Infinitely going limit

$\begin{aligned} x & = \lim_{n \to \infty} \left(1-\frac 1{2^2}\right) \left(1-\frac 1{3^2}\right) \left(1-\frac 1{4^2}\right) \cdots \left(1-\frac 1{n^2}\right) \\ & = \lim_{n \to \infty} \frac {2^2-1}{2^2} \times \frac {3^2-1}{3^2} \times \frac {4^2-1}{4^2} \times \cdots \times \frac {n^2-1}{n^2} \\ & = \lim_{n \to \infty} \frac {(2-1)(2+1)}{2^2} \times \frac {(3-1)(3+1)}{3^2} \times \frac {(4-1)(4+1)}{4^2} \times \cdots \times \frac {(n-1)(n+1)}{n^2} \\ & = \lim_{n \to \infty} \frac {1\times 3}{2^2} \times \frac {2\times 4}{3^2} \times \frac {3\times 5}{4^2} \times \cdots \times \frac {(n-3)(n-1)}{(n-2)^2} \times \frac {(n-2)n}{(n-1)^2} \times \frac {(n-1)(n+1)}{n^2} \\ & = \lim_{n \to \infty} \frac {2\times 3^2 \times 4^2 \times 5^2 \times \cdots \times (n-1)^2n(n+1)}{2^2\times 3^2 \times 4^2 \times 5^2 \times \cdots \times (n-1)^2n^2} \\ & = \lim_{n \to \infty} \frac {n+1}{2n} = \lim_{n \to \infty} \frac {1+\frac 1n}2 = \frac 12 \end{aligned}$

Therefore, $50x = \boxed{25}$ .