Limit of quotient of inverse cdfs I am trying to obtain $$\lim_{x\to0}\frac{\Phi^{-1}(1-x)}{\Phi^{-1}(1-x/n)}$$ where $\Phi^{-1}$ is the inverse cdf of the standard normal distribution and $n>0$. As there is an indeterminate form ($\infty/\infty$), I am applying l'Hôpital's rule, but the resulting expression (I mean, of the derivatives of both the numerator and denominator) is of the form $\infty/\infty$ as well. Would you give me any advice on how to proceed?
 A: This answer is not complete and needs correction. I am stuck at a point and will update this answer soon. However, I am retaining this answer in hope that someone might state more clearly the transition from equation (a) to equation (b).
$$1-x = \Phi(t) = \frac{1}{2} + \frac{1}{2}\text{erf}\left(\frac{t}{\sqrt{2}}\right) \implies \Phi^{-1}(1-x) = \sqrt{2}\ \text{erf}^{-1}(1-2x)$$
Similarly,
$$\Phi^{-1}\left(1-\frac{x}{n}\right) = \sqrt{2}\ \text{erf}^{-1}\left(1-\frac{2x}{n}\right)$$
where,
$$\text{erf}^{-1}(u) = \sum_{k=0}^{\infty}\frac{c_k(\sqrt{\pi}/2)^{2k+1}}{2k+1}u^{2k+1}$$
Note that $\text{erf}^{-1}(0) = 0$.
$$\begin{align}\lim_{x\rightarrow 0} \frac{\Phi^{-1}(1-x)}{\Phi^{-1}\left(1-\frac{x}{n}\right)} &= \lim_{x\rightarrow 0} \frac{\sqrt{2}\ \text{erf}^{-1}(1-2x)}{\sqrt{2}\ \text{erf}^{-1}\left(1-\frac{2x}{n}\right)} = \lim_{x\rightarrow 0} \frac{ \text{erf}^{-1}(1-2x)}{\ \text{erf}^{-1}\left(1-\frac{2x}{n}\right)}-1+1 \\\\
&= \lim_{x\rightarrow 0} \frac{ \text{erf}^{-1}(1-2x) - \text{erf}^{-1}\left(1-\frac{2x}{n}\right)}{\text{erf}^{-1}\left(1-\frac{2x}{n}\right)}+1 \\\\
&= \lim_{x\rightarrow 0} \frac{\sum_{k=0}^{\infty}\frac{c_k(\sqrt{\pi}/2)^{2k+1}}{2k+1}\left((1-2x)^{2k+1} - \left(1-\frac{2x}{n}\right)^{2k+1}\right)}{\text{erf}^{-1}\left(1-\frac{2x}{n}\right)} + 1\\\\
&= \lim_{x\rightarrow 0} \frac{\sum_{k=0}^{\infty}\frac{c_k(\sqrt{\pi}/2)^{2k+1}}{2k+1}\left(2x(\frac{1}{n}-1)(\sum_{i=0}^{2k}(1-2x)^{i}(1-\frac{2x}{n})^{2k-i})\right)}{\text{erf}^{-1}\left(1-\frac{2x}{n}\right)} + 1 \\\\
&= \frac{\sum_{k=0}^{\infty}\frac{c_k(\sqrt{\pi}/2)^{2k+1}}{2k+1}\left((\rightarrow 0)(\rightarrow (2k+1)\right)}{\rightarrow \infty} + 1 \tag{a} \\\\
&= \rightarrow 0 + 1 = 1 \tag{b} \end{align}$$
Note: How did I get ($\rightarrow 0$)? It turns out that the numerator can be upper bounded by $2x(1/n-1)$ times derivative of $\text{ erf}^{-1}(u)$ evaluated at $u=(1-2x/\max(1,n))$, where $x \rightarrow 0$. Now, although the numerator has form of $(\rightarrow 0)(\rightarrow \infty)$ and the denominator is tending to $\infty$ as $x \rightarrow 0$. Near $0^+$, the rate(speed) with which the denominator is tending to $\infty$ is much faster than the rate at which numerator is tending to $\infty$ due to the multiplication of $(\rightarrow 0)$ in the numerator. I observed this by plotting the function for different values of $n$ and the function is approaching $0$ as $x \rightarrow 0$. However, concretely proving this seems to me to be non trivial.
Some equations that might be useful to solve.
$$\frac{\partial \text{ erf}^{-1}(u)}{\partial u} =  \sum_{k=0}^{\infty}c_k(\sqrt{\pi}/2)^{2k+1}u^{2k} = \frac{\sqrt{\pi}}{2}e^{(\text{erf}^{-1}(u))^2}$$
$$\lim_{u\rightarrow 0}u\frac{\partial \text{ erf}^{-1}(u)}{\partial u} =  \lim_{u\rightarrow 0} \sum_{k=0}^{\infty}c_k(\sqrt{\pi}/2)^{2k+1}u^{2k+1} = \lim_{u\rightarrow 0}u\frac{\sqrt{\pi}}{2}e^{(\text{erf}^{-1}(u))^2} = (\rightarrow 0)(\rightarrow \sqrt{\pi}/2)$$
Check this wiki page for more information on $\text{erf}^{-1}$ and its taylor series.
Check this wolfram page for a closed formula for the derivative of $\text{erf}^{-1}(u)$. It is easy to derive it from the definition of $\text{erf}^{-1}(u)$ in the integral form.
With simulations using different values of $n$, I get convergence to $1$.

A: Here is another much simpler way to solve.
Note that,
$$\frac{\partial \Phi^{-1}(x)}{\partial x} = \frac{1}{\phi(\Phi^{-1}(x))} \ \ \text{ 
 and  } \ \ \frac{\partial \Phi^{-1}(x/n)}{\partial x} = \frac{1}{n\phi(\Phi^{-1}(x/n))}$$
Also,
$$\frac{\partial \phi(\Phi^{-1}(x))}{\partial x} = \frac{-\Phi^{-1}(x)\phi(\Phi^{-1}(x))}{\phi(\Phi^{-1}(x))} = -\Phi^{-1}(x) $$
$$ \frac{\partial \phi(\Phi^{-1}(x/n))}{\partial x} = \frac{-\Phi^{-1}(x/n)\phi(\Phi^{-1}(x/n))}{n\phi(\Phi^{-1}(x/n))} = -\frac{\Phi^{-1}(x/n)}{n}$$
$$\begin{align} L &= \lim_{x \rightarrow 0} \frac{\Phi^{-1}(1-x)}{\Phi^{-1}(1-x/n)} = \lim_{x \rightarrow 0} \frac{-\Phi^{-1}(x)}{-\Phi^{-1}(x/n)} = \color{red}{\lim_{x \rightarrow 0} \frac{\Phi^{-1}(x)}{\Phi^{-1}(x/n)}} = \frac{\rightarrow -\infty}{\rightarrow -\infty}\\\\
&= \lim_{x \rightarrow 0} \frac{\frac{\partial \Phi^{-1}(x)}{\partial x}}{\frac{\partial \Phi^{-1}(x/n)}{\partial x}} = \lim_{x \rightarrow 0} \frac{n\phi(\Phi^{-1}(x/n))}{\phi(\Phi^{-1}(x))} = \frac{\rightarrow 0}{\rightarrow 0} \\\\
&= \lim_{x \rightarrow 0} \frac{n\frac{\partial \phi(\Phi^{-1}(x/n))}{\partial x}}{\frac{\partial \phi(\Phi^{-1}(x))}{\partial x}} = \color{red}{\lim_{x \rightarrow 0} \frac{\Phi^{-1}(x/n)}{\Phi^{-1}(x)}} = 1/L \end{align}$$
Therefore,
$$L^2 = 1 \implies L = \pm 1$$
Clearly, $L$ cannot be negative, so $L=1$.
