The continuous compounding formula is $$A = Pe^{rt}$$
However, $e = (1+\frac{1}{n})^n$ provided that $x\rightarrow0$
Then is it correct to rewrite $A = Pe^{rt}$ as $A = P((1+\frac{1}{n})^n)^{rt}$ ?
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1the definition of e requires that you take the limit as n goes to infinity – Benjamin Wang Dec 20 '20 at 17:52
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No, but it is correct to write $Pe^{rt}=\lim\limits_{x\to\infty}\left(1+\frac1x\right)^{rt/x}P$. – Dec 20 '20 at 18:06
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Rather than directly defining $e$ as $$ e=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n \, , $$ it is better to first define the exponential function $\exp$ or $e^x$ as $$ \exp(x)=e^x=\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n \, , $$ in which case $e=\exp(1)$. It is not $e$ which is fundamental so much as the exponential function. If we define the exponential function in this way, then we can write $A=Pe^{rt}$ as $$ P \cdot \lim_{n \to \infty}\left(1+\frac{rt}{n}\right)^n \, . $$
Joe
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Yes, you can write it showing $n$ in the exponent $$ Pe^{rt}=P\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{nrt} $$
Narasimham
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