Mixing
How can I find $lim_{nrightarrowinfty}(1+frac{x}n)^{sqrt{n}}$?
How can I find $$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}};?$$
I know $lim_{nrightarrowinfty}left(1+frac{x}nright)^{n} = exp (x)$ but I don't know how can I put the definition in this particular limit.
I know then, that $lim_{nrightarrowinfty}big(1+frac{x}nbig)=1$, but I don't think this is right to consider.
real-analysis sequences-and-series limits exponential-function
edited Nov 20 '18 at 15:22
amWhy
192k28224439
asked Nov 20 '18 at 14:46
Dada
7010
add a comment |
How can I find $$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}};?$$
I know $lim_{nrightarrowinfty}left(1+frac{x}nright)^{n} = exp (x)$ but I don't know how can I put the definition in this particular limit.
I know then, that $lim_{nrightarrowinfty}big(1+frac{x}nbig)=1$, but I don't think this is right to consider.
real-analysis sequences-and-series limits exponential-function
edited Nov 20 '18 at 15:22
amWhy
192k28224439
asked Nov 20 '18 at 14:46
Dada
7010
add a comment |
How can I find $$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}};?$$
I know $lim_{nrightarrowinfty}left(1+frac{x}nright)^{n} = exp (x)$ but I don't know how can I put the definition in this particular limit.
I know then, that $lim_{nrightarrowinfty}big(1+frac{x}nbig)=1$, but I don't think this is right to consider.
real-analysis sequences-and-series limits exponential-function
edited Nov 20 '18 at 15:22
amWhy
192k28224439
asked Nov 20 '18 at 14:46
Dada
7010
How can I find $$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}};?$$
I know $lim_{nrightarrowinfty}left(1+frac{x}nright)^{n} = exp (x)$ but I don't know how can I put the definition in this particular limit.
I know then, that $lim_{nrightarrowinfty}big(1+frac{x}nbig)=1$, but I don't think this is right to consider.
real-analysis sequences-and-series limits exponential-function
real-analysis sequences-and-series limits exponential-function
edited Nov 20 '18 at 15:22
amWhy
192k28224439
asked Nov 20 '18 at 14:46
Dada
7010
edited Nov 20 '18 at 15:22
amWhy
192k28224439
asked Nov 20 '18 at 14:46
Dada
7010
edited Nov 20 '18 at 15:22
amWhy
192k28224439
edited Nov 20 '18 at 15:22
amWhy
192k28224439
edited Nov 20 '18 at 15:22
amWhy
192k28224439
192k28224439
asked Nov 20 '18 at 14:46
Dada
7010
asked Nov 20 '18 at 14:46
Dada
7010
asked Nov 20 '18 at 14:46
Dada
7010
7010
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}}$$
From
$$lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{nrightarrowinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ **, we get
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = e^0=1$$
EDIT
I add the note bellow as my calculation was considered insufficiently justified
**and because the terms are positive, and we don't have an indeterminate case $0^0$ or $1^{infty}$ or $infty ^0,;$
answered Nov 20 '18 at 15:03
user376343
2,8582823
4
One needs an argument for "if $a_nto a$ and $b_nto b$ ($b_n,bgeqslant0, a_n,a>0$) then $a_n^{b_n}to a^b$".
– user587192
Nov 20 '18 at 15:12
Or, if you prefer, put $t={n over x}.$ The mentioned limit becomes $lim_{t to infty} (1+{1over t})^{t}.$
– user376343
Nov 20 '18 at 15:18
1
What do you use for the "we get..." step in the last line?
– user587192
Nov 20 '18 at 15:19
2
@user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $lim_n a_n^{b_n} $ into $(lim_n a_n)^{(lim_n b_n)}$ but this isn't necessarily justified.
– Jam
Nov 20 '18 at 15:26
1
From $$lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{ntoinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ we get $$lim_{ntoinfty}left(1+frac{x}nright)^{sqrt{n}} =lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}} color{red}{ =left[lim_{ntoinfty}left(1+frac{x}nright)^{{frac{n}{x}}}right]^{lim_{ntoinfty}{frac{x}{n}}{sqrt{n}}} } color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified.
– user587192
Nov 20 '18 at 15:44
add a comment |
Hint:
Compute the limit of the log: $;sqrt nlogBigl(1+ dfrac xnBigr)$ and use equivalence:
$$log(1+u)sim_0 u.$$
answered Nov 20 '18 at 14:52
Bernard
118k639112
I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $log(1+u)$ with $u$. Since $log(1+u)sim u$, $lim_{utoinfty}frac{log(1+u)}{u}=1$ so $lim_{ntoinfty}frac{log(1+x/n)/left(frac{x}{n}right)}{sqrt{n}/{left(frac{x}{n}right)}}=frac{lim(ldots)}{lim_{ntoinfty}sqrt{n}{left(frac{n}{x}right)}}$.
– Jam
Nov 20 '18 at 15:32
1
Should just be $sqrt n$ our front of the log, not $1/sqrt n$.
– Shalop
Nov 20 '18 at 16:25
1
@Shalop: Thanks for pointing the typo. I've fixed it. For your first comment, I don't think it's necessary: I specified I used equivalence, in the sense of asymptotic analysis, and use the rules thereof (compatibility with product and quotient).
– Bernard
Nov 20 '18 at 16:53
That wasn’t my comment though. It was jam
– Shalop
Nov 20 '18 at 19:00
add a comment |
Let $y=(1+frac{x}{n})^{sqrt{n}}$. Note that
$$
ln y=sqrt{n}ln(1+frac{x}{n})=frac{ln(1+frac{x}{n})}{1/sqrt{n}}.
$$
As $ntoinfty$, this is a limit of the form $0/0$. Thus, we may apply L'Hopital's rule to obtain that
$$
lim_{ntoinfty}ln y=lim_{ntoinfty}frac{-frac{x}{n^{2}}/(1+frac{x}{n})}{-frac{1}{2}n^{-3/2}}=lim_{ntoinfty}frac{2x}{sqrt{n}+frac{x}{sqrt{n}}}=0
$$
Thus,
$$
lim_{ntoinfty}y=lim_{ntoinfty}e^{ln(y)}=e^{left(lim_{ntoinfty}ln(y)right)}=e^{0}=1.
$$
answered Nov 20 '18 at 15:04
ervx
10.3k31338
add a comment |
Hint for $xgeq1$:
- The expression and therefore the limit is relatively easily seen to be greater than or equal to $1$
- The expression is for any $epsilon>0$ eventually smaller than $left(1+frac xnright)^{epsilon n}$
answered Nov 20 '18 at 14:52
Arthur
111k7105186
Point 1. is only true for $x>0$. When $x<0$, the expression approaches $1$ from below so I think this argument might need a slight modification
– Jam
Nov 20 '18 at 15:09
@Jam You're right. I've modified my hint.
– Arthur
Nov 20 '18 at 15:12
add a comment |
By L'Hopital,
$$
lim_{ytoinfty}sqrt{y}log(1+frac{x}{y})
=lim_{tto 0}frac{log(1+tx)}{sqrt{t}}
=lim_{tto 0}dfrac{frac{x}{1+tx}}{frac{1}{2sqrt{t}}}=0.
$$
Now one can use the continuity of the exponential function:
$$
lim_{ntoinfty}expleft[sqrt{n}log(1+frac{x}{n})right]
=expleft[lim_{ntoinfty}sqrt{n}log(1+frac{x}{n})right]=1.
$$
answered Nov 20 '18 at 15:11
user587192
1,782215
add a comment |
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5 Answers
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$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}}$$
From
$$lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{nrightarrowinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ **, we get
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = e^0=1$$
EDIT
I add the note bellow as my calculation was considered insufficiently justified
**and because the terms are positive, and we don't have an indeterminate case $0^0$ or $1^{infty}$ or $infty ^0,;$
answered Nov 20 '18 at 15:03
user376343
2,8582823
4
One needs an argument for "if $a_nto a$ and $b_nto b$ ($b_n,bgeqslant0, a_n,a>0$) then $a_n^{b_n}to a^b$".
– user587192
Nov 20 '18 at 15:12
Or, if you prefer, put $t={n over x}.$ The mentioned limit becomes $lim_{t to infty} (1+{1over t})^{t}.$
– user376343
Nov 20 '18 at 15:18
1
What do you use for the "we get..." step in the last line?
– user587192
Nov 20 '18 at 15:19
2
@user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $lim_n a_n^{b_n} $ into $(lim_n a_n)^{(lim_n b_n)}$ but this isn't necessarily justified.
– Jam
Nov 20 '18 at 15:26
1
From $$lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{ntoinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ we get $$lim_{ntoinfty}left(1+frac{x}nright)^{sqrt{n}} =lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}} color{red}{ =left[lim_{ntoinfty}left(1+frac{x}nright)^{{frac{n}{x}}}right]^{lim_{ntoinfty}{frac{x}{n}}{sqrt{n}}} } color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified.
– user587192
Nov 20 '18 at 15:44
add a comment |
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}}$$
From
$$lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{nrightarrowinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ **, we get
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = e^0=1$$
EDIT
I add the note bellow as my calculation was considered insufficiently justified
**and because the terms are positive, and we don't have an indeterminate case $0^0$ or $1^{infty}$ or $infty ^0,;$
answered Nov 20 '18 at 15:03
user376343
2,8582823
4
One needs an argument for "if $a_nto a$ and $b_nto b$ ($b_n,bgeqslant0, a_n,a>0$) then $a_n^{b_n}to a^b$".
– user587192
Nov 20 '18 at 15:12
Or, if you prefer, put $t={n over x}.$ The mentioned limit becomes $lim_{t to infty} (1+{1over t})^{t}.$
– user376343
Nov 20 '18 at 15:18
1
What do you use for the "we get..." step in the last line?
– user587192
Nov 20 '18 at 15:19
2
@user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $lim_n a_n^{b_n} $ into $(lim_n a_n)^{(lim_n b_n)}$ but this isn't necessarily justified.
– Jam
Nov 20 '18 at 15:26
1
From $$lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{ntoinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ we get $$lim_{ntoinfty}left(1+frac{x}nright)^{sqrt{n}} =lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}} color{red}{ =left[lim_{ntoinfty}left(1+frac{x}nright)^{{frac{n}{x}}}right]^{lim_{ntoinfty}{frac{x}{n}}{sqrt{n}}} } color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified.
– user587192
Nov 20 '18 at 15:44
add a comment |
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}}$$
From
$$lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{nrightarrowinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ **, we get
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = e^0=1$$
EDIT
I add the note bellow as my calculation was considered insufficiently justified
**and because the terms are positive, and we don't have an indeterminate case $0^0$ or $1^{infty}$ or $infty ^0,;$
answered Nov 20 '18 at 15:03
user376343
2,8582823
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}}$$
From
$$lim_{nrightarrowinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{nrightarrowinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ **, we get
$$lim_{nrightarrowinfty}left(1+frac{x}nright)^{sqrt{n}} = e^0=1$$
EDIT
I add the note bellow as my calculation was considered insufficiently justified
**and because the terms are positive, and we don't have an indeterminate case $0^0$ or $1^{infty}$ or $infty ^0,;$
answered Nov 20 '18 at 15:03
user376343
2,8582823
edited Nov 20 '18 at 18:55
answered Nov 20 '18 at 15:03
user376343
2,8582823
answered Nov 20 '18 at 15:03
user376343
2,8582823
answered Nov 20 '18 at 15:03
user376343
2,8582823
2,8582823
4
One needs an argument for "if $a_nto a$ and $b_nto b$ ($b_n,bgeqslant0, a_n,a>0$) then $a_n^{b_n}to a^b$".
– user587192
Nov 20 '18 at 15:12
Or, if you prefer, put $t={n over x}.$ The mentioned limit becomes $lim_{t to infty} (1+{1over t})^{t}.$
– user376343
Nov 20 '18 at 15:18
1
What do you use for the "we get..." step in the last line?
– user587192
Nov 20 '18 at 15:19
2
@user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $lim_n a_n^{b_n} $ into $(lim_n a_n)^{(lim_n b_n)}$ but this isn't necessarily justified.
– Jam
Nov 20 '18 at 15:26
1
From $$lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{ntoinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ we get $$lim_{ntoinfty}left(1+frac{x}nright)^{sqrt{n}} =lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}} color{red}{ =left[lim_{ntoinfty}left(1+frac{x}nright)^{{frac{n}{x}}}right]^{lim_{ntoinfty}{frac{x}{n}}{sqrt{n}}} } color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified.
– user587192
Nov 20 '18 at 15:44
add a comment |
4
One needs an argument for "if $a_nto a$ and $b_nto b$ ($b_n,bgeqslant0, a_n,a>0$) then $a_n^{b_n}to a^b$".
– user587192
Nov 20 '18 at 15:12
Or, if you prefer, put $t={n over x}.$ The mentioned limit becomes $lim_{t to infty} (1+{1over t})^{t}.$
– user376343
Nov 20 '18 at 15:18
1
What do you use for the "we get..." step in the last line?
– user587192
Nov 20 '18 at 15:19
2
@user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $lim_n a_n^{b_n} $ into $(lim_n a_n)^{(lim_n b_n)}$ but this isn't necessarily justified.
– Jam
Nov 20 '18 at 15:26
1
From $$lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{ntoinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ we get $$lim_{ntoinfty}left(1+frac{x}nright)^{sqrt{n}} =lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}} color{red}{ =left[lim_{ntoinfty}left(1+frac{x}nright)^{{frac{n}{x}}}right]^{lim_{ntoinfty}{frac{x}{n}}{sqrt{n}}} } color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified.
– user587192
Nov 20 '18 at 15:44
4
4
One needs an argument for "if $a_nto a$ and $b_nto b$ ($b_n,bgeqslant0, a_n,a>0$) then $a_n^{b_n}to a^b$".
– user587192
Nov 20 '18 at 15:12
One needs an argument for "if $a_nto a$ and $b_nto b$ ($b_n,bgeqslant0, a_n,a>0$) then $a_n^{b_n}to a^b$".
– user587192
Nov 20 '18 at 15:12
Or, if you prefer, put $t={n over x}.$ The mentioned limit becomes $lim_{t to infty} (1+{1over t})^{t}.$
– user376343
Nov 20 '18 at 15:18
Or, if you prefer, put $t={n over x}.$ The mentioned limit becomes $lim_{t to infty} (1+{1over t})^{t}.$
– user376343
Nov 20 '18 at 15:18
1
1
What do you use for the "we get..." step in the last line?
– user587192
Nov 20 '18 at 15:19
What do you use for the "we get..." step in the last line?
– user587192
Nov 20 '18 at 15:19
2
2
@user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $lim_n a_n^{b_n} $ into $(lim_n a_n)^{(lim_n b_n)}$ but this isn't necessarily justified.
– Jam
Nov 20 '18 at 15:26
@user376343 I think what user587 is driving at is that you've assumed that you can distribute the limit $lim_n a_n^{b_n} $ into $(lim_n a_n)^{(lim_n b_n)}$ but this isn't necessarily justified.
– Jam
Nov 20 '18 at 15:26
1
1
From $$lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{ntoinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ we get $$lim_{ntoinfty}left(1+frac{x}nright)^{sqrt{n}} =lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}} color{red}{ =left[lim_{ntoinfty}left(1+frac{x}nright)^{{frac{n}{x}}}right]^{lim_{ntoinfty}{frac{x}{n}}{sqrt{n}}} } color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified.
– user587192
Nov 20 '18 at 15:44
From $$lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]=e quad text{and} quad lim_{ntoinfty}{{frac{x}{n}}{sqrt{n}}}=0,$$ we get $$lim_{ntoinfty}left(1+frac{x}nright)^{sqrt{n}} =lim_{ntoinfty}left[left(1+frac{x}nright)^{{frac{n}{x}}}right]^{{frac{x}{n}}{sqrt{n}}} color{red}{ =left[lim_{ntoinfty}left(1+frac{x}nright)^{{frac{n}{x}}}right]^{lim_{ntoinfty}{frac{x}{n}}{sqrt{n}}} } color{blue}{= e^0} $$ You have the blue one, but you need the red part, which is not explicitly justified.
– user587192
Nov 20 '18 at 15:44
add a comment |
Hint:
Compute the limit of the log: $;sqrt nlogBigl(1+ dfrac xnBigr)$ and use equivalence:
$$log(1+u)sim_0 u.$$
answered Nov 20 '18 at 14:52
Bernard
118k639112
I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $log(1+u)$ with $u$. Since $log(1+u)sim u$, $lim_{utoinfty}frac{log(1+u)}{u}=1$ so $lim_{ntoinfty}frac{log(1+x/n)/left(frac{x}{n}right)}{sqrt{n}/{left(frac{x}{n}right)}}=frac{lim(ldots)}{lim_{ntoinfty}sqrt{n}{left(frac{n}{x}right)}}$.
– Jam
Nov 20 '18 at 15:32
1
Should just be $sqrt n$ our front of the log, not $1/sqrt n$.
– Shalop
Nov 20 '18 at 16:25
1
@Shalop: Thanks for pointing the typo. I've fixed it. For your first comment, I don't think it's necessary: I specified I used equivalence, in the sense of asymptotic analysis, and use the rules thereof (compatibility with product and quotient).
– Bernard
Nov 20 '18 at 16:53
That wasn’t my comment though. It was jam
– Shalop
Nov 20 '18 at 19:00
add a comment |
Hint:
Compute the limit of the log: $;sqrt nlogBigl(1+ dfrac xnBigr)$ and use equivalence:
$$log(1+u)sim_0 u.$$
answered Nov 20 '18 at 14:52
Bernard
118k639112
I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $log(1+u)$ with $u$. Since $log(1+u)sim u$, $lim_{utoinfty}frac{log(1+u)}{u}=1$ so $lim_{ntoinfty}frac{log(1+x/n)/left(frac{x}{n}right)}{sqrt{n}/{left(frac{x}{n}right)}}=frac{lim(ldots)}{lim_{ntoinfty}sqrt{n}{left(frac{n}{x}right)}}$.
– Jam
Nov 20 '18 at 15:32
1
Should just be $sqrt n$ our front of the log, not $1/sqrt n$.
– Shalop
Nov 20 '18 at 16:25
1
@Shalop: Thanks for pointing the typo. I've fixed it. For your first comment, I don't think it's necessary: I specified I used equivalence, in the sense of asymptotic analysis, and use the rules thereof (compatibility with product and quotient).
– Bernard
Nov 20 '18 at 16:53
That wasn’t my comment though. It was jam
– Shalop
Nov 20 '18 at 19:00
add a comment |
Hint:
Compute the limit of the log: $;sqrt nlogBigl(1+ dfrac xnBigr)$ and use equivalence:
$$log(1+u)sim_0 u.$$
answered Nov 20 '18 at 14:52
Bernard
118k639112
Hint:
Compute the limit of the log: $;sqrt nlogBigl(1+ dfrac xnBigr)$ and use equivalence:
$$log(1+u)sim_0 u.$$
answered Nov 20 '18 at 14:52
Bernard
118k639112
edited Nov 20 '18 at 16:49
answered Nov 20 '18 at 14:52
Bernard
118k639112
answered Nov 20 '18 at 14:52
Bernard
118k639112
answered Nov 20 '18 at 14:52
Bernard
118k639112
118k639112
I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $log(1+u)$ with $u$. Since $log(1+u)sim u$, $lim_{utoinfty}frac{log(1+u)}{u}=1$ so $lim_{ntoinfty}frac{log(1+x/n)/left(frac{x}{n}right)}{sqrt{n}/{left(frac{x}{n}right)}}=frac{lim(ldots)}{lim_{ntoinfty}sqrt{n}{left(frac{n}{x}right)}}$.
– Jam
Nov 20 '18 at 15:32
1
Should just be $sqrt n$ our front of the log, not $1/sqrt n$.
– Shalop
Nov 20 '18 at 16:25
1
@Shalop: Thanks for pointing the typo. I've fixed it. For your first comment, I don't think it's necessary: I specified I used equivalence, in the sense of asymptotic analysis, and use the rules thereof (compatibility with product and quotient).
– Bernard
Nov 20 '18 at 16:53
That wasn’t my comment though. It was jam
– Shalop
Nov 20 '18 at 19:00
add a comment |
I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $log(1+u)$ with $u$. Since $log(1+u)sim u$, $lim_{utoinfty}frac{log(1+u)}{u}=1$ so $lim_{ntoinfty}frac{log(1+x/n)/left(frac{x}{n}right)}{sqrt{n}/{left(frac{x}{n}right)}}=frac{lim(ldots)}{lim_{ntoinfty}sqrt{n}{left(frac{n}{x}right)}}$.
– Jam
Nov 20 '18 at 15:32
1
Should just be $sqrt n$ our front of the log, not $1/sqrt n$.
– Shalop
Nov 20 '18 at 16:25
1
@Shalop: Thanks for pointing the typo. I've fixed it. For your first comment, I don't think it's necessary: I specified I used equivalence, in the sense of asymptotic analysis, and use the rules thereof (compatibility with product and quotient).
– Bernard
Nov 20 '18 at 16:53
That wasn’t my comment though. It was jam
– Shalop
Nov 20 '18 at 19:00
I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $log(1+u)$ with $u$. Since $log(1+u)sim u$, $lim_{utoinfty}frac{log(1+u)}{u}=1$ so $lim_{ntoinfty}frac{log(1+x/n)/left(frac{x}{n}right)}{sqrt{n}/{left(frac{x}{n}right)}}=frac{lim(ldots)}{lim_{ntoinfty}sqrt{n}{left(frac{n}{x}right)}}$.
– Jam
Nov 20 '18 at 15:32
I'd like to add an addendum to this answer since I wasn't sure about justifying the exchange of $log(1+u)$ with $u$. Since $log(1+u)sim u$, $lim_{utoinfty}frac{log(1+u)}{u}=1$ so $lim_{ntoinfty}frac{log(1+x/n)/left(frac{x}{n}right)}{sqrt{n}/{left(frac{x}{n}right)}}=frac{lim(ldots)}{lim_{ntoinfty}sqrt{n}{left(frac{n}{x}right)}}$.
– Jam
Nov 20 '18 at 15:32
1
1
Should just be $sqrt n$ our front of the log, not $1/sqrt n$.
– Shalop
Nov 20 '18 at 16:25
Should just be $sqrt n$ our front of the log, not $1/sqrt n$.
– Shalop
Nov 20 '18 at 16:25
1
1
@Shalop: Thanks for pointing the typo. I've fixed it. For your first comment, I don't think it's necessary: I specified I used equivalence, in the sense of asymptotic analysis, and use the rules thereof (compatibility with product and quotient).
– Bernard
Nov 20 '18 at 16:53
@Shalop: Thanks for pointing the typo. I've fixed it. For your first comment, I don't think it's necessary: I specified I used equivalence, in the sense of asymptotic analysis, and use the rules thereof (compatibility with product and quotient).
– Bernard
Nov 20 '18 at 16:53
That wasn’t my comment though. It was jam
– Shalop
Nov 20 '18 at 19:00
That wasn’t my comment though. It was jam
– Shalop
Nov 20 '18 at 19:00
add a comment |
Let $y=(1+frac{x}{n})^{sqrt{n}}$. Note that
$$
ln y=sqrt{n}ln(1+frac{x}{n})=frac{ln(1+frac{x}{n})}{1/sqrt{n}}.
$$
As $ntoinfty$, this is a limit of the form $0/0$. Thus, we may apply L'Hopital's rule to obtain that
$$
lim_{ntoinfty}ln y=lim_{ntoinfty}frac{-frac{x}{n^{2}}/(1+frac{x}{n})}{-frac{1}{2}n^{-3/2}}=lim_{ntoinfty}frac{2x}{sqrt{n}+frac{x}{sqrt{n}}}=0
$$
Thus,
$$
lim_{ntoinfty}y=lim_{ntoinfty}e^{ln(y)}=e^{left(lim_{ntoinfty}ln(y)right)}=e^{0}=1.
$$
answered Nov 20 '18 at 15:04
ervx
10.3k31338
add a comment |
Let $y=(1+frac{x}{n})^{sqrt{n}}$. Note that
$$
ln y=sqrt{n}ln(1+frac{x}{n})=frac{ln(1+frac{x}{n})}{1/sqrt{n}}.
$$
As $ntoinfty$, this is a limit of the form $0/0$. Thus, we may apply L'Hopital's rule to obtain that
$$
lim_{ntoinfty}ln y=lim_{ntoinfty}frac{-frac{x}{n^{2}}/(1+frac{x}{n})}{-frac{1}{2}n^{-3/2}}=lim_{ntoinfty}frac{2x}{sqrt{n}+frac{x}{sqrt{n}}}=0
$$
Thus,
$$
lim_{ntoinfty}y=lim_{ntoinfty}e^{ln(y)}=e^{left(lim_{ntoinfty}ln(y)right)}=e^{0}=1.
$$
answered Nov 20 '18 at 15:04
ervx
10.3k31338
add a comment |
Let $y=(1+frac{x}{n})^{sqrt{n}}$. Note that
$$
ln y=sqrt{n}ln(1+frac{x}{n})=frac{ln(1+frac{x}{n})}{1/sqrt{n}}.
$$
As $ntoinfty$, this is a limit of the form $0/0$. Thus, we may apply L'Hopital's rule to obtain that
$$
lim_{ntoinfty}ln y=lim_{ntoinfty}frac{-frac{x}{n^{2}}/(1+frac{x}{n})}{-frac{1}{2}n^{-3/2}}=lim_{ntoinfty}frac{2x}{sqrt{n}+frac{x}{sqrt{n}}}=0
$$
Thus,
$$
lim_{ntoinfty}y=lim_{ntoinfty}e^{ln(y)}=e^{left(lim_{ntoinfty}ln(y)right)}=e^{0}=1.
$$
answered Nov 20 '18 at 15:04
ervx
10.3k31338
Let $y=(1+frac{x}{n})^{sqrt{n}}$. Note that
$$
ln y=sqrt{n}ln(1+frac{x}{n})=frac{ln(1+frac{x}{n})}{1/sqrt{n}}.
$$
As $ntoinfty$, this is a limit of the form $0/0$. Thus, we may apply L'Hopital's rule to obtain that
$$
lim_{ntoinfty}ln y=lim_{ntoinfty}frac{-frac{x}{n^{2}}/(1+frac{x}{n})}{-frac{1}{2}n^{-3/2}}=lim_{ntoinfty}frac{2x}{sqrt{n}+frac{x}{sqrt{n}}}=0
$$
Thus,
$$
lim_{ntoinfty}y=lim_{ntoinfty}e^{ln(y)}=e^{left(lim_{ntoinfty}ln(y)right)}=e^{0}=1.
$$
answered Nov 20 '18 at 15:04
ervx
10.3k31338
answered Nov 20 '18 at 15:04
ervx
10.3k31338
answered Nov 20 '18 at 15:04
ervx
10.3k31338
answered Nov 20 '18 at 15:04
ervx
10.3k31338
10.3k31338
add a comment |
add a comment |
Hint for $xgeq1$:
- The expression and therefore the limit is relatively easily seen to be greater than or equal to $1$
- The expression is for any $epsilon>0$ eventually smaller than $left(1+frac xnright)^{epsilon n}$
answered Nov 20 '18 at 14:52
Arthur
111k7105186
Point 1. is only true for $x>0$. When $x<0$, the expression approaches $1$ from below so I think this argument might need a slight modification
– Jam
Nov 20 '18 at 15:09
@Jam You're right. I've modified my hint.
– Arthur
Nov 20 '18 at 15:12
add a comment |
Hint for $xgeq1$:
- The expression and therefore the limit is relatively easily seen to be greater than or equal to $1$
- The expression is for any $epsilon>0$ eventually smaller than $left(1+frac xnright)^{epsilon n}$
answered Nov 20 '18 at 14:52
Arthur
111k7105186
Point 1. is only true for $x>0$. When $x<0$, the expression approaches $1$ from below so I think this argument might need a slight modification
– Jam
Nov 20 '18 at 15:09
@Jam You're right. I've modified my hint.
– Arthur
Nov 20 '18 at 15:12
add a comment |
Hint for $xgeq1$:
- The expression and therefore the limit is relatively easily seen to be greater than or equal to $1$
- The expression is for any $epsilon>0$ eventually smaller than $left(1+frac xnright)^{epsilon n}$
answered Nov 20 '18 at 14:52
Arthur
111k7105186
Hint for $xgeq1$:
- The expression and therefore the limit is relatively easily seen to be greater than or equal to $1$
- The expression is for any $epsilon>0$ eventually smaller than $left(1+frac xnright)^{epsilon n}$
answered Nov 20 '18 at 14:52
Arthur
111k7105186
edited Nov 20 '18 at 15:12
answered Nov 20 '18 at 14:52
Arthur
111k7105186
answered Nov 20 '18 at 14:52
Arthur
111k7105186
answered Nov 20 '18 at 14:52
Arthur
111k7105186
111k7105186
Point 1. is only true for $x>0$. When $x<0$, the expression approaches $1$ from below so I think this argument might need a slight modification
– Jam
Nov 20 '18 at 15:09
@Jam You're right. I've modified my hint.
– Arthur
Nov 20 '18 at 15:12
add a comment |
Point 1. is only true for $x>0$. When $x<0$, the expression approaches $1$ from below so I think this argument might need a slight modification
– Jam
Nov 20 '18 at 15:09
@Jam You're right. I've modified my hint.
– Arthur
Nov 20 '18 at 15:12
Point 1. is only true for $x>0$. When $x<0$, the expression approaches $1$ from below so I think this argument might need a slight modification
– Jam
Nov 20 '18 at 15:09
Point 1. is only true for $x>0$. When $x<0$, the expression approaches $1$ from below so I think this argument might need a slight modification
– Jam
Nov 20 '18 at 15:09
@Jam You're right. I've modified my hint.
– Arthur
Nov 20 '18 at 15:12
@Jam You're right. I've modified my hint.
– Arthur
Nov 20 '18 at 15:12
add a comment |
By L'Hopital,
$$
lim_{ytoinfty}sqrt{y}log(1+frac{x}{y})
=lim_{tto 0}frac{log(1+tx)}{sqrt{t}}
=lim_{tto 0}dfrac{frac{x}{1+tx}}{frac{1}{2sqrt{t}}}=0.
$$
Now one can use the continuity of the exponential function:
$$
lim_{ntoinfty}expleft[sqrt{n}log(1+frac{x}{n})right]
=expleft[lim_{ntoinfty}sqrt{n}log(1+frac{x}{n})right]=1.
$$
answered Nov 20 '18 at 15:11
user587192
1,782215
add a comment |
By L'Hopital,
$$
lim_{ytoinfty}sqrt{y}log(1+frac{x}{y})
=lim_{tto 0}frac{log(1+tx)}{sqrt{t}}
=lim_{tto 0}dfrac{frac{x}{1+tx}}{frac{1}{2sqrt{t}}}=0.
$$
Now one can use the continuity of the exponential function:
$$
lim_{ntoinfty}expleft[sqrt{n}log(1+frac{x}{n})right]
=expleft[lim_{ntoinfty}sqrt{n}log(1+frac{x}{n})right]=1.
$$
answered Nov 20 '18 at 15:11
user587192
1,782215
add a comment |
By L'Hopital,
$$
lim_{ytoinfty}sqrt{y}log(1+frac{x}{y})
=lim_{tto 0}frac{log(1+tx)}{sqrt{t}}
=lim_{tto 0}dfrac{frac{x}{1+tx}}{frac{1}{2sqrt{t}}}=0.
$$
Now one can use the continuity of the exponential function:
$$
lim_{ntoinfty}expleft[sqrt{n}log(1+frac{x}{n})right]
=expleft[lim_{ntoinfty}sqrt{n}log(1+frac{x}{n})right]=1.
$$
answered Nov 20 '18 at 15:11
user587192
1,782215
By L'Hopital,
$$
lim_{ytoinfty}sqrt{y}log(1+frac{x}{y})
=lim_{tto 0}frac{log(1+tx)}{sqrt{t}}
=lim_{tto 0}dfrac{frac{x}{1+tx}}{frac{1}{2sqrt{t}}}=0.
$$
Now one can use the continuity of the exponential function:
$$
lim_{ntoinfty}expleft[sqrt{n}log(1+frac{x}{n})right]
=expleft[lim_{ntoinfty}sqrt{n}log(1+frac{x}{n})right]=1.
$$
answered Nov 20 '18 at 15:11
user587192
1,782215
answered Nov 20 '18 at 15:11
user587192
1,782215
answered Nov 20 '18 at 15:11
user587192
1,782215
answered Nov 20 '18 at 15:11
user587192
1,782215
1,782215
add a comment |
add a comment |
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