<s>[INST] A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$ When I had the calculus class about the limit, one of my classmate felt confused about this limit:

$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$

What he thought that since $x^2 > x$ and $x^2 > 3x$ when $x \to \infty$ so the first square root must be $x$ and same for the second. Hence, the limit must be $0$.
It is obviously problematic.
And what I thought is that make prefect square under the limit, though I know the right solution is to rationalize the numerator.
After perfect-squaring, $$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}$$ I assert that since there is a perfect square and a square root. As $x \to \infty$, the constant does not matter. So
$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}= \lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(x+\frac{3}{2}\right)^2}\\ = \frac{1}{2}-\frac{3}{2} = -1 $$
But when I conquer this limit: this post.
Here is my argument:

$\sqrt{ax^2+bx+c} =O(\sqrt a(x+\frac{b}{2a}))$ when $x \to \infty$
$$\lim_{x\to\infty}\frac{\sqrt{x^2-2x+3}+\sqrt{4x^2+5x-6}}{x+\sqrt{x^2-1}} =\frac{x-1+2(x+5/4)}{x+x}  = \frac{3}{2}$$

Very concise get this answer but it gets downvoted.
I do not know what is wrong with my strategy. But in general case: for instance this problem about cubic root my strategy seems to work really efficient:

$$\lim_{x\to \infty} \sqrt[3]{x^3+6x^2+9x+1}-\sqrt[3]{x^3+5x^2+x+1}$$

My solution is:
$$\lim_{x\to \infty} \sqrt[3]{x^3+bx^2+cx+d} = \lim_{x\to \infty} \sqrt[3]{\left(x+\frac{b}{3}\right)^3}$$
So the limit becomes: $$\lim_{x\to \infty} \sqrt[3]{(x+2)^3+O(x)}-\sqrt[3]{\left(x+\frac{5}{3}\right)^3 + O(x)} =\lim_{x\to \infty}  (x+2) -\left(x+\frac{5}{3}\right) = \frac{1}{3} $$
This result gets verified by wolframalpha.
To put all into a nutshell, what is wrong with my solution to these three problem. Is there any counterexample to this substitution. Any help, you will be appreciated.
 [/INST] Let our problem be of the form $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)}$, where $p$ and $q$ are $n$th degree monic(coefficient of $x^n$ is $1$) polynomials. First of all, without loss of generality, we will rewrite $p$ and $q$ in the forms:
$$
p = (x + a)^n + \sum_{i=0}^{n-1} a_ix^i, q = (x + b)^n + \sum_{i=0}^{n-1} b_ix^i
$$ 
where $a_i$ and $b_i$ are real constants which may be zero.(This we did in the cubic and quadratic case).
Now, note that $x+a \leq \sqrt[n]{(x + a)^n + \sum_{i=0}^{n-1} a_ix^i} \leq \sqrt[n]{(x + a)^n} + \sqrt[n]{\sum_{i=0}^{n-1} a_ix^i}$.Note that $$
\sqrt[n]{\sum_{i=0}^{n-1} a_ix^i} \leq \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}
$$
However, the term on the left goes to $0$ as $x \to \infty$, because $i < n, $ so $x^i \to 0$. A similar logic follows for $q$. To complete the argument,
However, this is better expressed by: $x+a - ((x+b) + \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) \leq \sqrt[n]{p(x)} - \sqrt[n]{q(x)} \leq (x+a +\sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) -  (x+b))$.
On taking limits on both sides, we get $a-b$ on both sides, and using squeeze theorem, we get the result: $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)} = a-b$.
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