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 August 18th, 2019, 09:11 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 Hard problem Evaluate $\displaystyle L(s)=\lim_{n\rightarrow \infty } s^{n} \:$ , $\displaystyle s\in \mathbb{R}^{+}$.
August 18th, 2019, 11:08 AM   #2
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Quote:
 Originally Posted by idontknow Evaluate $\displaystyle L(s)=\lim_{n\rightarrow \infty } s^{n} \:$ , $\displaystyle s\in \mathbb{R}^{+}$.
This is pretty much straightforward so I'm not going to supply a proof, just the answer.
$\displaystyle \lim_{n \to \infty} s^n \to \begin{cases} \infty & 1 < s \\ 1 & 1 = s \\ 0 & 1 > s \end{cases}$

-Dan

 August 19th, 2019, 06:54 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 Check this one : $\displaystyle L(s)=s\cdot \lfloor s \rfloor$. L=s[s] . But it is not working for s>1 . Last edited by idontknow; August 19th, 2019 at 07:25 AM.
August 19th, 2019, 08:13 AM   #4
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Quote:
 Originally Posted by idontknow Check this one : $\displaystyle L(s)=s\cdot \lfloor s \rfloor$. L=s[s] . But it is not working for s>1 .
Assuming $\displaystyle 1/0=\infty$ , the equality holds true .
$\displaystyle L(s)=s\lfloor s \rfloor +\frac{s\lfloor s \rfloor -1}{\lfloor s^{-1} \rfloor }$$\displaystyle \lfloor s \rfloor .$

Last edited by idontknow; August 19th, 2019 at 08:24 AM.

 August 19th, 2019, 09:14 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 On that assumption, $\displaystyle L(s) = \frac{\lfloor s \rfloor}{\lfloor s^{-1} \rfloor}$ is simpler. Thanks from idontknow

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