          but for




                          • $x_n = frac{pi}2+2pi n to infty$


                          $$(sqrt{x_n^2+x_n}-sqrt{x_n^2-x_n})cos(x_n) to 1cdot 0=0$$



                          therefore, since there exist two subsequnces with different limit, the given limit doesn't exist.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 24 at 23:32









                          gimusi

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