$lim_{nrightarrowinfty}(1+1/a_n)^n=lim e^{n/a_n}$, where $lim_{nrightarrowinfty}{a_n}=infty$












1












$begingroup$


We know that $lim_{nrightarrowinfty}(1+1/n)^n = e$

The following that
$lim_{nrightarrowinfty}(1+1/a_n)^n=lim e^{n/a_n}$,

where $a_n$ is positive sequence and $lim_{nrightarrowinfty}{a_n}=infty$
is true?






real-analysis limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    More generally, $$lim A_n^{B_n} = (lim A_n)^{lim B_n}$$ unless it is one of the usual indeterminate forms like $0^0$ or $1^infty$.
    $endgroup$
    – GEdgar
    Jan 24 at 14:42












  • $begingroup$
    Thank you for asking my questions.~
    $endgroup$
    – Hs P
    Jan 24 at 15:53
















1












$begingroup$


We know that $lim_{nrightarrowinfty}(1+1/n)^n = e$

The following that
$lim_{nrightarrowinfty}(1+1/a_n)^n=lim e^{n/a_n}$,

where $a_n$ is positive sequence and $lim_{nrightarrowinfty}{a_n}=infty$
is true?






real-analysis limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    More generally, $$lim A_n^{B_n} = (lim A_n)^{lim B_n}$$ unless it is one of the usual indeterminate forms like $0^0$ or $1^infty$.
    $endgroup$
    – GEdgar
    Jan 24 at 14:42












  • $begingroup$
    Thank you for asking my questions.~
    $endgroup$
    – Hs P
    Jan 24 at 15:53














1












1








1


1



$begingroup$


We know that $lim_{nrightarrowinfty}(1+1/n)^n = e$

The following that
$lim_{nrightarrowinfty}(1+1/a_n)^n=lim e^{n/a_n}$,

where $a_n$ is positive sequence and $lim_{nrightarrowinfty}{a_n}=infty$
is true?






real-analysis limits







share|cite|improve this question











$endgroup$




We know that $lim_{nrightarrowinfty}(1+1/n)^n = e$

The following that
$lim_{nrightarrowinfty}(1+1/a_n)^n=lim e^{n/a_n}$,

where $a_n$ is positive sequence and $lim_{nrightarrowinfty}{a_n}=infty$
is true?






real-analysis limits




real-analysis limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 14:31









user1729

17.4k64193




17.4k64193










asked Jan 24 at 14:09









Hs PHs P

275




275








  • 1




    $begingroup$
    More generally, $$lim A_n^{B_n} = (lim A_n)^{lim B_n}$$ unless it is one of the usual indeterminate forms like $0^0$ or $1^infty$.
    $endgroup$
    – GEdgar
    Jan 24 at 14:42












  • $begingroup$
    Thank you for asking my questions.~
    $endgroup$
    – Hs P
    Jan 24 at 15:53














  • 1




    $begingroup$
    More generally, $$lim A_n^{B_n} = (lim A_n)^{lim B_n}$$ unless it is one of the usual indeterminate forms like $0^0$ or $1^infty$.
    $endgroup$
    – GEdgar
    Jan 24 at 14:42












  • $begingroup$
    Thank you for asking my questions.~
    $endgroup$
    – Hs P
    Jan 24 at 15:53








1




1




$begingroup$
More generally, $$lim A_n^{B_n} = (lim A_n)^{lim B_n}$$ unless it is one of the usual indeterminate forms like $0^0$ or $1^infty$.
$endgroup$
– GEdgar
Jan 24 at 14:42






$begingroup$
More generally, $$lim A_n^{B_n} = (lim A_n)^{lim B_n}$$ unless it is one of the usual indeterminate forms like $0^0$ or $1^infty$.
$endgroup$
– GEdgar
Jan 24 at 14:42














$begingroup$
Thank you for asking my questions.~
$endgroup$
– Hs P
Jan 24 at 15:53




$begingroup$
Thank you for asking my questions.~
$endgroup$
– Hs P
Jan 24 at 15:53










1 Answer
1






active

oldest

votes


















2












$begingroup$

Yes, that is true. Hint: You can show that
$underset{xto a}{lim}{ f(x)}=1$ and $underset{xto a}{lim}{ g(x)}=+infty$
implies that
$$underset{xto a}{lim}{ f(x)^{g(x)}}=e^{underset{xto a}{lim}{
left(f(x)-1right)g(x)}}$$
whenever $underset{xto a}{lim}{
left(f(x)-1right)g(x)}$ exists.



To see this you can Write the following:
$$f(x)^{g(x)}=left(left(1+(f(x)-1)right)^{displaystyle{frac{1}{f(x)-1}}}right)^{(f(x)-1)g(x)}$$
then you can use the assertion of GEdgar's comment. You have to recall that



$$underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}} = e $$



and, so we get



$$ underset{xto a }{lim}{ bigg(1+(f(x)-1)bigg)^{displaystyle{frac{1}{f(x)-1}}}}
= underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}}=e
$$
using the change of variables $u=f(x)-1$. Then, $uto 0$ as $xto a$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for asking my questions. I'll dive into your answer~!!
    $endgroup$
    – Hs P
    Jan 24 at 15:54











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Yes, that is true. Hint: You can show that
$underset{xto a}{lim}{ f(x)}=1$ and $underset{xto a}{lim}{ g(x)}=+infty$
implies that
$$underset{xto a}{lim}{ f(x)^{g(x)}}=e^{underset{xto a}{lim}{
left(f(x)-1right)g(x)}}$$
whenever $underset{xto a}{lim}{
left(f(x)-1right)g(x)}$ exists.



To see this you can Write the following:
$$f(x)^{g(x)}=left(left(1+(f(x)-1)right)^{displaystyle{frac{1}{f(x)-1}}}right)^{(f(x)-1)g(x)}$$
then you can use the assertion of GEdgar's comment. You have to recall that



$$underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}} = e $$



and, so we get



$$ underset{xto a }{lim}{ bigg(1+(f(x)-1)bigg)^{displaystyle{frac{1}{f(x)-1}}}}
= underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}}=e
$$
using the change of variables $u=f(x)-1$. Then, $uto 0$ as $xto a$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for asking my questions. I'll dive into your answer~!!
    $endgroup$
    – Hs P
    Jan 24 at 15:54
















2












$begingroup$

Yes, that is true. Hint: You can show that
$underset{xto a}{lim}{ f(x)}=1$ and $underset{xto a}{lim}{ g(x)}=+infty$
implies that
$$underset{xto a}{lim}{ f(x)^{g(x)}}=e^{underset{xto a}{lim}{
left(f(x)-1right)g(x)}}$$
whenever $underset{xto a}{lim}{
left(f(x)-1right)g(x)}$ exists.



To see this you can Write the following:
$$f(x)^{g(x)}=left(left(1+(f(x)-1)right)^{displaystyle{frac{1}{f(x)-1}}}right)^{(f(x)-1)g(x)}$$
then you can use the assertion of GEdgar's comment. You have to recall that



$$underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}} = e $$



and, so we get



$$ underset{xto a }{lim}{ bigg(1+(f(x)-1)bigg)^{displaystyle{frac{1}{f(x)-1}}}}
= underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}}=e
$$
using the change of variables $u=f(x)-1$. Then, $uto 0$ as $xto a$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for asking my questions. I'll dive into your answer~!!
    $endgroup$
    – Hs P
    Jan 24 at 15:54














2












2








2





$begingroup$

Yes, that is true. Hint: You can show that
$underset{xto a}{lim}{ f(x)}=1$ and $underset{xto a}{lim}{ g(x)}=+infty$
implies that
$$underset{xto a}{lim}{ f(x)^{g(x)}}=e^{underset{xto a}{lim}{
left(f(x)-1right)g(x)}}$$
whenever $underset{xto a}{lim}{
left(f(x)-1right)g(x)}$ exists.



To see this you can Write the following:
$$f(x)^{g(x)}=left(left(1+(f(x)-1)right)^{displaystyle{frac{1}{f(x)-1}}}right)^{(f(x)-1)g(x)}$$
then you can use the assertion of GEdgar's comment. You have to recall that



$$underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}} = e $$



and, so we get



$$ underset{xto a }{lim}{ bigg(1+(f(x)-1)bigg)^{displaystyle{frac{1}{f(x)-1}}}}
= underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}}=e
$$
using the change of variables $u=f(x)-1$. Then, $uto 0$ as $xto a$.






share|cite|improve this answer











$endgroup$



Yes, that is true. Hint: You can show that
$underset{xto a}{lim}{ f(x)}=1$ and $underset{xto a}{lim}{ g(x)}=+infty$
implies that
$$underset{xto a}{lim}{ f(x)^{g(x)}}=e^{underset{xto a}{lim}{
left(f(x)-1right)g(x)}}$$
whenever $underset{xto a}{lim}{
left(f(x)-1right)g(x)}$ exists.



To see this you can Write the following:
$$f(x)^{g(x)}=left(left(1+(f(x)-1)right)^{displaystyle{frac{1}{f(x)-1}}}right)^{(f(x)-1)g(x)}$$
then you can use the assertion of GEdgar's comment. You have to recall that



$$underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}} = e $$



and, so we get



$$ underset{xto a }{lim}{ bigg(1+(f(x)-1)bigg)^{displaystyle{frac{1}{f(x)-1}}}}
= underset{uto 0}{lim}{ (1+u)^{displaystyle{frac{1}{u}}}}=e
$$
using the change of variables $u=f(x)-1$. Then, $uto 0$ as $xto a$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 26 at 14:15

























answered Jan 24 at 14:28









Hector BlandinHector Blandin

1,988816




1,988816








  • 1




    $begingroup$
    Thank you for asking my questions. I'll dive into your answer~!!
    $endgroup$
    – Hs P
    Jan 24 at 15:54














  • 1




    $begingroup$
    Thank you for asking my questions. I'll dive into your answer~!!
    $endgroup$
    – Hs P
    Jan 24 at 15:54








1




1




$begingroup$
Thank you for asking my questions. I'll dive into your answer~!!
$endgroup$
– Hs P
Jan 24 at 15:54




$begingroup$
Thank you for asking my questions. I'll dive into your answer~!!
$endgroup$
– Hs P
Jan 24 at 15:54


















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