Let $a$ be fixed. I am trying to show $P(B_t\le t-a)=\frac{\int_{\infty}^a \exp(-x^2/(2(t-a)^2))}{\sqrt{2\pi}(t-a)}dx $ tends to 1 as $t\to \infty.$
However, I am getting that the integral tends to about .84 https://www.wolframalpha.com/input/?i=1%2F2(1%2B+Erf(1%2Fsqrt2), but shouldn't the limit tend to 1 since Brownian motion is almost surely finite?
The probability $P(B_t\le t-a)$ (for a standard Brownian motion $B$) is equal to $ \Phi(\sqrt{t}-at^{-1/2})$. This is not the same as the integral you have written down. The latter is $P(B_{t-a}\le a)$, which is in turn equal to $\Phi(a/(t-a))$, at least for $t>a$..