Finding the limit for recurrence relation $ x_{n+1} = \sqrt{x_n + \frac 14} - \frac 1 2 $

$$ 0 < x_{n+1} = \sqrt{x_n + \frac 14} - \frac 1 2 = \frac{x_n}{\sqrt{x_n + \frac 14} + \frac 1 2} < x_n $$ shows that the sequence is decreasing and bounded below by zero, therefore convergent. The limit $L$ must satisfy $$ L = \sqrt{L + \frac 14} - \frac 1 2 $$ which implies that $L= 0$.

Then Stolz-Cesaro shows that $$ \lim_{n \to \infty}\frac{1}{n x_n} = \lim_{n \to \infty}\left(\frac{1}{x_{n+1}} - \frac{1}{x_n} \right) $$ provided that the latter limit exists. But $$ \frac{1}{x_{n+1}} - \frac{1}{x_n} = \frac{1}{\sqrt{x_n + \frac 14} - \frac 1 2} - \frac{1}{x_n} = \frac{1}{\sqrt{x_n + \frac 14} + \frac 1 2} $$ converges to one since $x_n$ converges to zero.