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 July 31st, 2018, 04:14 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 588 Thanks: 86 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: 20,917 Thanks: 2200 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: 635 Thanks: 401 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: 20,917 Thanks: 2200 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: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Thanks from topsquark August 1st, 2018, 07:14 AM   #6
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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: 20,917 Thanks: 2200 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: somewhere Posts: 588 Thanks: 86 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: 20,917 Thanks: 2200 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 $n$. Tags basic, limit Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zylo Calculus 13 May 31st, 2017 12:53 PM nbg273 Calculus 7 February 18th, 2017 01:48 AM DerGiLLster Calculus 1 September 16th, 2015 06:44 PM Tutu Calculus 4 June 26th, 2012 04:08 AM sjeddie Calculus 3 August 28th, 2008 07:45 AM

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