calculus - 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?

Answer

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 $L=1$.