Mills' Ratio for Gaussian Q Function

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, $$ \mathop {\lim }\limits_{x \to \infty } \frac{{Q(x)}}{{\varphi '(x)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{Q'(x)}}{{\varphi ''(x)}} = 0. $$

The fact that the limit is $1$ can be simply obtained from the given bounds, as explained by Johan. And the bounds are more important than the limit stuff, since the bounds give practical estimates.

Let us look at your attempt through L'Hospital's Rule. The idea was good, but there were some problems in execution.

The result that you mention obtaining cannot be correct. You say that the limit as $x \to \infty$ of $Q(x)/\varphi(x)$ is $1/x$. But the limit, if it exists, of $Q(x)/\varphi(x)$ must be a "number" (let us temporarily assign honorary number status to $+\infty$). So in particular the limit cannot be $1/x$, that is a function of $x$.

However, a L'Hospital's Rule argument of the type you attempted can be made to work.

There are various ways to set up the calculation. Some work more quickly than others. We will use a somewhat inefficient method, and mention a better way at the end. We will find $$\lim_{x\to\infty} \frac{xQ(x)}{\varphi(x)}$$

By L'Hospital's Rule, this is the same as $$\lim_{x\to\infty} \frac{xQ'(x)+Q(x)}{\varphi'(x)}$$

Calculate.
$$xQ'(x)=-x\varphi(x) +Q(x)$$ and $$\varphi'(x)=-x\varphi(x)$$ So L'Hospital's Rule says that our limit is $$\lim_{x\to\infty}\frac{-x\varphi(x) +Q(x)}{-x\varphi(x)}$$ Almost there! By dividing, we can see that the function we are trying to take the limit of is $$1 -\frac{Q(x)}{\varphi(x)}$$

If we can show that $$\lim_{x \to \infty}\frac{Q(x)}{\varphi(x)}=0$$ we will be finished. Use L'Hospital's Rule again. We find that $$\lim_{x \to \infty}\frac{Q(x)}{\varphi(x)}=\lim_{x\to\infty}\frac{-\varphi(x)}{-x\varphi(x)}$$ Do the obvious cancellation. We now want $$\lim_{x\to\infty}\frac{1}{x}$$ and this is $0$.

It is more efficient to use L'Hospital's Rule to show that $$\lim_{x\to\infty} \frac{\frac{\varphi(x)}{x}}{Q(x)}=1$$ (Here "top" is $\varphi(x)/x$ and "bottom" is $Q(x)$.) A single calculation is enough. Try it!