limit of $limlimits_{x to infty}((x+1)^a - x^a)$ [on hold]
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Hello everybody, i couldn't figure out how to demonstrate this equation without L'Hopital i wonder if its possible.
$a in (0,1)$
$limlimits_{x to infty}((x+1)^a - x^a) = 0$
calculus
asked Nov 18 at 19:59
Eray Xx
85
put on hold as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-3
down vote
favorite
Hello everybody, i couldn't figure out how to demonstrate this equation without L'Hopital i wonder if its possible.
$a in (0,1)$
$limlimits_{x to infty}((x+1)^a - x^a) = 0$
calculus
asked Nov 18 at 19:59
Eray Xx
85
put on hold as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 at 20:05
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Hello everybody, i couldn't figure out how to demonstrate this equation without L'Hopital i wonder if its possible.
$a in (0,1)$
$limlimits_{x to infty}((x+1)^a - x^a) = 0$
calculus
asked Nov 18 at 19:59
Eray Xx
85
Hello everybody, i couldn't figure out how to demonstrate this equation without L'Hopital i wonder if its possible.
$a in (0,1)$
$limlimits_{x to infty}((x+1)^a - x^a) = 0$
calculus
calculus
asked Nov 18 at 19:59
Eray Xx
85
asked Nov 18 at 19:59
Eray Xx
85
asked Nov 18 at 19:59
Eray Xx
85
asked Nov 18 at 19:59
Eray Xx
85
asked Nov 18 at 19:59
Eray Xx
85
85
put on hold as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 at 11:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 at 20:05
add a comment |
1
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 at 20:05
1
1
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 at 20:05
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 at 20:05
add a comment |
2 Answers
2
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2
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With Bernoulli's inequality
$$0leq x^a(1+frac1x)^a-x^aleq x^aleft(1+dfrac{a}{x}right)-x^ato0$$
as $xtoinfty$.
answered Nov 18 at 20:05
Nosrati
1
add a comment |
up vote
2
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By applying a Taylor expansion, as $u to 0$, one gets
$$
left( 1+uright)^a=1+au+o(u)
$$ Then one may write, as $x to infty$,
$$
(x+1)^a - x^a=x^a left[left(1+frac 1xright)^a-1 right]=x^a left[left(1+frac {a}x+oleft(frac1x right)right)-1 right]=ax^{a-1}+oleft(x^{a-1} right)
$$ which tends to $0$ since $a in (0,1)$.
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
With Bernoulli's inequality
$$0leq x^a(1+frac1x)^a-x^aleq x^aleft(1+dfrac{a}{x}right)-x^ato0$$
as $xtoinfty$.
answered Nov 18 at 20:05
Nosrati
1
add a comment |
up vote
2
down vote
With Bernoulli's inequality
$$0leq x^a(1+frac1x)^a-x^aleq x^aleft(1+dfrac{a}{x}right)-x^ato0$$
as $xtoinfty$.
answered Nov 18 at 20:05
Nosrati
1
add a comment |
up vote
2
down vote
up vote
2
down vote
With Bernoulli's inequality
$$0leq x^a(1+frac1x)^a-x^aleq x^aleft(1+dfrac{a}{x}right)-x^ato0$$
as $xtoinfty$.
answered Nov 18 at 20:05
Nosrati
1
With Bernoulli's inequality
$$0leq x^a(1+frac1x)^a-x^aleq x^aleft(1+dfrac{a}{x}right)-x^ato0$$
as $xtoinfty$.
answered Nov 18 at 20:05
Nosrati
1
answered Nov 18 at 20:05
Nosrati
1
answered Nov 18 at 20:05
Nosrati
1
answered Nov 18 at 20:05
Nosrati
1
1
add a comment |
add a comment |
up vote
2
down vote
By applying a Taylor expansion, as $u to 0$, one gets
$$
left( 1+uright)^a=1+au+o(u)
$$ Then one may write, as $x to infty$,
$$
(x+1)^a - x^a=x^a left[left(1+frac 1xright)^a-1 right]=x^a left[left(1+frac {a}x+oleft(frac1x right)right)-1 right]=ax^{a-1}+oleft(x^{a-1} right)
$$ which tends to $0$ since $a in (0,1)$.
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
add a comment |
up vote
2
down vote
By applying a Taylor expansion, as $u to 0$, one gets
$$
left( 1+uright)^a=1+au+o(u)
$$ Then one may write, as $x to infty$,
$$
(x+1)^a - x^a=x^a left[left(1+frac 1xright)^a-1 right]=x^a left[left(1+frac {a}x+oleft(frac1x right)right)-1 right]=ax^{a-1}+oleft(x^{a-1} right)
$$ which tends to $0$ since $a in (0,1)$.
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
add a comment |
up vote
2
down vote
up vote
2
down vote
By applying a Taylor expansion, as $u to 0$, one gets
$$
left( 1+uright)^a=1+au+o(u)
$$ Then one may write, as $x to infty$,
$$
(x+1)^a - x^a=x^a left[left(1+frac 1xright)^a-1 right]=x^a left[left(1+frac {a}x+oleft(frac1x right)right)-1 right]=ax^{a-1}+oleft(x^{a-1} right)
$$ which tends to $0$ since $a in (0,1)$.
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
By applying a Taylor expansion, as $u to 0$, one gets
$$
left( 1+uright)^a=1+au+o(u)
$$ Then one may write, as $x to infty$,
$$
(x+1)^a - x^a=x^a left[left(1+frac 1xright)^a-1 right]=x^a left[left(1+frac {a}x+oleft(frac1x right)right)-1 right]=ax^{a-1}+oleft(x^{a-1} right)
$$ which tends to $0$ since $a in (0,1)$.
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
answered Nov 18 at 20:06
Olivier Oloa
107k17175293
107k17175293
add a comment |
add a comment |
1
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 at 20:05