# Limit containing inverse normal

I am having trouble taking the following limit that contains an inverse normal distribution as alpha approaches 1:

$\lim_{\alpha \to 1} \frac{\mu + \sigma \frac{\phi^{-1}(\alpha)}{1-\alpha}}{\mu + \sigma \phi^{-1}(\alpha)}$

First I tried re-arranging the $(1-\alpha)$ in the numerator to get:

$\lim_{\alpha \to 1} \frac{\mu(1-\alpha) + \sigma {\phi^{-1}(\alpha)}}{\mu + \sigma \phi^{-1}(\alpha)}$

Then I thought maybe I need to use L'Hopital's rule, but I have no idea how to do that with an inverse normal imbedded in my function. I feel that I'm probably missing something simple and my calculus is too rusty. Any hints for how to compute this limit? I know it should converge to 1, I'm just not seeing why.

Many thanks.

The limit equals $\infty$. Indeed, note that
$$\lim_{\alpha\uparrow 1} \frac{\mu + \sigma \Phi^{-1}(\alpha)/(1-\alpha)}{\mu+\sigma \Phi^{-1}(\alpha)}= \lim_{\alpha\uparrow 1} \frac{(1-\alpha)\mu + \sigma \Phi^{-1}(\alpha)}{(1-\alpha)\mu+\sigma (1-\alpha)\Phi^{-1}(\alpha)}= \lim_{\alpha\uparrow 1} \frac{1}{1-\alpha}=\infty.$$
• If you have something of the form $(a_n+b_n)/(c_n+d_n)$ and $a_n/b_n\to 0$ and $c_n/d_n\to 0$ as $n\to\infty$ then $(a_n+b_n)/(c_n+d_n)$ behaves asymptotically like $b_n/d_n$. This is true for random objects, also. See Slutsky's theorem. – JPi Apr 11 '14 at 18:08