How to find $ \lim_{n\to\infty}(1- \frac{1-e^{-\frac{x(e-1)}{n}}}{1-e})^n$?

Solution 1:

Notice that as $n\to+\infty$ we have

$$e^{\frac{-x(e-1)}{n}} \sim 1 - \frac{x(e-1)}{n} + O(1/n^2)$$

hence

$$\lim_{n\to+\infty} \left(1 - \dfrac{x(e-1)}{(1-e)n}\right)^n = \lim_{n\to+\infty}\left(1 + \dfrac{x}{n}\right)^n = e^x$$

Solution 2:

I'll call the expression

$$ f(n) = \left(1-\frac{1-e^{-{x(e-1)/n}}}{1-e}\right)^{\!n} = \left(\frac{e - e^{-x(e-1)/n}}{e-1}\right)^{\!n}$$

For a given negative real $x$, we can consider this $f$ a function of real values $n$ with $n>-x(e-1)$.

Since the behavior as $n \to \infty$ looks like a "$1^\infty$" case, we can try taking the logarithm and then applying L'Hopital's rule.

$$ \begin{align*} \ln f(n) &= \ln \left(\frac{e - e^{-x(e-1)/n}}{e-1}\right)^{\!n} \\ \ln f(n) &= n \left[ \ln \left(e - e^{-x(e-1)/n}\right) - \ln (e-1)\right] \\ \ln f(n) &= \frac{\ln \left(e - e^{-x(e-1)/n}\right) - \ln (e-1)}{1/n} \\ \lim_{n \to \infty} \ln f(n) &= \lim_{n \to \infty} \frac{\frac{1}{e - e^{-x(e-1)/n}} \left(-e^{-x(e-1)/n}\right) \frac{x(e-1)}{n^2}}{-1/n^2} \\ \lim_{n \to \infty} \ln f(n) &= \lim_{n \to \infty} \frac{x(e-1) e^{-x(e-1)/n}}{e-e^{-x(e-1)/n}} \\ \lim_{n \to \infty} \ln f(n) &= \frac{x (e-1) e^0}{e-e^0} \\ \lim_{n \to \infty} \ln f(n) &= x \\ \ln \left(\lim_{n \to \infty} f(n)\right) &= x \\ \lim_{n \to \infty} f(n) &= e^x \end{align*} $$