Proving $lim_{xto-infty}x^{-k} = 0$












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Please prove $lim_{xto-infty}x^{-k} = 0$ for k a natural number using the definition of this limit. I am having problems discovering the proof.






calculus limits infinity







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$endgroup$








  • 1




    $begingroup$
    I assume that $kinmathbb{N}$. In that case since $x^ktopm infty$ (as all polynomials of positive degree do) then $x^{-k}=1/x^kto 0$.
    $endgroup$
    – freakish
    Jan 3 at 21:39












  • $begingroup$
    Please use the definition of this limit at infinity.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:43






  • 1




    $begingroup$
    Hint: prove it for $k=1$ and then use $|x^{-k}|<|x^{-1}|$ for $x<-1$ and $k>1$.
    $endgroup$
    – SmileyCraft
    Jan 3 at 21:50










  • $begingroup$
    But this requires the use of the squeeze theorem for functions as x→-∞, which has not been proven by me yet.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:55
















0












$begingroup$


Please prove $lim_{xto-infty}x^{-k} = 0$ for k a natural number using the definition of this limit. I am having problems discovering the proof.






calculus limits infinity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I assume that $kinmathbb{N}$. In that case since $x^ktopm infty$ (as all polynomials of positive degree do) then $x^{-k}=1/x^kto 0$.
    $endgroup$
    – freakish
    Jan 3 at 21:39












  • $begingroup$
    Please use the definition of this limit at infinity.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:43






  • 1




    $begingroup$
    Hint: prove it for $k=1$ and then use $|x^{-k}|<|x^{-1}|$ for $x<-1$ and $k>1$.
    $endgroup$
    – SmileyCraft
    Jan 3 at 21:50










  • $begingroup$
    But this requires the use of the squeeze theorem for functions as x→-∞, which has not been proven by me yet.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:55














0












0








0





$begingroup$


Please prove $lim_{xto-infty}x^{-k} = 0$ for k a natural number using the definition of this limit. I am having problems discovering the proof.






calculus limits infinity







share|cite|improve this question











$endgroup$




Please prove $lim_{xto-infty}x^{-k} = 0$ for k a natural number using the definition of this limit. I am having problems discovering the proof.






calculus limits infinity




calculus limits infinity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 21:42







Eric Brown

















asked Jan 3 at 21:30









Eric BrownEric Brown

11011




11011








  • 1




    $begingroup$
    I assume that $kinmathbb{N}$. In that case since $x^ktopm infty$ (as all polynomials of positive degree do) then $x^{-k}=1/x^kto 0$.
    $endgroup$
    – freakish
    Jan 3 at 21:39












  • $begingroup$
    Please use the definition of this limit at infinity.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:43






  • 1




    $begingroup$
    Hint: prove it for $k=1$ and then use $|x^{-k}|<|x^{-1}|$ for $x<-1$ and $k>1$.
    $endgroup$
    – SmileyCraft
    Jan 3 at 21:50










  • $begingroup$
    But this requires the use of the squeeze theorem for functions as x→-∞, which has not been proven by me yet.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:55














  • 1




    $begingroup$
    I assume that $kinmathbb{N}$. In that case since $x^ktopm infty$ (as all polynomials of positive degree do) then $x^{-k}=1/x^kto 0$.
    $endgroup$
    – freakish
    Jan 3 at 21:39












  • $begingroup$
    Please use the definition of this limit at infinity.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:43






  • 1




    $begingroup$
    Hint: prove it for $k=1$ and then use $|x^{-k}|<|x^{-1}|$ for $x<-1$ and $k>1$.
    $endgroup$
    – SmileyCraft
    Jan 3 at 21:50










  • $begingroup$
    But this requires the use of the squeeze theorem for functions as x→-∞, which has not been proven by me yet.
    $endgroup$
    – Eric Brown
    Jan 3 at 21:55








1




1




$begingroup$
I assume that $kinmathbb{N}$. In that case since $x^ktopm infty$ (as all polynomials of positive degree do) then $x^{-k}=1/x^kto 0$.
$endgroup$
– freakish
Jan 3 at 21:39






$begingroup$
I assume that $kinmathbb{N}$. In that case since $x^ktopm infty$ (as all polynomials of positive degree do) then $x^{-k}=1/x^kto 0$.
$endgroup$
– freakish
Jan 3 at 21:39














$begingroup$
Please use the definition of this limit at infinity.
$endgroup$
– Eric Brown
Jan 3 at 21:43




$begingroup$
Please use the definition of this limit at infinity.
$endgroup$
– Eric Brown
Jan 3 at 21:43




1




1




$begingroup$
Hint: prove it for $k=1$ and then use $|x^{-k}|<|x^{-1}|$ for $x<-1$ and $k>1$.
$endgroup$
– SmileyCraft
Jan 3 at 21:50




$begingroup$
Hint: prove it for $k=1$ and then use $|x^{-k}|<|x^{-1}|$ for $x<-1$ and $k>1$.
$endgroup$
– SmileyCraft
Jan 3 at 21:50












$begingroup$
But this requires the use of the squeeze theorem for functions as x→-∞, which has not been proven by me yet.
$endgroup$
– Eric Brown
Jan 3 at 21:55




$begingroup$
But this requires the use of the squeeze theorem for functions as x→-∞, which has not been proven by me yet.
$endgroup$
– Eric Brown
Jan 3 at 21:55










2 Answers
2






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oldest

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2












$begingroup$

For all $k > 0$ there exists a $pinmathbb{N}$ such that $frac{1}{p} < k$. Then



$$0 < frac{1}{x^k} = left(frac{1}{x}right)^k < left(frac{1}{x}right)^{1/p}$$



Since $frac{1}{x}to 0$ and $f(u) = u^{1/p}$ is continuous at $0$, we have



$$lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{uto 0}u^{1/p} = 0^{1/p} = 0$$



Therefore



$$lim_{xto -infty}frac{1}{x^k} = 0 k > 0$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The limit is equivalent to $lim_{xto -infty} frac{x^k}{x^{2k}}$, an indeterminate form of $frac{infty}{infty}$. By L'Hospitals rule, that is equivalent to $lim_{xto -infty} frac{kx^{k-1}}{2kx^{2k-1}}$. A little bit of algebraic manipulation shows that that is equal to $lim_{xto -infty} frac{1}{2}*x^{-k}$. Taking the 1/2 out of the limit, we have $lim_{xto -infty} x^{-k}=frac{1}{2}*lim_{xto -infty} x^{-k}$. Treating $lim_{xto -infty} x^{-k}$ as a variable in a linear equation to be solved for, we have $lim_{xto -infty} x^{-k}=0$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
      $endgroup$
      – Eric Brown
      Jan 5 at 2:18











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    For all $k > 0$ there exists a $pinmathbb{N}$ such that $frac{1}{p} < k$. Then



    $$0 < frac{1}{x^k} = left(frac{1}{x}right)^k < left(frac{1}{x}right)^{1/p}$$



    Since $frac{1}{x}to 0$ and $f(u) = u^{1/p}$ is continuous at $0$, we have



    $$lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{uto 0}u^{1/p} = 0^{1/p} = 0$$



    Therefore



    $$lim_{xto -infty}frac{1}{x^k} = 0 k > 0$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      For all $k > 0$ there exists a $pinmathbb{N}$ such that $frac{1}{p} < k$. Then



      $$0 < frac{1}{x^k} = left(frac{1}{x}right)^k < left(frac{1}{x}right)^{1/p}$$



      Since $frac{1}{x}to 0$ and $f(u) = u^{1/p}$ is continuous at $0$, we have



      $$lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{uto 0}u^{1/p} = 0^{1/p} = 0$$



      Therefore



      $$lim_{xto -infty}frac{1}{x^k} = 0 k > 0$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        For all $k > 0$ there exists a $pinmathbb{N}$ such that $frac{1}{p} < k$. Then



        $$0 < frac{1}{x^k} = left(frac{1}{x}right)^k < left(frac{1}{x}right)^{1/p}$$



        Since $frac{1}{x}to 0$ and $f(u) = u^{1/p}$ is continuous at $0$, we have



        $$lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{uto 0}u^{1/p} = 0^{1/p} = 0$$



        Therefore



        $$lim_{xto -infty}frac{1}{x^k} = 0 k > 0$$






        share|cite|improve this answer









        $endgroup$



        For all $k > 0$ there exists a $pinmathbb{N}$ such that $frac{1}{p} < k$. Then



        $$0 < frac{1}{x^k} = left(frac{1}{x}right)^k < left(frac{1}{x}right)^{1/p}$$



        Since $frac{1}{x}to 0$ and $f(u) = u^{1/p}$ is continuous at $0$, we have



        $$lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{xto -infty}left(frac{1}{x}right)^{1/p} = lim_{uto 0}u^{1/p} = 0^{1/p} = 0$$



        Therefore



        $$lim_{xto -infty}frac{1}{x^k} = 0 k > 0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 21:56









        WolfyWolfy

        2,31811138




        2,31811138























            0












            $begingroup$

            The limit is equivalent to $lim_{xto -infty} frac{x^k}{x^{2k}}$, an indeterminate form of $frac{infty}{infty}$. By L'Hospitals rule, that is equivalent to $lim_{xto -infty} frac{kx^{k-1}}{2kx^{2k-1}}$. A little bit of algebraic manipulation shows that that is equal to $lim_{xto -infty} frac{1}{2}*x^{-k}$. Taking the 1/2 out of the limit, we have $lim_{xto -infty} x^{-k}=frac{1}{2}*lim_{xto -infty} x^{-k}$. Treating $lim_{xto -infty} x^{-k}$ as a variable in a linear equation to be solved for, we have $lim_{xto -infty} x^{-k}=0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              $endgroup$
              – Eric Brown
              Jan 5 at 2:18
















            0












            $begingroup$

            The limit is equivalent to $lim_{xto -infty} frac{x^k}{x^{2k}}$, an indeterminate form of $frac{infty}{infty}$. By L'Hospitals rule, that is equivalent to $lim_{xto -infty} frac{kx^{k-1}}{2kx^{2k-1}}$. A little bit of algebraic manipulation shows that that is equal to $lim_{xto -infty} frac{1}{2}*x^{-k}$. Taking the 1/2 out of the limit, we have $lim_{xto -infty} x^{-k}=frac{1}{2}*lim_{xto -infty} x^{-k}$. Treating $lim_{xto -infty} x^{-k}$ as a variable in a linear equation to be solved for, we have $lim_{xto -infty} x^{-k}=0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              $endgroup$
              – Eric Brown
              Jan 5 at 2:18














            0












            0








            0





            $begingroup$

            The limit is equivalent to $lim_{xto -infty} frac{x^k}{x^{2k}}$, an indeterminate form of $frac{infty}{infty}$. By L'Hospitals rule, that is equivalent to $lim_{xto -infty} frac{kx^{k-1}}{2kx^{2k-1}}$. A little bit of algebraic manipulation shows that that is equal to $lim_{xto -infty} frac{1}{2}*x^{-k}$. Taking the 1/2 out of the limit, we have $lim_{xto -infty} x^{-k}=frac{1}{2}*lim_{xto -infty} x^{-k}$. Treating $lim_{xto -infty} x^{-k}$ as a variable in a linear equation to be solved for, we have $lim_{xto -infty} x^{-k}=0$






            share|cite|improve this answer









            $endgroup$



            The limit is equivalent to $lim_{xto -infty} frac{x^k}{x^{2k}}$, an indeterminate form of $frac{infty}{infty}$. By L'Hospitals rule, that is equivalent to $lim_{xto -infty} frac{kx^{k-1}}{2kx^{2k-1}}$. A little bit of algebraic manipulation shows that that is equal to $lim_{xto -infty} frac{1}{2}*x^{-k}$. Taking the 1/2 out of the limit, we have $lim_{xto -infty} x^{-k}=frac{1}{2}*lim_{xto -infty} x^{-k}$. Treating $lim_{xto -infty} x^{-k}$ as a variable in a linear equation to be solved for, we have $lim_{xto -infty} x^{-k}=0$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 1:22









            H HuangH Huang

            89110




            89110












            • $begingroup$
              This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              $endgroup$
              – Eric Brown
              Jan 5 at 2:18


















            • $begingroup$
              This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              $endgroup$
              – Eric Brown
              Jan 5 at 2:18
















            $begingroup$
            This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
            $endgroup$
            – Eric Brown
            Jan 5 at 2:18




            $begingroup$
            This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
            $endgroup$
            – Eric Brown
            Jan 5 at 2:18


















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