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.
 A: 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?
A: Discarding the low order terms (which become negligible when $n$ grows), the expression has the same limit as
$$\frac{\sqrt3 n}{n+n}.$$
A: $$
\lim_{n\to\infty}\frac{\sqrt{3n^2 + n-1}}{n+\sqrt{n^2-1}}
$$
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 $n\to\infty, k\in\Bbb N$?
A: A somewhat heuristic and informal argument: (note: a guide to formalize it follows)
As $n \to \infty$, the term of greatest degree dominates the growth of a polynomial. In other words, in a cubic $f(x) = ax^3 + bx^2 + cx + d$, as $x$ grows larger and larger, $f(x) \approx ax^3$. This idea generalizes to all polynomials and like expressions (expressions of "rational" degree, like $x^{1/2}$ or $x^\pi$ or whatever, and sums thereof, for example).
With that idea in mind, we examine:
$$\sqrt{3n^2+n-1} \approx \sqrt{3n^2} = n \sqrt 3$$
Similarly,
$$\sqrt{n^2 - 1} \approx \sqrt{n^2} = n$$
Thus,
$$ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}} = \lim\limits_{n \to \infty}\frac{n \sqrt 3}{n+n} = \lim\limits_{n \to \infty} \frac{n \sqrt 3}{2n} = \lim\limits_{n \to \infty} \frac{\sqrt 3}{2}=\frac{\sqrt 3}{2}$$

A more formal, alternate argument can be obtained by dividing through the numerator and denominator by $n$, as touched on in other answers, but I feel my approach is more intuitive.

You can also make the original argument more rigorous by considering what it means to be "asymptotically equivalent," as noted in the comments by Bernard. Essentially, two functions $f,g$ are asymptotically equivalent if
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$$
We also know, provided the limits exist, we can "distribute" the limit into the fraction as so:
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to \infty} f(x)}{\lim\limits_{x \to \infty} g(x)}$$
which means with the above that two functions are asymptotically equivalent if
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x)$$
In that light, this essentially gives us a means to make substitutions into the limit given. Since
$$\lim_{x \to \infty} \sqrt{3n^2+n-1} = \lim_{x \to \infty} \sqrt{3n^2}$$
we can make that replacement in our limit, for example. Of course, spelling out all the details is something I'll leave to you. ;)
