Suppose I have the following lower and upper bound for the Gaussian Q Function:

$$ \frac{x}{x^2 + 1} \varphi(x) < Q(x) < \frac{1}{x} \varphi(x), $$

where $Q(x) = \int\_x^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \, \mathrm{d}u$ and $\varphi(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2 / 2}$.

How do I show $Q(x) \sim \varphi(x)/x$ as $x \to \infty$? Apparently, this fact can be shown simply from the upper and lower bounds.

Also, I am unsure how the limit $\lim\_{x \to \infty} \frac{Q(x)}{\varphi(x)} = \frac{1}{x}$ (which I can verify through L'Hopital's Rule) proves $Q(x) \sim \varphi(x)/x$, since the claim $\lim\_{x \to \infty} \frac{A}{B} = C \iff \lim\_{x \to \infty} A = \left(\lim\_{x \to \infty} B \right) \times C$ doesn't seem to be legitimate. 

Thanks
['https://math.stackexchange.com/questions/45694', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/8140/']
You are supposed to show that $\frac{Q(x)}{\varphi(x)/x}$ converges to $1$. (That is the meaning of the curly line.)

To do this divide by $\varphi(x)/x$ everywhere in your inequality and use the squeeze rule.
First, as Johan mentioned in his answer, 
$$
Q(x)\sim \frac{{\varphi (x)}}{x} \;\; {\rm as} \; x \to \infty
$$
means that
$$
\frac{{Q(x)}}{{\varphi (x)/x}} \to 1 \;\; {\rm as} \; x \to \infty.
$$

Concerning the last paragraph of your question, it is certainly wrong to write $\lim \_{x \to \infty } \frac{{Q(x)}}{{\varphi (x)}} = \frac{1}{x}$; rather, the asymptotic equivalence $Q(x)\sim \varphi (x)/x$ can be proved as follows.
$$
\mathop {\lim }\limits\_{x \to \infty } \frac{{Q(x)}}{{\varphi (x)/x}} = \mathop {\lim }\limits\_{x \to \infty } \frac{{xQ(x)}}{{\varphi (x)}} = \mathop {\lim }\limits\_{x \to \infty } \frac{{(xQ(x))'}}{{\varphi '(x)}} = \mathop {\lim }\limits\_{x \to \infty } \frac{{Q(x) + xQ'(x)}}{{\varphi '(x)}}.
$$
Now,
$$
Q'(x) = \frac{d}{{dx}}\int\_x^\infty {\frac{1}{{\sqrt {2\pi } }}e^{ - u^2 /2} du} = - \frac{1}{{\sqrt {2\pi } }}e^{ - x^2 /2} 
$$
and
$$
\varphi '(x) = \frac{d}{{dx}}\frac{1}{{\sqrt {2\pi } }}e^{ - x^2 /2} = \frac{1}{{\sqrt {2\pi } }}e^{ - x^2 /2} ( - x).
$$
Hence $xQ'(x) = \varphi '(x)$, and so it remains to show that
$$
\mathop {\lim }\limits\_{x \to \infty } \frac{{Q(x)}}{{\varphi '(x)}} = 0.
$$
Indeed,
$$
\mathop {\lim }\limits\_{x \to \infty } \frac{{Q(x)}}{{\varphi '(x)}} = \mathop {\lim }\limits\_{x \to \infty } \frac{{Q'(x)}}{{\varphi ''(x)}} = 0.
$$