How to compute the limit of the following function
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
asked Dec 29 '18 at 10:04
Math95Math95
113
add a comment |
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
asked Dec 29 '18 at 10:04
Math95Math95
113
add a comment |
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
asked Dec 29 '18 at 10:04
Math95Math95
113
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
asked Dec 29 '18 at 10:04
Math95Math95
113
asked Dec 29 '18 at 10:04
Math95Math95
113
asked Dec 29 '18 at 10:04
Math95Math95
113
asked Dec 29 '18 at 10:04
Math95Math95
113
asked Dec 29 '18 at 10:04
Math95Math95
113
113
add a comment |
add a comment |
1 Answer
1
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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 $limlimits_{x to infty} f(x) = 0$.
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
@ClaudeLeibovici Merci Claude! I'll update the answer.
– mathcounterexamples.net
Dec 29 '18 at 10:23
Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
– Claude Leibovici
Dec 29 '18 at 10:26
@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!
– mathcounterexamples.net
Dec 29 '18 at 10:30
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?
– Math95
Dec 30 '18 at 14:46
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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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 $limlimits_{x to infty} f(x) = 0$.
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
@ClaudeLeibovici Merci Claude! I'll update the answer.
– mathcounterexamples.net
Dec 29 '18 at 10:23
Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
– Claude Leibovici
Dec 29 '18 at 10:26
@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!
– mathcounterexamples.net
Dec 29 '18 at 10:30
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?
– Math95
Dec 30 '18 at 14:46
add a comment |
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 $limlimits_{x to infty} f(x) = 0$.
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
@ClaudeLeibovici Merci Claude! I'll update the answer.
– mathcounterexamples.net
Dec 29 '18 at 10:23
Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
– Claude Leibovici
Dec 29 '18 at 10:26
@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!
– mathcounterexamples.net
Dec 29 '18 at 10:30
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?
– Math95
Dec 30 '18 at 14:46
add a comment |
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 $limlimits_{x to infty} f(x) = 0$.
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
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 $limlimits_{x to infty} f(x) = 0$.
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
edited Dec 29 '18 at 10:23
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
answered Dec 29 '18 at 10:15
mathcounterexamples.netmathcounterexamples.net
25.3k21953
25.3k21953
@ClaudeLeibovici Merci Claude! I'll update the answer.
– mathcounterexamples.net
Dec 29 '18 at 10:23
Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
– Claude Leibovici
Dec 29 '18 at 10:26
@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!
– mathcounterexamples.net
Dec 29 '18 at 10:30
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?
– Math95
Dec 30 '18 at 14:46
add a comment |
@ClaudeLeibovici Merci Claude! I'll update the answer.
– mathcounterexamples.net
Dec 29 '18 at 10:23
Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
– Claude Leibovici
Dec 29 '18 at 10:26
@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!
– mathcounterexamples.net
Dec 29 '18 at 10:30
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?
– Math95
Dec 30 '18 at 14:46
@ClaudeLeibovici Merci Claude! I'll update the answer.
– mathcounterexamples.net
Dec 29 '18 at 10:23
@ClaudeLeibovici Merci Claude! I'll update the answer.
– mathcounterexamples.net
Dec 29 '18 at 10:23
Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
– Claude Leibovici
Dec 29 '18 at 10:26
Hi Jean-Pierre ! Interesting website for sure. Are you working in university ? Cheers.
– Claude Leibovici
Dec 29 '18 at 10:26
@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!
– mathcounterexamples.net
Dec 29 '18 at 10:30
@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!
– mathcounterexamples.net
Dec 29 '18 at 10:30
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?
– Math95
Dec 30 '18 at 14:46
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?
– Math95
Dec 30 '18 at 14:46
add a comment |
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