= Compounding limit for e
{title2=$\lim_{n\to\infty}(1+1/n)^n=e$}
Start at one and apply $n$ updates, each multiplying the current value by $1+1/n$. The final value is $(1+1/n)^n$: one update gives $2$, two give $2.25$, ten give approximately $2.59374$, and one hundred give approximately $2.70481$. Increasing the number of steps, while reducing each increment proportionally, approaches <Euler's number> rather than producing unbounded accumulation.
\Image[/compound-growth-limit.png]
{title=Repeated smaller proportional updates approach Euler's number over one fixed interval}
The <natural logarithm> makes the <limit> transparent. For $u>0$, its <integral> definition gives $u/(1+u)\leq\ln(1+u)\leq u$, by bounding $1/t$ on $[1,1+u]$. With $u=1/n$,
$$
\frac1{1+1/n}\leq n\ln(1+1/n)\leq1.
$$
The <squeeze theorem> gives a logarithmic <limit> of one; continuity of the <exponential function> gives $(1+1/n)^n\to e$. This links a discrete multiplicative rule with continuous <exponential growth>, without assuming that a finite number of steps equals the limiting value.
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