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.