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.