How to show that lim $(1+1/n^k) = 1$ for $ 0<k<1$












1












$begingroup$


We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.



But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?






real-analysis limits







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  • 4




    $begingroup$
    You forgot a power outside of your parentheses.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:06






  • 1




    $begingroup$
    Yes, so I modified it
    $endgroup$
    – Hs P
    Jan 24 at 2:19
















1












$begingroup$


We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.



But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?






real-analysis limits







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    You forgot a power outside of your parentheses.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:06






  • 1




    $begingroup$
    Yes, so I modified it
    $endgroup$
    – Hs P
    Jan 24 at 2:19














1












1








1





$begingroup$


We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.



But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?






real-analysis limits







share|cite|improve this question











$endgroup$




We know that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = e$ for $k=1$.



But for $0< k< 1$ how do I show that $lim_{nrightarrow+infty}(1+frac1{n^k})^n = 1$ for $ 0<k<1$?






real-analysis limits




real-analysis limits






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 24 at 2:15









Namaste

1




1










asked Jan 24 at 2:00









Hs PHs P

275




275








  • 4




    $begingroup$
    You forgot a power outside of your parentheses.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:06






  • 1




    $begingroup$
    Yes, so I modified it
    $endgroup$
    – Hs P
    Jan 24 at 2:19














  • 4




    $begingroup$
    You forgot a power outside of your parentheses.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:06






  • 1




    $begingroup$
    Yes, so I modified it
    $endgroup$
    – Hs P
    Jan 24 at 2:19








4




4




$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06




$begingroup$
You forgot a power outside of your parentheses.
$endgroup$
– lightxbulb
Jan 24 at 2:06




1




1




$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19




$begingroup$
Yes, so I modified it
$endgroup$
– Hs P
Jan 24 at 2:19










1 Answer
1






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$begingroup$

Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$




[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]






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  • $begingroup$
    Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
    $endgroup$
    – Hs P
    Jan 24 at 2:27













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1 Answer
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$begingroup$

Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$




[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
    $endgroup$
    – Hs P
    Jan 24 at 2:27


















1












$begingroup$

Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$




[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
    $endgroup$
    – Hs P
    Jan 24 at 2:27
















1












1








1





$begingroup$

Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$




[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]






share|cite|improve this answer









$endgroup$



Well, $lim_{n rightarrow infty} frac{1}{n^k} = 0$ for fixed $k in (0,1)$, so $lim_{n rightarrow infty} (1-frac{1}{k})$ $ = lim_{n rightarrow infty} 1 + lim_{n rightarrow infty} frac{1}{n^k}$ $ = 1 +0 = 1.$




[Note that $lim_{n rightarrow infty} (1+frac{1}{n})^n = e$ i.e., there is the exponent $n$, and for $k<1$,
$lim_{n rightarrow infty} (1+frac{1}{n^k})^n =$ $lim_{n rightarrow infty} ((1+frac{1}{n^k})^{n^k})^{n^{1-k}} = e^{n^{1-k}} = infty $ ]







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 2:10









MikeMike

4,396412




4,396412












  • $begingroup$
    Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
    $endgroup$
    – Hs P
    Jan 24 at 2:27




















  • $begingroup$
    Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
    $endgroup$
    – Hs P
    Jan 24 at 2:27


















$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27






$begingroup$
Thank you! Maybe you assume that lim$(1+1/n)^{a_{n}}=$ lim$e^{a_{n}/n}$. Is this assumption true?
$endgroup$
– Hs P
Jan 24 at 2:27




















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