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












0














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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  • 1




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












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






  • 1




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










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
















0














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




















  • 1




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












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






  • 1




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










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














0












0








0







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















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













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edited Jan 3 at 21:42

























asked Jan 3 at 21:30









Eric Brown

10511




10511








  • 1




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












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






  • 1




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










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














  • 1




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












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






  • 1




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










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








1




1




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






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














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




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




1




1




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




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












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




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










2 Answers
2






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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



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






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    0














    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





















    • This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
      – Eric Brown
      yesterday











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    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


























      2














      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
























        2












        2








        2






        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












        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









        Wolfy

        2,31311038




        2,31311038























            0














            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





















            • This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              – Eric Brown
              yesterday
















            0














            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





















            • This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              – Eric Brown
              yesterday














            0












            0








            0






            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












            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 Huang

            437




            437












            • This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              – Eric Brown
              yesterday


















            • This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
              – Eric Brown
              yesterday
















            This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
            – Eric Brown
            yesterday




            This works only after you have proven L'Hospitals rule as x decreases to negative infinity.
            – Eric Brown
            yesterday


















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