I need to show that $F_{X}(x)=1-e^{-x \lambda} $ is a cumulative distribution function and I have:
i) $F_{X}(x)$ is monotone (that is $F_{X}(a)<F_{X}(b)$ for $a<b$ $F_{X}(a)<F_{X}(b) \Rightarrow 1-e^{-a \lambda}<1-e^{-b \lambda} \Rightarrow e^{-a \lambda}>e^{-b \lambda} \Rightarrow a \lambda<b \lambda \Rightarrow a<b$.
ii) $F_{X}(x)$ is continuos from the right (that is $\lim_{0<h\rightarrow 0}F_{X}(x+h)=F_{X}(x)$) $\lim_{0<h\rightarrow 0}F_{X}(x+h)=\lim_{h\rightarrow 0}(1-e^{-(x+h)\lambda})=1-e^{-x\lambda}$.
iii) $\lim_{x\rightarrow - \infty}F_{X}(x)=0$ and $\lim_{x\rightarrow \infty}F_{X}(x)=1 $ $\lim_{x\rightarrow \infty}F_{X}(x)=\left ( 1-e^{-x\lambda} \right )=1$ But I cannot find the other limit and I don't know why.
This is only a CDF for a variable restricted to $[0,\infty)$.
In that case you'd define $F(x)=0 \;\;\forall x<0$
With this you have a valid CDF as your arguments show.