limit of $limlimits_{x to infty}((x+1)^a - x^a)$ [closed]
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 '18 at 19:59
Eray Xx
85
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 '18 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 |
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 '18 at 19:59
Eray Xx
85
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 '18 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 '18 at 20:05
add a comment |
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 '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 '18 at 19:59
Eray Xx
85
asked Nov 18 '18 at 19:59
Eray Xx
85
asked Nov 18 '18 at 19:59
Eray Xx
85
asked Nov 18 '18 at 19:59
Eray Xx
85
asked Nov 18 '18 at 19:59
Eray Xx
85
85
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 '18 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.
closed as off-topic by TheSimpliFire, Holo, Did, José Carlos Santos, Brahadeesh Nov 25 '18 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 '18 at 20:05
add a comment |
1
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 '18 at 20:05
1
1
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 '18 at 20:05
I'd use MVT.${}$
– Lord Shark the Unknown
Nov 18 '18 at 20:05
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 20:05
Nosrati
26.5k62353
add a comment |
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 '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
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 '18 at 20:05
Nosrati
26.5k62353
add a comment |
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 '18 at 20:05
Nosrati
26.5k62353
add a comment |
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 '18 at 20:05
Nosrati
26.5k62353
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 '18 at 20:05
Nosrati
26.5k62353
answered Nov 18 '18 at 20:05
Nosrati
26.5k62353
answered Nov 18 '18 at 20:05
Nosrati
26.5k62353
answered Nov 18 '18 at 20:05
Nosrati
26.5k62353
26.5k62353
add a comment |
add a comment |
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 '18 at 20:06
Olivier Oloa
107k17175293
add a comment |
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 '18 at 20:06
Olivier Oloa
107k17175293
add a comment |
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 '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 '18 at 20:06
Olivier Oloa
107k17175293
answered Nov 18 '18 at 20:06
Olivier Oloa
107k17175293
answered Nov 18 '18 at 20:06
Olivier Oloa
107k17175293
answered Nov 18 '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 '18 at 20:05