Normal density's rate of convergence to 0 as mean goes to infinity while x and standard deviation are fixed Consider the density of the Normal distribution given by
$$f(x; \mu, \sigma) = \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right)$$
It is obvious that, with a fixed $x$ and $\sigma$, the Normal density tends to $0$ as $\mu \to \infty$ or $\mu \to -\infty$. However, is there a way to characterize the rate of this convergence?
Similarly, what about the rate of convergence as $\mu \to -\infty$ or $\mu \to \infty$ for the distribution function
$$\int \limits_0^{\infty} \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right) \ d x,$$ where $\sigma > 0$?
It seems that knowing these rates of convergences would be helpful in answering the following question, which I would appreciate any insights on if people have them.
Does the product
$$\left[\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right)\right] \left[\dfrac{1}{\int \limits_0^{\infty} \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right) \ d x}\right],$$ where $\sigma > 0$ is the same value in both factors and $x > 0$ in the first factor, tend to either $0$ or to $\infty$ as $\mu \to -\infty$? Intuitively, it seems that the answer would depend on the value of $\sigma$, but I'm not sure how to formally characterize this dependence.
Many thanks to all for any insights.

Let us introduce some notation: Let $Z \sim N(0,1)$ with density $\varphi(z) = f(z; 0, 1)$ and distribution function $\Phi$.
The denominator of the second term of the product you seek to study may be written as:
$$ \mathbb P[ Z/\sigma + \mu \geq 0] = \mathbb P[ Z \geq - \mu/\sigma] = 1 - \Phi(-\mu/\sigma).$$
Now by Mill's inequality (see e.g., this math.se post), it holds that:
$$ 1-\Phi(-\mu/\sigma) = \frac{\varphi(-\mu/\sigma)}{-\mu/\sigma}(1+o(1)) \text{ as } \mu/\sigma \to -\infty.$$
Hence the product you are interested in is equal to:
$$ \frac{\varphi((x-\mu)/\sigma)/\sigma}{1 - \Phi(-\mu/\sigma)} = -\mu\exp(-x^2/(2\sigma^2)) \exp(\mu x/\sigma)(1+o(1)) \text{ as } \mu/\sigma \to -\infty.$$
Since $x > 0$, we thus conclude that the ratio converges to $0$ as $\mu \to -\infty$.
