How to compute the limit of the following function












1












$begingroup$


I'm looking for a way to compute the following limit:



$$limlimits_{xto infty} left(|x|^beta-|x-c|^betaright),quad cinmathbb{R},,betain(0,1).$$



My hypothesis is that this limit is equal to $0$ and the range of $beta$ plays an important role because for $betain[1,infty)$ the claim fails. I've tried L'Hospital's Rule and the squeezing theorem. But so far, I didn't succeed. Any help is appreciated.






real-analysis limits







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$endgroup$

















    1












    $begingroup$


    I'm looking for a way to compute the following limit:



    $$limlimits_{xto infty} left(|x|^beta-|x-c|^betaright),quad cinmathbb{R},,betain(0,1).$$



    My hypothesis is that this limit is equal to $0$ and the range of $beta$ plays an important role because for $betain[1,infty)$ the claim fails. I've tried L'Hospital's Rule and the squeezing theorem. But so far, I didn't succeed. Any help is appreciated.






    real-analysis limits







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm looking for a way to compute the following limit:



      $$limlimits_{xto infty} left(|x|^beta-|x-c|^betaright),quad cinmathbb{R},,betain(0,1).$$



      My hypothesis is that this limit is equal to $0$ and the range of $beta$ plays an important role because for $betain[1,infty)$ the claim fails. I've tried L'Hospital's Rule and the squeezing theorem. But so far, I didn't succeed. Any help is appreciated.






      real-analysis limits







      share|cite|improve this question









      $endgroup$




      I'm looking for a way to compute the following limit:



      $$limlimits_{xto infty} left(|x|^beta-|x-c|^betaright),quad cinmathbb{R},,betain(0,1).$$



      My hypothesis is that this limit is equal to $0$ and the range of $beta$ plays an important role because for $betain[1,infty)$ the claim fails. I've tried L'Hospital's Rule and the squeezing theorem. But so far, I didn't succeed. Any help is appreciated.






      real-analysis limits




      real-analysis limits






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 29 '18 at 10:04









      Math95Math95

      164




      164






















          1 Answer
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          0












          $begingroup$

          For $x$ large enough the quantities $x$ and $x-c$ are positive. Therefore we can ignore the absolute values.



          $$f(x) = vert x vert^beta - vert x - cvert^beta=x^beta left(1 - left(1-frac{c}{x}right)^betaright)$$



          And



          $$left(1-frac{c}{x}right)^beta = 1 - frac{cbeta}{x} +o(frac{1}{x})$$



          Hence
          $$f(x) =x^beta left(1 - left(1 - frac{cbeta}{x} +o(frac{1}{x})right)right) = frac{cbeta}{x^{1-beta}} + o(frac{1}{x^{1-beta}})$$



          Proving according to your intution that $limlimits_{x to infty} f(x) = 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ClaudeLeibovici Merci Claude! I'll update the answer.
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:23










          • $begingroup$
            Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 29 '18 at 10:26










          • $begingroup$
            @ClaudeLeibovici Thanks! No, I'm an engineer who turn (badly ;-)) to sales activities. But I always was found of mathematics and studied by myself as a hobby. I was "agrégé" (also as a hobby) in 2012 but finally didn't switch to teaching. Your works in simulation were probably great!
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:30










          • $begingroup$
            Thank you for your answer. In the second row of your formulas, you are using the approximation (+errorterm) of the functionvalue with the help of the tangent, right? In the last step, I would use the following: $x^{beta}=o(x)$ and the errorterm in the approximation above is $g(x)=o(frac{1}{x})$. This implies that $f(x)=frac{cbeta}{x^{1-beta}}+o(frac{1}{x}x)$ which clearly shows the claim, right? For $xto-infty$ the limit is the same because of point-symmetry with respect to $frac{c}{2}$. Am I right?
            $endgroup$
            – Math95
            Dec 30 '18 at 14:46














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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          For $x$ large enough the quantities $x$ and $x-c$ are positive. Therefore we can ignore the absolute values.



          $$f(x) = vert x vert^beta - vert x - cvert^beta=x^beta left(1 - left(1-frac{c}{x}right)^betaright)$$



          And



          $$left(1-frac{c}{x}right)^beta = 1 - frac{cbeta}{x} +o(frac{1}{x})$$



          Hence
          $$f(x) =x^beta left(1 - left(1 - frac{cbeta}{x} +o(frac{1}{x})right)right) = frac{cbeta}{x^{1-beta}} + o(frac{1}{x^{1-beta}})$$



          Proving according to your intution that $limlimits_{x to infty} f(x) = 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ClaudeLeibovici Merci Claude! I'll update the answer.
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:23










          • $begingroup$
            Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 29 '18 at 10:26










          • $begingroup$
            @ClaudeLeibovici Thanks! No, I'm an engineer who turn (badly ;-)) to sales activities. But I always was found of mathematics and studied by myself as a hobby. I was "agrégé" (also as a hobby) in 2012 but finally didn't switch to teaching. Your works in simulation were probably great!
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:30










          • $begingroup$
            Thank you for your answer. In the second row of your formulas, you are using the approximation (+errorterm) of the functionvalue with the help of the tangent, right? In the last step, I would use the following: $x^{beta}=o(x)$ and the errorterm in the approximation above is $g(x)=o(frac{1}{x})$. This implies that $f(x)=frac{cbeta}{x^{1-beta}}+o(frac{1}{x}x)$ which clearly shows the claim, right? For $xto-infty$ the limit is the same because of point-symmetry with respect to $frac{c}{2}$. Am I right?
            $endgroup$
            – Math95
            Dec 30 '18 at 14:46


















          0












          $begingroup$

          For $x$ large enough the quantities $x$ and $x-c$ are positive. Therefore we can ignore the absolute values.



          $$f(x) = vert x vert^beta - vert x - cvert^beta=x^beta left(1 - left(1-frac{c}{x}right)^betaright)$$



          And



          $$left(1-frac{c}{x}right)^beta = 1 - frac{cbeta}{x} +o(frac{1}{x})$$



          Hence
          $$f(x) =x^beta left(1 - left(1 - frac{cbeta}{x} +o(frac{1}{x})right)right) = frac{cbeta}{x^{1-beta}} + o(frac{1}{x^{1-beta}})$$



          Proving according to your intution that $limlimits_{x to infty} f(x) = 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ClaudeLeibovici Merci Claude! I'll update the answer.
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:23










          • $begingroup$
            Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 29 '18 at 10:26










          • $begingroup$
            @ClaudeLeibovici Thanks! No, I'm an engineer who turn (badly ;-)) to sales activities. But I always was found of mathematics and studied by myself as a hobby. I was "agrégé" (also as a hobby) in 2012 but finally didn't switch to teaching. Your works in simulation were probably great!
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:30










          • $begingroup$
            Thank you for your answer. In the second row of your formulas, you are using the approximation (+errorterm) of the functionvalue with the help of the tangent, right? In the last step, I would use the following: $x^{beta}=o(x)$ and the errorterm in the approximation above is $g(x)=o(frac{1}{x})$. This implies that $f(x)=frac{cbeta}{x^{1-beta}}+o(frac{1}{x}x)$ which clearly shows the claim, right? For $xto-infty$ the limit is the same because of point-symmetry with respect to $frac{c}{2}$. Am I right?
            $endgroup$
            – Math95
            Dec 30 '18 at 14:46
















          0












          0








          0





          $begingroup$

          For $x$ large enough the quantities $x$ and $x-c$ are positive. Therefore we can ignore the absolute values.



          $$f(x) = vert x vert^beta - vert x - cvert^beta=x^beta left(1 - left(1-frac{c}{x}right)^betaright)$$



          And



          $$left(1-frac{c}{x}right)^beta = 1 - frac{cbeta}{x} +o(frac{1}{x})$$



          Hence
          $$f(x) =x^beta left(1 - left(1 - frac{cbeta}{x} +o(frac{1}{x})right)right) = frac{cbeta}{x^{1-beta}} + o(frac{1}{x^{1-beta}})$$



          Proving according to your intution that $limlimits_{x to infty} f(x) = 0$.






          share|cite|improve this answer











          $endgroup$



          For $x$ large enough the quantities $x$ and $x-c$ are positive. Therefore we can ignore the absolute values.



          $$f(x) = vert x vert^beta - vert x - cvert^beta=x^beta left(1 - left(1-frac{c}{x}right)^betaright)$$



          And



          $$left(1-frac{c}{x}right)^beta = 1 - frac{cbeta}{x} +o(frac{1}{x})$$



          Hence
          $$f(x) =x^beta left(1 - left(1 - frac{cbeta}{x} +o(frac{1}{x})right)right) = frac{cbeta}{x^{1-beta}} + o(frac{1}{x^{1-beta}})$$



          Proving according to your intution that $limlimits_{x to infty} f(x) = 0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 10:23

























          answered Dec 29 '18 at 10:15









          mathcounterexamples.netmathcounterexamples.net

          26.9k22158




          26.9k22158












          • $begingroup$
            @ClaudeLeibovici Merci Claude! I'll update the answer.
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:23










          • $begingroup$
            Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 29 '18 at 10:26










          • $begingroup$
            @ClaudeLeibovici Thanks! No, I'm an engineer who turn (badly ;-)) to sales activities. But I always was found of mathematics and studied by myself as a hobby. I was "agrégé" (also as a hobby) in 2012 but finally didn't switch to teaching. Your works in simulation were probably great!
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:30










          • $begingroup$
            Thank you for your answer. In the second row of your formulas, you are using the approximation (+errorterm) of the functionvalue with the help of the tangent, right? In the last step, I would use the following: $x^{beta}=o(x)$ and the errorterm in the approximation above is $g(x)=o(frac{1}{x})$. This implies that $f(x)=frac{cbeta}{x^{1-beta}}+o(frac{1}{x}x)$ which clearly shows the claim, right? For $xto-infty$ the limit is the same because of point-symmetry with respect to $frac{c}{2}$. Am I right?
            $endgroup$
            – Math95
            Dec 30 '18 at 14:46




















          • $begingroup$
            @ClaudeLeibovici Merci Claude! I'll update the answer.
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:23










          • $begingroup$
            Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
            $endgroup$
            – Claude Leibovici
            Dec 29 '18 at 10:26










          • $begingroup$
            @ClaudeLeibovici Thanks! No, I'm an engineer who turn (badly ;-)) to sales activities. But I always was found of mathematics and studied by myself as a hobby. I was "agrégé" (also as a hobby) in 2012 but finally didn't switch to teaching. Your works in simulation were probably great!
            $endgroup$
            – mathcounterexamples.net
            Dec 29 '18 at 10:30










          • $begingroup$
            Thank you for your answer. In the second row of your formulas, you are using the approximation (+errorterm) of the functionvalue with the help of the tangent, right? In the last step, I would use the following: $x^{beta}=o(x)$ and the errorterm in the approximation above is $g(x)=o(frac{1}{x})$. This implies that $f(x)=frac{cbeta}{x^{1-beta}}+o(frac{1}{x}x)$ which clearly shows the claim, right? For $xto-infty$ the limit is the same because of point-symmetry with respect to $frac{c}{2}$. Am I right?
            $endgroup$
            – Math95
            Dec 30 '18 at 14:46


















          $begingroup$
          @ClaudeLeibovici Merci Claude! I'll update the answer.
          $endgroup$
          – mathcounterexamples.net
          Dec 29 '18 at 10:23




          $begingroup$
          @ClaudeLeibovici Merci Claude! I'll update the answer.
          $endgroup$
          – mathcounterexamples.net
          Dec 29 '18 at 10:23












          $begingroup$
          Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
          $endgroup$
          – Claude Leibovici
          Dec 29 '18 at 10:26




          $begingroup$
          Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
          $endgroup$
          – Claude Leibovici
          Dec 29 '18 at 10:26












          $begingroup$
          @ClaudeLeibovici Thanks! No, I'm an engineer who turn (badly ;-)) to sales activities. But I always was found of mathematics and studied by myself as a hobby. I was "agrégé" (also as a hobby) in 2012 but finally didn't switch to teaching. Your works in simulation were probably great!
          $endgroup$
          – mathcounterexamples.net
          Dec 29 '18 at 10:30




          $begingroup$
          @ClaudeLeibovici Thanks! No, I'm an engineer who turn (badly ;-)) to sales activities. But I always was found of mathematics and studied by myself as a hobby. I was "agrégé" (also as a hobby) in 2012 but finally didn't switch to teaching. Your works in simulation were probably great!
          $endgroup$
          – mathcounterexamples.net
          Dec 29 '18 at 10:30












          $begingroup$
          Thank you for your answer. In the second row of your formulas, you are using the approximation (+errorterm) of the functionvalue with the help of the tangent, right? In the last step, I would use the following: $x^{beta}=o(x)$ and the errorterm in the approximation above is $g(x)=o(frac{1}{x})$. This implies that $f(x)=frac{cbeta}{x^{1-beta}}+o(frac{1}{x}x)$ which clearly shows the claim, right? For $xto-infty$ the limit is the same because of point-symmetry with respect to $frac{c}{2}$. Am I right?
          $endgroup$
          – Math95
          Dec 30 '18 at 14:46






          $begingroup$
          Thank you for your answer. In the second row of your formulas, you are using the approximation (+errorterm) of the functionvalue with the help of the tangent, right? In the last step, I would use the following: $x^{beta}=o(x)$ and the errorterm in the approximation above is $g(x)=o(frac{1}{x})$. This implies that $f(x)=frac{cbeta}{x^{1-beta}}+o(frac{1}{x}x)$ which clearly shows the claim, right? For $xto-infty$ the limit is the same because of point-symmetry with respect to $frac{c}{2}$. Am I right?
          $endgroup$
          – Math95
          Dec 30 '18 at 14:46




















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