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

I've started it this way.


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


$\Box$ 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 $x_n, y_n$


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.

  • $\begingroup$ What's $c$ in your lemma? It feels like it's just randomly introduced? $\endgroup$
    – Jam
    Commented Dec 6, 2018 at 19:30
  • $\begingroup$ @Jam, the exercise to prove this lemma comes right before the exercise on the limit (which in in the question section). No constraints for $c$ are given except for the fact that from some $N$ $y_n \ge c > 0$. This lemma is actually one of the several of the same kind in the exercise before limit. $\endgroup$
    – roman
    Commented Dec 6, 2018 at 19:32
  • $\begingroup$ I may be mistaken but I think you need more constraints on $c$. It seems like there's nothing stopping us from having $c=1/\varepsilon>0$, in which case $x_ny_n>c\varepsilon=1$ doesn't tell us much. $\endgroup$
    – Jam
    Commented Dec 6, 2018 at 19:39
  • $\begingroup$ In other words, you've shown that $x$ is unbounded above and $y$ is bounded from below but how do we know that all $y$ have the same lower bound? And how do we know that the lower bound of $y$ doesn't get small very quickly (counteracting the size of $\varepsilon$)? $\endgroup$
    – Jam
    Commented Dec 6, 2018 at 19:42
  • $\begingroup$ @Jam, I have updated the question and added more context of where that lemma came from $\endgroup$
    – roman
    Commented Dec 6, 2018 at 22:10

1 Answer 1


All you need here is:

  1. $\sqrt{n^2-1}\ge n-1$ if $n\ge 1$ (to see this, just square both sides);
  2. $\sqrt n\le n/2$ if $n\ge 4$ (to see this, just square both sides).

So $\sqrt{n^2-1}-\sqrt n \ge n/2-1$ if $n\ge 4$.

And $\lim_{n\rightarrow\infty}(n/2-1) = +\infty$.


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