Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$
I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$
Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$
To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,
begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}
hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.
Here I again use the same inequality:
begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}
If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$
Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.
real-analysis calculus limits proof-verification
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
add a comment |
I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$
Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$
To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,
begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}
hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.
Here I again use the same inequality:
begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}
If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$
Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.
real-analysis calculus limits proof-verification
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
1
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
– Barry Cipra
2 days ago
@ Barry Cipra can you please suggest edits to my answer?
– Arjun Banerjee
2 days ago
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
– Barry Cipra
2 days ago
add a comment |
I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$
Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$
To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,
begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}
hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.
Here I again use the same inequality:
begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}
If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$
Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.
real-analysis calculus limits proof-verification
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
I'm trying to evaluate the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$
Since there is a square root of a product namely $sqrt{(x-a)(x-b)}$ I
think it's useful to use $text{GM}letext{AM}$ for $(x-a)$ & $(x-b)$:
$sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}tag*{}$
To see whether the function $f(x)=x-sqrt{(x-a)(x-b)}$ is increasing
or decreasing,
begin{align}f'(x)&=1-dfrac{2x-(a+b)}{2sqrt{(x-a)(x-b)}}\&=1-dfrac{dfrac{x-(a+b)}2}{sqrt{(x-a)(x-b)}}\&=1-dfrac{text{AM}}{GM}\&le0
becausetext{ GM}letext{AM}end{align}
hence $f(x)=x-sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the
existence of the limit, $f(x)$ must be bounded below.
Here I again use the same inequality:
begin{align}\&sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}=x-dfrac{a+b}{2}\&implies
f(x)=x-sqrt{(x-a)(x-b)}gedfrac{a+b}{2}end{align}
If I'm right, then the limit exists and
$displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}=dfrac{a+b}2$
Another query is that in the inequality $sqrt{(x-a)(x-b)}ledfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.
real-analysis calculus limits proof-verification
real-analysis calculus limits proof-verification
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
edited 2 days ago
Arjun Banerjee
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
asked 2 days ago
Arjun BanerjeeArjun Banerjee
48010
48010
1
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
– Barry Cipra
2 days ago
@ Barry Cipra can you please suggest edits to my answer?
– Arjun Banerjee
2 days ago
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
– Barry Cipra
2 days ago
add a comment |
1
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
– Barry Cipra
2 days ago
@ Barry Cipra can you please suggest edits to my answer?
– Arjun Banerjee
2 days ago
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
– Barry Cipra
2 days ago
1
1
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
– Barry Cipra
2 days ago
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
– Barry Cipra
2 days ago
@ Barry Cipra can you please suggest edits to my answer?
– Arjun Banerjee
2 days ago
@ Barry Cipra can you please suggest edits to my answer?
– Arjun Banerjee
2 days ago
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
– Barry Cipra
2 days ago
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
– Barry Cipra
2 days ago
add a comment |
3 Answers
3
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You are correct, but the limit is easier to evaluate if you consider
$$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
for $x> max(a,b,0)$.
answered 2 days ago
Robert ZRobert Z
93.8k1061132
add a comment |
Alternatively, do the change of variable $x = 1/t$
$$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
and apply Taylor:
$$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
add a comment |
Hint:
Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,
Rationalizing the numerator,
$$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$
As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are correct, but the limit is easier to evaluate if you consider
$$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
for $x> max(a,b,0)$.
answered 2 days ago
Robert ZRobert Z
93.8k1061132
add a comment |
You are correct, but the limit is easier to evaluate if you consider
$$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
for $x> max(a,b,0)$.
answered 2 days ago
Robert ZRobert Z
93.8k1061132
add a comment |
You are correct, but the limit is easier to evaluate if you consider
$$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
for $x> max(a,b,0)$.
answered 2 days ago
Robert ZRobert Z
93.8k1061132
You are correct, but the limit is easier to evaluate if you consider
$$x-sqrt{(x-a)(x-b)}=frac{x^2-(x-a)(x-b)}{x+sqrt{(x-a)(x-b)}}=
frac{(a+b)-ab/x}{1+sqrt{(1-a/x)(1-b/x)}}$$
for $x> max(a,b,0)$.
answered 2 days ago
Robert ZRobert Z
93.8k1061132
answered 2 days ago
Robert ZRobert Z
93.8k1061132
answered 2 days ago
Robert ZRobert Z
93.8k1061132
answered 2 days ago
Robert ZRobert Z
93.8k1061132
93.8k1061132
add a comment |
add a comment |
Alternatively, do the change of variable $x = 1/t$
$$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
and apply Taylor:
$$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
add a comment |
Alternatively, do the change of variable $x = 1/t$
$$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
and apply Taylor:
$$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
add a comment |
Alternatively, do the change of variable $x = 1/t$
$$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
and apply Taylor:
$$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
Alternatively, do the change of variable $x = 1/t$
$$lim_{xto+infty}x - sqrt{(x-a)(x-b)} = lim_{tto 0^+}frac{1 - sqrt{1 - (a + b)t + abt^2}}t$$
and apply Taylor:
$$sqrt{1 - (a + b)t + abt^2} = 1 - frac{a + b}2 t + cdots$$
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
answered 2 days ago
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.1k42771
34.1k42771
add a comment |
add a comment |
Hint:
Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,
Rationalizing the numerator,
$$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$
As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
add a comment |
Hint:
Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,
Rationalizing the numerator,
$$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$
As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
add a comment |
Hint:
Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,
Rationalizing the numerator,
$$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$
As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
Hint:
Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$,
Rationalizing the numerator,
$$lim_{tto0^+}dfrac{1-sqrt{1-(a+b)t+abt^2}}t=lim_{tto0^+}dfrac{1-(1-(a+b)t+abt^2)}tcdotlim_{tto0^+}dfrac1{1+sqrt{1-(a+b)t+abt^2}}$$
As $tto0,tne0,dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
answered 2 days ago
lab bhattacharjeelab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
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1
It's worth noting that your AM/GM argument, while very nice and well presented, only proves that the limit exists and is greater than or equal to $(a+b)/2$. Something more is required to show that it's actually equal to $(a+b)/2$.
– Barry Cipra
2 days ago
@ Barry Cipra can you please suggest edits to my answer?
– Arjun Banerjee
2 days ago
I was just about to add that I'd actually like to see your argument completed, because nothing obvious comes to (my) mind that doesn't effectively replace the entire proof with the approach given in Robert Z's answer. It's worth thinking about!
– Barry Cipra
2 days ago