why does (1+x/n)^n approximate e^x?

[edgy goblet]

as n approaches infinite of course

i get that if x = 1 then it approximates e

[frigid fern]

$a_n = (1+\frac{1}{n})^n$

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Donkey

[frigid fern]

consider $\lim_{n\to\infty}[a_n = (1+\frac{1}{n})^n]$

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Donkey

[frigid fern]

take log both sides

$\lim_{n\to\infty}\ln(a_n) = \lim_{n\to\infty}n\ln(1+\frac{1}{n})$

[bronze wave]

Donkey

[frigid fern]

consider t = $\frac{1}{n}$, then $n\to\infty\implies t\to0$ so $\lim_{n\to\infty}n\ln(1+\frac{1}{n}) = \lim_{t\to0}\frac{\ln(1+t)}{t} = 1 = \ln(a_n)$ which can be proven by definition of derivative

[bronze wave]

Donkey

[frigid fern]

so $a_n = e$

[bronze wave]

Donkey

[frigid fern]

as for $e^x$, you js write $(1+\frac{x}{n})^n = e^{n\ln(1+\frac{x}{n})}$

[bronze wave]

Donkey

[frigid fern]

similar argument again