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In a book i read they try to proove a formula: $$e = \lim\limits_{x\rightarrow \infty} \left( 1 + \frac{1}{x}\right)^x$$

Well they do it like this:

$$ \begin{split} y &= \lim \limits_{x \rightarrow \infty} \left(1 + \frac{1}{x}\right)^x\\ \ln y &= \lim \limits_{x \rightarrow \infty} \left[\ln \left(1 + \frac{1}{x} \right)^x \right]\\ \ln y &= \lim \limits_{x \rightarrow \infty} \left[x \ln \left(1 + \frac{1}{x} \right) \right]\\ \ln y &= \lim \limits_{x \rightarrow \infty} \left[\frac{\ln \left(1 + \frac{1}{x} \right)}{\frac{1}{x}} \right]~ ~ ~ ~ ;~ ~ ~ ~ t = \frac{1}{x}\\ \ln y &= \lim \limits_{t \rightarrow 0} \left[\frac{\ln \left(1 + t \right)}{t} \right]\\ \ln y &= \lim \limits_{t \rightarrow 0} \left[\frac{\ln \left(1 + t \right) - \overbrace{\ln 1}^{=0}}{t} \right] \end{split} $$ Here book states that part in the squared brackets is a derivative, but this isn't quite right, because of "$\ln$". I don't get it.

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There should be a minus sign in front of the $\ln 1$, but it is a derivative: $f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$. Here $x=1$, replace $h$ with $t$ and $f$ is $\ln$ (also, you should have $t\to 0$ rather than $x\to \infty$ after the substitution). – L. F. Feb 4 '13 at 19:34
Oh sorry yes it is was a typo. And thank you now i kind of understand yes! – 71GA Feb 4 '13 at 19:35
You got an obvious typo in the 2nd last line - still limit for $x$, but you mean the one for $t$. – gnometorule Feb 4 '13 at 19:50
up vote 2 down vote accepted

The definition of the derivative is:

$$f'(x) = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$$

If you let $f(x) = \ln x$ and calculate the derivative at $x = 1$ you will get what you have in the brackets.

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