How to prove that geometric distributions converge to an exponential distribution? 
How to prove that geometric distributions converge to an exponential distribution?

To solve this, I am trying to define an indexing $n$/$m$ and to send $m$ to infinity, but I get zero, not some relevant distribution. What is the technique or approach one must use here?

The geometric distribution is given by $$P(n) = (1-p)^n p = q^np$$
Here $P(n)$ is the probability of a succes after $n$ failures.
If we define the chance of succes $p=X/n$, where $X$ is a bernouilli random variable then we get after substitution in the above equation:
$$P(n) = \left(1-\frac{X}{n}\right)^n \left(\frac{X}{n}\right) $$
In taking the limit we get:
$$ \lim_{n\rightarrow \infty} dP(n) = \lim_{n\rightarrow \infty} \left(1-\frac{X}{n}\right)^n \left(\frac{X}{n}\right) = e^{-X} dX$$
Where we see $e^{-X}$ which is the exponential probability density function.
