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 July 31st, 2018, 04:14 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27 Basic Limit Check whether I derived correctly $\displaystyle y' \xi \rightarrow \Delta y = y(n+\xi ) - y(n)$ Set $\displaystyle y=e^n$ and $\displaystyle \xi =\frac{1}{n}$ $\displaystyle \frac{1}{n} (e^n )' \rightarrow e^{n+\frac{1}{n} } -e^n$ or $\displaystyle \frac{e^n }{n} \rightarrow e^n (e^{\frac{1}{n} } -1 )$ $\displaystyle \frac{1}{n} \rightarrow e^{ \frac{1}{n} } -1$ $\displaystyle 1+\frac{1}{n} \rightarrow e^{ \frac{1}{n} }$ $\displaystyle (1+\frac{1}{n} )^n \rightarrow e$ Now if we set side by side the limit for $\displaystyle n\rightarrow \infty$ we have $\displaystyle \lim_{n\rightarrow \infty} (1+\frac{1}{n} )^n =e$ Last edited by skipjack; July 31st, 2018 at 05:07 AM.
 July 31st, 2018, 05:22 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,712 Thanks: 1805 You shouldn't do it like that. If your $e$ were replaced by 2, at what stage would you be going wrong? You end up seemingly showing something that is used as a definition of $e$. Thanks from greg1313 and topsquark
 July 31st, 2018, 04:35 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 472 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics This reasoning is not correct. The problems start on line 1. $y'\xi \to \nabla y$ is not true. What is true is that as $\xi \to 0$, both $y' \xi \to 0$ and $\nabla y \to 0$. But this is true for any differentiable function. This error manifests continuously throughout the "proof". For example you have $1/n \to e^{1/n} - 1$ which has not been proved. All you know is that $1/n \to 0$ and $e^{1/n} \to 1$ which is not the same. Thanks from topsquark Last edited by skipjack; July 31st, 2018 at 10:19 PM.
 July 31st, 2018, 10:27 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,712 Thanks: 1805 I disagree, SDK. The problems lie elsewhere, especially in making $\xi$ a function of $n$. Thanks from topsquark
 August 1st, 2018, 06:05 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra Thanks from topsquark
August 1st, 2018, 07:14 AM   #6
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Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by skipjack I disagree, SDK. The problems lie elsewhere, especially in making $\xi$ a function of $n$.
What is the problem with that? As near as I can tell the intention is to evaluate a limit as $\xi \to 0$. Any limit which exists, does so with respect to any choice of sequence, $\xi_n \to 0$ and choosing $1/n$ is just one of many choices which would all be fine.

 August 1st, 2018, 07:39 AM #7 Global Moderator   Joined: Dec 2006 Posts: 19,712 Thanks: 1805 The post was using the derivative of $y$ with respect to $n$. For that to work, $n$ should have some fixed value for which the derivative of $y$ is given by a limit as $\xi$ tends to zero.
 August 1st, 2018, 10:50 AM #8 Senior Member   Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27 If $\displaystyle \lim_{\xi_n \rightarrow 0 } \xi_n = \lim_{n \rightarrow \infty } n^{-1}=0$ Then we can write $\displaystyle \xi$ as a function of $\displaystyle n$ $\displaystyle \xi = \xi (n) \rightarrow 0$
 August 1st, 2018, 01:52 PM #9 Global Moderator   Joined: Dec 2006 Posts: 19,712 Thanks: 1805 I've just explained why $n$ is fixed when you use $\xi$. You don't want $\xi$ to be fixed, so it shouldn't be a function of $n$.

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