Evaluation of the limit $displaystylelim_{xtoinfty}x-sqrt{(x-a)(x-b)}$
$begingroup$
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 Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
$endgroup$
1
$begingroup$
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 $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24
$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28
$begingroup$
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!
$endgroup$
– Barry Cipra
Jan 5 at 14:30
add a comment |
$begingroup$
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 Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
$endgroup$
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 Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
asked Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
edited Jan 5 at 13:41
Arjun Banerjee
asked Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
asked Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
asked Jan 5 at 13:38
Arjun BanerjeeArjun Banerjee
54210
54210
1
$begingroup$
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 $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24
$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28
$begingroup$
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!
$endgroup$
– Barry Cipra
Jan 5 at 14:30
add a comment |
1
$begingroup$
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 $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24
$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28
$begingroup$
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!
$endgroup$
– Barry Cipra
Jan 5 at 14:30
1
1
$begingroup$
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 $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24
$begingroup$
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 $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24
$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28
$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28
$begingroup$
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!
$endgroup$
– Barry Cipra
Jan 5 at 14:30
$begingroup$
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!
$endgroup$
– Barry Cipra
Jan 5 at 14:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
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add a comment |
$begingroup$
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 Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
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add a comment |
$begingroup$
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 Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
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votes
$begingroup$
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 Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
$endgroup$
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 Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
answered Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
answered Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
answered Jan 5 at 13:41
Robert ZRobert Z
99.6k1068140
99.6k1068140
add a comment |
add a comment |
$begingroup$
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 Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
$endgroup$
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 Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
answered Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
answered Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
answered Jan 5 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42871
34.5k42871
add a comment |
add a comment |
$begingroup$
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 Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
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 Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 5 at 16:52
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
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$begingroup$
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 $(a+b)/2$.
$endgroup$
– Barry Cipra
Jan 5 at 14:24
$begingroup$
@ Barry Cipra can you please suggest edits to my answer?
$endgroup$
– Arjun Banerjee
Jan 5 at 14:28
$begingroup$
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!
$endgroup$
– Barry Cipra
Jan 5 at 14:30