calculus - Proof verification for $lim_{ntoinfty}(sqrt{n^2-1} - sqrt n) = +infty$

Show that:
$$
\lim_{n\to\infty}\left(\sqrt{n^2-1} - \sqrt n\right) = +\infty
$$

I've started it this way.

Lemma:

Let $x_n$ and $y_n$ be two sequences. Claim:

If:
$$
\begin{cases}
&\lim_{n\to\infty} x_n =+\infty \\

&\exists N\in \Bbb N, \ \forall n >N:y_n\ge c > 0
\end{cases}
$$
Then:
$$
\lim_{n\to\infty}(x_ny_n) = +\infty
$$

Proof:

$\Box$ Start with definition of limit for this case:
$$
\forall\varepsilon>0,\ \exists N_1\in\Bbb N: \forall n > N_1 \implies x_n >\varepsilon
$$
Also:
$$
\exists N_2\in\Bbb N:\forall n>N_2 \implies y_n \ge c > 0
$$

Let:
$$
N = \max\{N_1, N_2\}
$$

Then starting from this $N$ we obtain:
$$
x_n\cdot y_n > c\cdot \varepsilon
$$

And we have that:
$$
\forall\varepsilon>0,\ \exists N =\max\{N_1, N_2\}\in\Bbb N: \forall n > N \implies x_n y_n > c\varepsilon
$$

Thus:
$$
\lim_{n\to\infty}(x_ny_n) = +\infty \ \Box
$$

Now back to the initial problem. Let:
$$
z_n = \sqrt{n^2-1} - \sqrt n = \frac{n^2 - n - 1}{\sqrt{n^2 - 1} + \sqrt{n}}
$$

Define:
$$
x_n = n - 1 - {1\over n} \\
y_n = \frac{n}{\sqrt{n^2 - 1} + \sqrt{n}}
$$

Obviously $y_n \ge c > 0$ for some $N$ and $n>N$. Also $x_n \to +\infty$, then by lemma:
$$
\lim_{n\to\infty}z_n = \lim_{n\to\infty}{x_ny_n} = +\infty
$$

I know this is a bit overkill, but i wanted to use that exact lemma for the proof. Apart from that, is it valid?

BTW here is a visualization for $x_n, y_n$

Update

Since it is not clear where the lemma comes from here is the problem from the problem book right before the limit.

Let:
$$
\lim_{n\to\infty}x_n = a\ , \text{where}\ a = +\infty \ \text{or} \ a = -\infty
$$
Prove that if for all $n$ starting from some $N$ $y_n \ge c > 0$ then

$$
\lim_{n\to\infty}x_ny_n = a
$$
And if for all $n$ starting from some $N$ $y_n \le c < 0$ then
$$
\lim_{n\to\infty}x_ny_n = -a
$$

No other constraints are given.