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Different proof of the derivative of e^x

  1. Aug 20, 2010 #1
  2. jcsd
  3. Aug 20, 2010 #2
    On the third-to-last line you could have just subtracted out the n terms. This would allow you to not make the mistake of assuming infinity - infinity = 0.
     
  4. Aug 20, 2010 #3

    Mute

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    No, that's not a valid proof. You've made several errors. For example, the claim

    [tex]\lim_{y\rightarrow 0} \frac{z}{y} = \lim_{n\rightarrow \infty} zn[/tex]
    is valid due to the change of variables 1/y = n, but you then go and plug this in to a formula which already has a variable labelled n which is different and indepedent from1/y. So, you should more appropriately have labelled it 1/y = m, so m and n remain independent. Your entire result relies on using the same symbol for two different variables.

    Another error is that you cannot multiply "rule B" by y. Given 1/y = m you can conclude 1 = my, but you cannot make the claim that
    [tex]\lim_{y \rightarrow 0} 1/y = \lim_{m\rightarrow \infty} m \Rightarrow 1 = \lim_{y\rightarrow 0}\lim_{m \rightarrow \infty} my[/tex]

    The implication is not correct.
     
    Last edited: Aug 20, 2010
  5. Aug 20, 2010 #4
    I guess my original question was are these assertions correct independently, even if i didn't prove them?

    Is this true?
    [tex]\lim_{a\rightarrow 0} \frac{b}{a} = \lim_{d\rightarrow \infty} bd[/tex]

    [tex] 1 = \lim_{y\rightarrow 0}\lim_{m \rightarrow \infty} my[/tex]
     
  6. Aug 20, 2010 #5

    Mute

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    If you define 1/a = d, then yes, this is true. However, if you plug this into an expression which has a variable already labelled a or d, then you have to use a different letter to denote one of these variables, because they are not the same. This is a mistake you made in your proof: you set 1/y = n, but you already had an n in your expression, so you should have let 1/y = m. Your proof would have then had to work out different, because your manipulations depended on the confusion between the n from "1/y = n" and the n that was already present in your expression.

    If m and y are independent, no. The right hand side is an indeterminate expression. If m depends on y somehow, then it could be true, but then you would only have one limit on the right hand side because m is a function of y. If m = 1/y, then you get
    [tex]\lim_{y \rightarrow 0} m(y)y = y/y = 1,[/tex]
    but if m = ln(y), you get
    [tex]\lim_{y \rightarrow 0} m(y)y = y\ln(y) = 0.[/tex]
     
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