The definition of $e$ is if $n$ is equal to $\infty$
By that definition can I algebraic manipulate the equation and say that $r/\infty=e^{(1/\infty)}-1$?
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4................. no. – fleablood Mar 31 '17 at 06:12
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i don't know about algebraic manipulation but yes that is correct – Saketh Malyala Mar 31 '17 at 06:14
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$$\begin{array}{ll} \rm no: & r/\infty = e^{1/\infty}-1 \ \rm yes: & \lim_{n\to\infty}(1+r/n)^n=e^r \end{array}$$ – anon Mar 31 '17 at 06:30
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It doesn't quite make sense to divide by $\infty$, but conceptually it's accurate. $$\lim_{x\to\infty} \dfrac{r}{x}=0$$ while $$\lim_{x\to\infty} {e^{\frac{1}{x}}}=1$$
So the expression you have is technically correct, although you should really use limits to express it.
S.S.
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Considering $$A=P\left(1+\frac r n \right)^{nt}$$ take logarithms $$\log(A)=\log(P)+nt \log\left(1+\frac r n \right)$$ When $n$ is large, using equivalents which make $$\log\left(1+\frac r n \right)\sim \frac r n$$ and then $$\log(A)\sim\log(P)+nt\times\frac r n=\log(P)+rt\implies A=Pe^{rt}$$
Claude Leibovici
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