Does there exists a good asymptotic formula for $$A(x) := \prod_{p\leq x}(1-\frac 1p).$$

By using a heuristic argument one can guess: $$A(x) \sim \frac{1}{2\,\mathrm{ln}(x)}.$$

Here is the argument:

$$\frac{x^2}{2\, \mathrm{ln}(x) } \sim \pi(x^2)-\pi(x) = \sum_{p_1<p_2<\dots<p_k\leq x}(-1)^k\left\lfloor \frac{x^2}{p_1p_2\cdots p_k}\right\rfloor \\ \stackrel{??}{\simeq}\sum_{p_1<p_2<\dots<p_k\leq x}(-1)^k\left( \frac{x^2}{p_1p_2\cdots p_k}\right) = x^2 \prod_{p\leq x}(1-\frac{1}{p}).$$

Is this conjectural asymptotic formula true? If so, can this heuristic argument be completed to an exact proof? The later question is more important for me!

Thanks!