Finding and Proving the limit as x approaches infinity.












0












$begingroup$


I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.



I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.






calculus







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
    $endgroup$
    – T. Bongers
    Nov 30 '18 at 16:27










  • $begingroup$
    I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
    $endgroup$
    – Scott Dye
    Nov 30 '18 at 19:00
















0












$begingroup$


I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.



I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.






calculus







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
    $endgroup$
    – T. Bongers
    Nov 30 '18 at 16:27










  • $begingroup$
    I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
    $endgroup$
    – Scott Dye
    Nov 30 '18 at 19:00














0












0








0





$begingroup$


I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.



I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.






calculus







share|cite|improve this question











$endgroup$




I need to evaluate the limit as x approaches infinity of $dfrac{1-x^2}{x-2}$ and then prove it.



I have found the limit to be negative infinity. Now to prove it I believe I need to find two subsequences that diverge to two different results and therefore prove that the limit actually does not exist. Am I thinking correctly? If so I am stuck on what subsequences to choose.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 16:24









amWhy

192k28225439




192k28225439










asked Nov 30 '18 at 16:21









Scott DyeScott Dye

11




11








  • 4




    $begingroup$
    What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
    $endgroup$
    – T. Bongers
    Nov 30 '18 at 16:27










  • $begingroup$
    I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
    $endgroup$
    – Scott Dye
    Nov 30 '18 at 19:00














  • 4




    $begingroup$
    What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
    $endgroup$
    – T. Bongers
    Nov 30 '18 at 16:27










  • $begingroup$
    I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
    $endgroup$
    – Scott Dye
    Nov 30 '18 at 19:00








4




4




$begingroup$
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
$endgroup$
– T. Bongers
Nov 30 '18 at 16:27




$begingroup$
What you've written is not consistent. You said that the limit is negative infinity (correct), but that you're trying to prove the limit does not exist. How are these compatible?
$endgroup$
– T. Bongers
Nov 30 '18 at 16:27












$begingroup$
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
$endgroup$
– Scott Dye
Nov 30 '18 at 19:00




$begingroup$
I understand and that is part of my question. The proof part is to show that for every epsilon > 0 there exists M>0 s.t. the absolute value f(x) - L < epsilon. Since this is not bounded below and approaches negative infinity does the Limit actually not exist.
$endgroup$
– Scott Dye
Nov 30 '18 at 19:00










1 Answer
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$begingroup$

Hint:
For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
$$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$



I would rather try the following approach (first divide by $x$)




$$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$







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    0












    $begingroup$

    Hint:
    For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
    $$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$



    I would rather try the following approach (first divide by $x$)




    $$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$







    share|cite|improve this answer









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      0












      $begingroup$

      Hint:
      For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
      $$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$



      I would rather try the following approach (first divide by $x$)




      $$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$







      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:
        For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
        $$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$



        I would rather try the following approach (first divide by $x$)




        $$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$







        share|cite|improve this answer









        $endgroup$



        Hint:
        For some constant $ainmathbb R$ and some constant $ninmathbb R_{>0}$
        $$lim_{xto infty}Bigl(frac{a}{x^n}Bigr)=0$$



        I would rather try the following approach (first divide by $x$)




        $$lim_{xto infty}Bigl(frac{1-x^2}{x-2}Bigr)=lim_{xto infty}Biggl(frac{frac{1}{x}-x}{1-frac{2}{x}}Biggr)=frac{lim_{xto infty}bigl(frac{1}{x}-xbigr)}{lim_{xto infty}bigl(1-frac{2}{x}bigr)}=frac{-infty}{1}=-infty$$








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 16:57









        Dr. MathvaDr. Mathva

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