Different proof of the derivative of e^x

[Natel]

Here is a different way I (think I) proved that the derivative of e^x is e^x:https://docs.google.com/document/ed...lWuaUol92k7cBaOjdBzoSM&hl=en&authkey=CPuduaIB

Are my "rules" with infinitives correct or do they not work in other cases?

 

[dmatador]

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.

 

[Mute]

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

\lim_{y\rightarrow 0} \frac{z}{y} = \lim_{n\rightarrow \infty} zn
is valid due to the change of variables 1/y = n, but you then go and plug this into 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
\lim_{y \rightarrow 0} 1/y = \lim_{m\rightarrow \infty} m \Rightarrow 1 = \lim_{y\rightarrow 0}\lim_{m \rightarrow \infty} my

The implication is not correct.

 

[Natel]

I guess my original question was are these assertions correct independently, even if i didn't prove them?

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

1 = \lim_{y\rightarrow 0}\lim_{m \rightarrow \infty} my

 

[Mute]

I guess my original question was are these assertions correct independently, even if i didn't prove them?

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

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.

1 = \lim_{y\rightarrow 0}\lim_{m \rightarrow \infty} my

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
\lim_{y \rightarrow 0} m(y)y = y/y = 1,
but if m = ln(y), you get
\lim_{y \rightarrow 0} m(y)y = y\ln(y) = 0.