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 $e$.

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. 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 $n$.

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  
August 1st, 2018, 07:14 AM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  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 $n$.


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