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I'm reading a proof and I don't understand one step.

1−(1−λ/n)^tn. Now, taking the limit as n→∞, P(Yn/n≤t)⟶1−e^(−λt). 

How does that work out? In this case, t is time(in seconds), n is the number of Bernoulli(p) trials per second and Yn is the time until the first success.

Further, if this converges to an exponential as n -> infinity, why is this significant?

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    sorry,didn't see that2012-11-04

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$\lim_{n \to \infty} (1-\lambda / n)^n = e^{-\lambda}$, so $\lim_{n \to \infty} (1-\lambda / n)^{t n} = e^{-\lambda t}$.

Don't know if you also want a proof of the first limit or already know it.

This is significant because the exponential distribution has many useful properties, such as memory independence. It is also simpler. It also won the 2011 competition as the distribution of the year.