<s>[INST] Compute $ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$ 
Compute $$ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$$

I did the following:
$$ \lim\limits_{n \to \infty}\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n^2}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}} = \frac{\sqrt{3}}{\sqrt{1}}=\sqrt3$$
However, the correct answer is different. Why am I wrong?
Thank you for your help.
 [/INST] $$
\lim_{n\to\infty}\frac{\sqrt{3n^2 + n-1}}{n+\sqrt{n^2-1}}
$$
Factor out $n$, thus:
$$
\frac{\sqrt{3n^2 + n-1}}{n+\sqrt{n^2-1}} = \frac{n\sqrt{3 + {1\over n} - {1\over n^2}}}{n\left(1 + \sqrt{1- {1\over n^2}}\right)} = \frac{\sqrt{3 + {1\over n} - {1\over n^2}}}{1 + \sqrt{1- {1\over n^2}}}
$$
So your limit becomes:
$$
\lim_{n\to\infty}\frac{\sqrt{3 + {1\over n} - {1\over n^2}}}{1 + \sqrt{1- {1\over n^2}}}
$$
Now what happens to ${1\over n^k}$ when $n\to\infty, k\in\Bbb N$?
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