geometric CDF converges to exponential cdf 
Let $X_p\sim Geo(p), Y_p=(1-p)X_p$. Show that in the limit $p\to1$ the CDF of $Y_p$ converges pointwise to the CDF of the Exp(1) distribution.

I tried the following: Substitute $1-p=\dfrac{1}{t}$, then $p\to 1$ is the same as $t\to\infty$. So $F_Y(n)=P(Y\le n)=P((1-p)X\le n)=P(X\le nt)=1-(1-p)^{nt}=1-(1-(1-\dfrac{1}{t}))^{nt}=1-(\dfrac{1}{t})^{nt}$
which goes to 1 for $n\to\infty$. It should be $1-\exp(-n)$ though. Where did I go wrong?
 A: The transformation and the condition that gives the desired result (for a geometric distribution without $0$ in its support) is
$$Y_p= pX_p ,\;\;  p\rightarrow 0 \Rightarrow Y_p \rightarrow \text {Exp(1)}$$
For every nonnegative real number $y$, we have
$$F_Y(y)=P(Y_p\le y)=P(pX_p\le y)=P\left(X_p\le \frac yp\right)=1-\left[1-p\right]^{y/p}$$
Set $t\equiv  1/p$. Then
$$\lim_{p\rightarrow 0}\left (1-\left[1-p\right]^{y/p}\right) = \lim_{t\rightarrow \infty}\left (1-\left(1-\frac 1t\right)^{yt}\right) = 1-e^{-y} $$ 
DISCUSSION 
So why this result is not "symmetric"? After all it "looks like" the scenario  $\{Y_p= pX_p ,\;\;  p\rightarrow 0\}$ is "equivalent" to the scenario $\{Y_{1-p}= (1-p)X_p ,\;\;  p\rightarrow 1\}$. Informal intuition can be perhaps gained by remembering that $p = P(X_p=1)$, which is the minimum value in the support of $X_p$. So, as $p\rightarrow 0$ the probability of observing the value unity goes to zero, "spreading" the probability mass to the right and to all other values of the support, leaving to the $Y_p$ transformation the job of determining the new probability mass allocation, while,  as $p\rightarrow 1$ this probability goes to unity, concentrating the probability mass to the value unity, rendering the $Y_{1-p}$ transformation ineffective in preserving a probability distribution.
