Objects are shuffled. What is the probability that exactly one object remains in its original position?

To find the number of "good" permutations fix one card and derange the rest $(n-1)$ cards. This can be done in $!n$ ways. ($!n$ is the number of derangements of $n$ objects). Then the number of "good" permutations is $n\cdot !(n-1)$. Hence we have:

$$\lim_{n \to \infty} p_n = \lim_{n \to \infty} \frac{n\cdot !(n-1)}{n!} = \lim_{n \to \infty} \frac{!(n-1)}{(n-1)!} = e^{-1}$$

The last limit can be seen in the link I added above.

By the inclusion-exclusion principle, in the symmetric group $S_n$ there are $$ n!\sum_{k=0}^{n}\frac{(-1)^k}{k!} $$ permutations without fixed points (see derangements). It follows that in the same group there are $$ n\cdot (n-1)!\sum_{k=0}^{n-1}\frac{(-1)^k}{k!} $$ elements with exactly one fixed point, and the limit probability is $\color{red}{\large\frac{1}{e}}$ in both cases.