# Convergence in distribution and find limit

For $n \geqslant 1$ , let [Fn:R ->[0,1]] ,be $Fn(x) = \left\{ \begin{gathered} 0{\text{ for x < 0}} \\ 1 - {(1 - x)^n}{\text{ for 0}} \leqslant {\text{x}} \leqslant {\text{1}} \\ 1{\text{ for x > 1}} \\ \end{gathered} \right.$

(1) find the $\mathop {\lim }\limits_{n \to \infty } Fn(x)$

Solution $\mathop {\lim }\limits_{n \to \infty } Fn(x) = \left\{ \begin{gathered} 0{\text{ for x < 0}} \\ 1{\text{ for }}x \geqslant 0{\text{ }} \\ \end{gathered} \right.$

It True or false  ?  If true how we show


$\mathop {\lim }\limits_{n \to \infty } [1 - {(1 - x)^n}] = 1$ Thank you very much.

Yes is correct; If $0<x<1\implies0<1-x<1$, then $$y=(1-x)^n\implies \ln{y}=n\ln{(1-x)}\implies\ln{(\lim_{n\to \infty}y)}=\lim_{n\to \infty}n\ln{(1-x)}\to-\infty\implies\lim_{n\to \infty}y\to 0$$