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.

By dividing the numerator and the denominator by $n$, you should have
$$\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}}=\frac{\sqrt{3+\frac{1}{n}-\frac{1}{n^2}}}{1+\sqrt{1-\frac{1}{n^2}}}.$$
Instead at your denominator we have $\frac{n}{n^2}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}$.
Then what is the final limit?
