I have this ugly proof that some sequence converges to something and I really don't like it because it seems too hard for no reason... can anyone help me make this more simple?
Here it is : We are given a sequence $p_n$ with $n p_n \to \lambda > 0$. We show that $ \left( 1 - p_n \right)^n \to e^{-\lambda}. $ So my proof begins by noticing that $ (1-p_n)^n = \left( 1 + \frac{-np_n}n \right)^n $ and that the family of functions $s_n(x) = \left(1+\frac xn \right)^n$ is equicontinuous at $x$ (I think I can pull some argument stating that the derivative of $s_n(x)$ is always bounded (because it also converges to $e^x$) when considering an interval of arbitrary but fixed length centered at $x$, so that in this interval, all functions $s_n$ can have the same Lipschitz constant, therefore giving equicontinuity, correct me if I'm wrong) so that if I note by $x_n$ the sequence $-n p_n$, $x = -\lambda$ and $s(x) = e^x$, I wish to show that $ \forall \varepsilon > 0, \quad \exists N \quad s.t. \quad \forall n > N, \quad |s_n(x_n) - s(x)| < \varepsilon. $ so now, $ |s_n(x_n) - s(x)| \le |s_n(x_n) - s_n(x_m)| + |s_n(x_m) - s_n(x)| + |s_n(x) - s(x)|. $ Since $x_n$ is convergent, it is also Cauchy, so that for all $\delta > 0$, there exists an $N_1$ such that for all $n,m > N_1$, $|x_m - x_n| < \delta$ and $|x_m - x| < \delta$. Since the family $\{s_n\}$ is equicontinuous, for all $\varepsilon > 0$, there exists a $\delta > 0$ such that $|x - y| < \delta \Rightarrow |s_n(x) - s_n(y)| < \frac{\varepsilon}3$, so that the first two terms are taken care of by $N_1$ by taking $(x,y) = (x_n, x_m)$ and $(x,y) = (x_m, x)$ respectively. The last term is also bounded by $\varepsilon/3$ when $n > N_2$ because $s_n \to s$ pointwisely. Taking $N = \max \{ N_1, N_2 \}$ we're done.
I feel like my proof is a little too harsh on this easy-looking problem... anyone has a better proof? What I mean by "better" is using less powerful or advanced tools, something simple. I can't seem to find anything better right now.