Let's define $p_nn = \lambda_n$, and $\lim_{n\to\infty} \lambda_n = \lambda>0$. $p_n \in [0,1] \forall n$
I know that
$\lim_{n\to\infty} \left( 1+\frac{\lambda}{n}\right)^n = e^{\lambda}$, (1).
I need to show that
$\lim_{n\to\infty} \left( 1-\frac{\lambda_n}{n}\right)^n=e^{-\lambda}.$
I have tried
$\lim_{n\to\infty} \left(1-\frac{\lambda _{n}}{n}\right)^n = \lim_{n\to\infty} \left(1-\frac{\lambda _{n}}{n}\right)^{\lambda_n} \left(\frac{n- \lambda _{n}}{n}\right)^{n-\lambda_n}$
$= \lim_{n\to\infty} \frac{\left(1-\frac{\lambda _{n}}{n}\right)^{\lambda_n}}{\left(\frac{n}{n- \lambda _{n}}\right)^{n-\lambda_n}}=\lim_{n\to\infty} \frac{\left(1-\frac{\lambda _{n}}{n}\right)^{\lambda_n}} {\left( 1+ \frac{\lambda_n}{n- \lambda _{n}}\right)^{n-\lambda_n}}$.
Now I would like to use (1) on the denominator and say that
$\lim_{n\to\infty} \frac{\left(1-\frac{\lambda _{n}}{n}\right)^{\lambda_n}} {\left( 1+ \frac{\lambda_n}{n- \lambda _{n}}\right)^{n-\lambda_n}} = \frac{1^\lambda}{e^\lambda} = e^{-\lambda}$
But It doesn't seem quite right. Am I correct to doubt my solution? How should I proceed?
Thank you in advance!