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Consider some right-continuous function $f:\mathbb{R\cup\{-\infty,\infty\}}\to [0,1]$. I have to evaluate (i) $\lim_{b \to 0^+} f(\frac{a}{b})$, and (ii) $\lim_{b \to 0^-} f(\frac{a}{b})$ where $a \in \mathbb{R}^-$.

QUESTION: In general, for which (i) and/or (ii) am I allowed to push the limit inside and evaluate (i) $f(\lim_{b\to 0^+} \frac{a}{b})=f(-\infty)$, and (ii) $f(\lim_{b\to 0^-} \frac{a}{b} )= f(\infty)$?

I was told you can't do this for discontinuous functions (this function has jumps when limit approaches from left), which makes me concerned about doing this, even though the domain is the extended reals.

FYI: This $f$ is actually the cumulative distribution function of the standard normal probability density function (the standard normal, $\mathscr{N}(0,1)$ CDF). ((related to the erf(x) or even the complementary error function)).

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up vote 1 down vote accepted

Assuming that your definition of right semi-continuity is the same used for functions with domain in $\mathbb{R}$. For your function holds: $$\lim_{\alpha \rightarrow \beta^+} f(\alpha)= f(\beta), \qquad \beta\in\mathbb{R}\cup \{-\infty, + \infty\}.$$

So you are guaranteed you can exchange function and limit if and only if you are approaching the limit point from the right.

In case $i)$ you have that $\frac{a}{b} \rightarrow -\infty$ from the right, so you can exchange them.

While in case $ii)$ $\frac{a}{b} \rightarrow +\infty$ from the left, so you are not guaranteed to get the correct value.

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(+1) I have a related question. Is this proposition true or false: $(f\text{ is Right continuous})\Rightarrow \neg (f\text{ is Left continuous})$ – Jase Oct 17 '12 at 16:10
If I understood the logic symbol you used, then the proposition is false. For example any continuous function is both left continuous and right continuous. Does it answer your question? – Giovanni De Gaetano Oct 17 '12 at 16:23
Yes, it does, Thank YOU! – Jase Oct 17 '12 at 16:34
You're welcome! :) If your satisfied with the answer would you mind accepting it? – Giovanni De Gaetano Oct 17 '12 at 16:37

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