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July 31st, 2018, 04:14 AM   #1
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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.
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July 31st, 2018, 05:22 AM   #2
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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 $e$.
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July 31st, 2018, 04:35 PM   #3
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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.
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Last edited by skipjack; July 31st, 2018 at 10:19 PM.
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July 31st, 2018, 10:27 PM   #4
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I disagree, SDK. The problems lie elsewhere, especially in making $\xi$ a function of $n$.
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August 1st, 2018, 06:05 AM   #5
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August 1st, 2018, 07:14 AM   #6
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Quote:
Originally Posted by skipjack View Post
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.
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August 1st, 2018, 07:39 AM   #7
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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.
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August 1st, 2018, 10:50 AM   #8
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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 $\displaystyle n$
$\displaystyle \xi = \xi (n) \rightarrow 0$
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August 1st, 2018, 01:52 PM   #9
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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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