Showing that the shortfall-to-quantile ratio of a normal distribution goes to one

I dont get why $$\lim_{x \to \infty} \frac{\mu \{1 - \Phi(x)\} + \sigma \phi(x)}{(\mu + \sigma x) \{1 - \Phi(x)\} } = \lim_{x \to \infty} \frac{1}{1 - \sigma \frac{1 - \Phi(x)}{(\mu + \sigma x)\phi(x)}},$$. This is the only part that I'm stuck with, otherwise I don't see how to evaluate the limit. Thanks for reading

$$\lim_{x \to \infty} \frac{\sigma - \sigma \Phi(x)}{(\mu + \sigma x)\phi(x)}$$
$$\stackrel{0/0}{=} \lim_{x \to \infty} \frac{-\sigma \phi(x)}{\sigma\phi(x) + (\mu + \sigma x)\phi'(x) }$$
$$=\lim_{x \to \infty}\frac{-\sigma \phi(x)}{\sigma\phi(x) - (\mu + \sigma x)x\phi(x) } = 0$$
(Used $$\phi'(x) = -x\phi(x)$$ and $$\Phi'(x) =\phi(x)$$, just like in your first L'Hospital application to get your equality. And L'Hospital works the second time too as $$\lim_{x \to \infty}\phi(x) = 0$$, $$\lim_{x \to \infty}x \phi(x) = 0$$.)