Consider a $t\in(0,1)$. Consider, for $\Delta>0$ the random variable $X_t^{(\Delta)}$ defined as
$$ \mathbb{P}[X_t^{(\Delta)}=1]=\left(1-\lambda\,\Delta\right)^{\left\lfloor t/\Delta\right\rfloor},\quad \mathbb{P}[X_t^{(\Delta)}=0]=1-\left(1-\lambda\,\Delta\right)^{\left\lfloor t/\Delta\right\rfloor}, $$ with $\lambda\in(0,1)$. Clearly when $\Delta\rightarrow 0$
$$ \mathbb{P}[X_t^{(\Delta)}=1]=\left(1-\lambda\,\Delta\right)^{\left\lfloor t/\Delta\right\rfloor}\rightarrow e^{-\lambda\,t} $$
and
$$ \mathbb{P}[X_t^{(\Delta)}=0]=1-\left(1-\lambda\,\Delta\right)^{\left\lfloor t/\Delta\right\rfloor}\rightarrow 1-e^{-\lambda\,t}. $$
The problem is: does the sequence $X_t^{(\Delta)}$ converge (in probability? distribution?) or not? My initial guess was that the sequence converges to a random variable $X_t$ that is $1$ with probability $e^{-\lambda\,t}$ and $0$ with probability $1-e^{-\lambda\,t}$, but it seems wrong since
\begin{eqnarray} \mathbb{P}\left[\left|X_t^{(\Delta)}-X_t\right|>\varepsilon\right] &=&\mathbb{P}\left[X_t^{(\Delta)}=1\right]\,\mathbb{P}\left[X_t=0\right]+ \mathbb{P}\left[X_t^{(\Delta)}=0\right]\,\mathbb{P}\left[X_t=1\right]\\ &=&\left(1-\lambda\,\Delta\right)^{\left\lfloor t/\Delta\right\rfloor}\,(1-e^{-\lambda\,t})+(1-\left(1-\lambda\,\Delta\right)^{\left\lfloor t/\Delta\right\rfloor})\,e^{-\lambda\,t}\\ &\rightarrow& 2\,e^{-\lambda\,t}\,(1-e^{-\lambda\,t})\neq 0 \end{eqnarray}