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This question already has an answer here:

What we are allowed to use - 1) The fact that limit of $(1+1/n)^n$ exists and assumed to be some real number $e$ 2) Subsequencial properties of limits of sequences 3) Basic properties of limit

In the previous question $x(n) = (1+1/n^2)^{(2(n^2))}$

We just used the fact that the given sequence is square of the subsequence of the sequence $(1+1/n)^n$ And shall thus converge to $e^2$

I'd expect we'd be required to use something similar in this question but I am unable to, can't see a valid subsequence forming.

Edit - This question is not a duplicate of the question suggested as that question assumes that a lot more theorems have been proven, especially, Limits of log.

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marked as duplicate by user99914, Jimmy R., Claude Leibovici, N. F. Taussig, Watson Feb 18 '16 at 12:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Try to rewrite you expression in the new variable $u=n/2$. $\endgroup$ – matovitch Feb 18 '16 at 12:00
  • $\begingroup$ write $(1 + 2/n) = (n + 2)/(n + 1) \times (n + 1)/n$ and then on raising each factor to power $n$ we can see that each factor tends to $e$ and hence the overall product tends to $e^{2}$. $\endgroup$ – Paramanand Singh Feb 19 '16 at 4:08
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The limit that defines the constant $e$ is $$ e=\lim_{n\to\infty}\left(1+\frac1n\right)^n $$ Since $x^2$ is a continuous function, we get $$ e^2=\lim_{n\to\infty}\left(1+\frac2n+\frac1{n^2}\right)^n $$ It is simple to see that $$ \left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n\le\left(1+\frac2n\right)^n\left(1+\frac1{n^2}\right)^n $$ which is the same as $$ \left(1+\frac2n+\frac1{n^2}\right)^n\left(1-\frac1{n^2+1}\right)^n\le\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n $$ which implies by Bernoulli's Inequality $$ \left(1+\frac2n+\frac1{n^2}\right)^n\left(1-\frac{n}{n^2+1}\right)\le\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n $$ The Squeeze Theorem says $$ \lim_{n\to\infty}\left(1+\frac2n\right)^n=e^2 $$

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Let $n=2m$. Then $\lim_{n\to+\infty}x(n)=\lim_{m\to+\infty}x(2m)$ and $$\left(1+\frac{2}{n}\right)^n=\left(1+\frac{2}{2m}\right)^{2m}=\left(\left(1+\frac{1}{m}\right)^{m}\right)^2=\left(1+\frac{1}{m}\right)^{m}\left(1+\frac{1}{m}\right)^{m}$$

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  • $\begingroup$ How can we just assume that n=2m, n is not necessarily even. $\endgroup$ – Lelouch Feb 18 '16 at 12:25
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    $\begingroup$ $m$ is not necessarily an integer. As we are taking the limit towards infinity, it does not matter. $\endgroup$ – nippon Feb 18 '16 at 12:38
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Considering $\displaystyle \sqrt{x(2n)}=\left( 1+\frac{1}{n} \right)^{n}$,

as we know $\displaystyle \lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^{n}=e$,

so $\displaystyle \lim_{n\to \infty} \sqrt{x(2n)}=e$,

$\displaystyle \lim_{n\to \infty} x(2n)=e^{2}$

$\displaystyle \lim_{n\to \infty} \left( 1+\frac{2}{n} \right)^{n}=e^{2}$

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  • $\begingroup$ How is the first step valid? $\endgroup$ – Lelouch Feb 18 '16 at 12:27
  • $\begingroup$ $\sqrt{x(2n)}=\sqrt{(1+\frac{2}{2n})^{2n}}=(1+\frac{1}{n})^{n}$ $\endgroup$ – Ng Chung Tak Feb 18 '16 at 12:32
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Consider $$x_n=\Big(1+\frac 2n\Big)^n$$ Take logarithms $$\log(x_n)=n \log\Big(1+\frac 2n\Big)$$ Now, using that, for small $y$, $\log(1+y)\simeq y$, then $$\log(x_n)\simeq n \times \frac 2n=2$$

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  • $\begingroup$ Could you prove this by sequencial properties? Limits of log hasn't been proven yet. $\endgroup$ – Lelouch Feb 18 '16 at 12:26

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