A question about limit and sequence Let $\{p_n\},\{q_n\}$ be sequences of positive integers such that 
$$
\lim_{n \to \infty}\frac{p_n}{q_n}=\sqrt{2}.
$$
Show that $$\lim_{n\to\infty}\dfrac{1}{q_n}=0.$$
I have no clue in proving this. Could somebody give me some hints?
 A: If not, then there is an $\varepsilon>0$, such that 
$$
\frac{1}{q_n}\ge\varepsilon,
$$
for infinitely many $n$. So there are subsequences $p_{n_k},q_{n_k}$, $k\in\mathbb N$,
such that
$$
\frac{p_{n_k}}{q_{n_k}}\to\sqrt{2}\quad\text{and}\quad q_n\le\frac{1}{\varepsilon}.
$$
But 
$$
q_{n_k}\le\frac{1}{\varepsilon}\quad\Longrightarrow\quad q_{n_k}\in \{1,\ldots,\lfloor\varepsilon^{-1}\rfloor\},
$$
Let
$$
d=\min\left\{\left|\sqrt{2}-\frac{p}{q}\right|: p\in\mathbb N\,\,\text{and}\,\,
q\in \{1,\ldots,\lfloor\varepsilon^{-1}\rfloor\}\right\}.
$$
Clearly $d>0$, as
$$
d=\min\left\{\left|\sqrt{2}-\frac{p}{q}\right|:
q\in \{1,\ldots,\lfloor\varepsilon^{-1}\rfloor\} \,\,\text{and}\,\,
p\in\{1,2,\ldots,2q\}
\right\},
$$
since if $p>2q$, then $\big|\sqrt{2}-\frac{p}{q}\big|>\big|\sqrt{2}-\frac{p}{p}\big|$.
Thus
$$
\left|\frac{p_{n_k}}{q_{n_k}}-\sqrt{2}\right|\ge d.
$$
Thus $\frac{p_{n_k}}{q_{n_k}}\not\to\sqrt{2}$.
A: The idea is to try and approximate $\sqrt{2}$ with a rational number. Because $\sqrt{2}$ is irrational, the integer making up the denominator of the rational approximation must continue to grow endlessly in order for the approximation to improve. 
